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For this Question: Given a positive integer N, your task is to check if the given number prime or not.<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: def IFprime(N):
is_prime = 'Yes'
if N == 1: is_prime = 'No'
else:
for i in range(2, N//2+1):
if N % i == 0:
is_prime = 'No'
break
print(is_prime)
IFprime(int(input())), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to check if the given number prime or not.<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: static void Ifprime(int N)
{
if(N==1){
System.out.print("No");
return;
}
boolean prime = true;
for(int i=2; i*i<=N ;i++){
if(N%i==0){prime=false;break;}
}
if(prime==true){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code:
void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: public static void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: function patternMaking(N)
{
for(let j=1; j <= 2*N - 1; j++) {
let k;
if(j<= N) {
k = j;
} else {
k = 2*N - j;
}
let res = "";
for(let i=1; i<=2*k-1; i++) {
if(i<=k) {
res += i + " ";
} else {
res += 2*k - i + " ";
}
}
console.log(res);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: def pattern_making(n):
for i in range(1, n+1):
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=n-1
while i>=1 :
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=i-1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: def ispangram(str):
alphabet = "abcdefghijklmnopqrstuvwxyz"
for char in alphabet:
if char not in str.lower():
return False
return True
N = int(input())
arr = []
for i in range(N):
arr.append(input())
for i in range(N):
if(ispangram(arr[i]) == True):
print(1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: function pangrams(s) {
// write code here
// do not console.log it
// return 1 or 0
let alphabet = "abcdefghijklmnopqrstuvwxyz";
let regex = /\s/g;
let lowercase = s.toLowerCase().replace(regex, "");
for(let i = 0; i < alphabet.length; i++){
if(lowercase.indexOf(alphabet[i]) === -1){
return 0;
}
}
return 1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
String s = sc.next();
int check = 1;
int p =0;
for(char ch = 'a';ch<='z';ch++){
p=0;
for(int i = 0;i<s.length();i++){
if(s.charAt(i)==ch){p=1;}
}
if(p==0){check=0;}
}
System.out.println(check);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[26],B[26];
signed main()
{
int t;
cin>>t;
string p;
getline(cin,p);
while(t>0)
{
t--;
string s;
getline(cin,s);
int n=s.size();
memset(A,0,sizeof(A));
int ch=1;
for(int i=0;i<n;i++)
{
int x=s[i]-'a';
if(x>=0 && x<26)
{
A[x]++;
}
x=s[i]-'A';
if(x>=0 && x<26)
{
A[x]++;
}
}
for(int i=0;i<26;i++)
{
if(A[i]==0) ch=0;
}
cout<<ch<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade βAβ
If the percentage is <80 and >=60 then print Grade βBβ
If the percentage is <60 and >=40 then print Grade βCβ
else print Grade βDβThe input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade βAβ
If the percentage is <80 and >=60 then print Grade βBβ
If the percentage is <60 and >=40 then print Grade βCβ
else print Grade βDβThe input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In an 8X8 chessboard. Given the positions of the Queen as (X, Y) and the King as (P, Q) .
Your task is to check whether the queen can attack the king in one move or not.
The queen is the most powerful piece in the game of chess. It can move any number of squares vertically, horizontally or diagonally .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>QueenAttack()</b> that takes integers X, Y, P, and Q as arguments.
Constraints:-
1 <= X, Y, P, Q <= 8
Note:- King and Queen can not be in the same positionReturn 1 if the king is in the check position else return 0.Sample Input:-
1 1 5 5
Sample Output:-
1
Sample Input:-
3 4 6 6
Sample Output:-
0, I have written this Solution Code: def QueenAttack(X, Y, P, Q):
if X==P or Y==Q or abs(X-P)==abs(Y-Q):
return 1
return 0, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In an 8X8 chessboard. Given the positions of the Queen as (X, Y) and the King as (P, Q) .
Your task is to check whether the queen can attack the king in one move or not.
The queen is the most powerful piece in the game of chess. It can move any number of squares vertically, horizontally or diagonally .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>QueenAttack()</b> that takes integers X, Y, P, and Q as arguments.
Constraints:-
1 <= X, Y, P, Q <= 8
Note:- King and Queen can not be in the same positionReturn 1 if the king is in the check position else return 0.Sample Input:-
1 1 5 5
Sample Output:-
1
Sample Input:-
3 4 6 6
Sample Output:-
0, I have written this Solution Code: int QueenAttack(int X, int Y, int P, int Q){
if(X==P || Y==Q || abs(X-P)==abs(Y-Q) ){
return 1;
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In an 8X8 chessboard. Given the positions of the Queen as (X, Y) and the King as (P, Q) .
Your task is to check whether the queen can attack the king in one move or not.
The queen is the most powerful piece in the game of chess. It can move any number of squares vertically, horizontally or diagonally .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>QueenAttack()</b> that takes integers X, Y, P, and Q as arguments.
Constraints:-
1 <= X, Y, P, Q <= 8
Note:- King and Queen can not be in the same positionReturn 1 if the king is in the check position else return 0.Sample Input:-
1 1 5 5
Sample Output:-
1
Sample Input:-
3 4 6 6
Sample Output:-
0, I have written this Solution Code: static int QueenAttack(int X, int Y, int P, int Q){
if(X==P || Y==Q || Math.abs(X-P)==Math.abs(Y-Q) ){
return 1;
}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In an 8X8 chessboard. Given the positions of the Queen as (X, Y) and the King as (P, Q) .
Your task is to check whether the queen can attack the king in one move or not.
The queen is the most powerful piece in the game of chess. It can move any number of squares vertically, horizontally or diagonally .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>QueenAttack()</b> that takes integers X, Y, P, and Q as arguments.
Constraints:-
1 <= X, Y, P, Q <= 8
Note:- King and Queen can not be in the same positionReturn 1 if the king is in the check position else return 0.Sample Input:-
1 1 5 5
Sample Output:-
1
Sample Input:-
3 4 6 6
Sample Output:-
0, I have written this Solution Code: int QueenAttack(int X, int Y, int P, int Q){
if(X==P || Y==Q || abs(X-P)==abs(Y-Q) ){
return 1;
}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list of N nodes. The task is to reverse the list by changing links between nodes (i.e if the list is 1->2->3->4 then it becomes 1<-2<-3<-4) and return the head of the modified list.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b> ReverseLinkedList</b> that takes head node as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data<= 100Return the head of the modified linked list.Input-1:
6
1 2 3 4 5 6
Output-1:
6 5 4 3 2 1
Explanation:
After reversing the list, elements are as 6->5->4->3->2->1.
Input-2:
5
1 2 8 4 5
Output-2:
5 4 8 2 1, I have written this Solution Code: public static Node ReverseLinkedList(Node head) {
Node prev = null;
Node current = head;
Node next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Is it possible to get a sum of <b>B</b> when throwing a die with six faces 1, 2, β¦, 6 <b>A</b> times?The input consists of two space-separated integers.
A B
<b>Constraints</b>
1≤A≤100
1≤B≤1000
A and B are integers.If it is possible to get a sum of B, print Yes; otherwise, print No.<b>Sample Input 1</b>
2 11
<b>Sample Output 1</b>
Yes
<b>Sample Input 2</b>
2 13
<b>Sample Output 2</b>
No, I have written this Solution Code: #include <iostream>
using namespace std;
int main() {
int A, B;
cin >> A >> B;
if(A <= B && B <= 6 * A) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of size n containing distinct integers, where n is odd. Find the element which has the same number of lesser elements and the same number of greater elements in the array.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array.
Constraints
1 <= N <= 20
-10000 <= Arr[i] <= 10000Single integer n which has the same number of lesser elements and the same number of greater elements.Input:
3
3 1 2
Output:
2
Explanation:-
2 has one greater element 3 and one smaller element 1
Input:
5
2 3 4 9 1
Output:
3, I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
l.sort()
m=len(l)//2
print(l[m]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of size n containing distinct integers, where n is odd. Find the element which has the same number of lesser elements and the same number of greater elements in the array.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array.
Constraints
1 <= N <= 20
-10000 <= Arr[i] <= 10000Single integer n which has the same number of lesser elements and the same number of greater elements.Input:
3
3 1 2
Output:
2
Explanation:-
2 has one greater element 3 and one smaller element 1
Input:
5
2 3 4 9 1
Output:
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split("\\s+");
int N=Integer.parseInt(str[0]);
str=br.readLine().split("\\s+");
int arr[]=new int[N];
for(int i=0;i<N;i++){
arr[i]=Integer.parseInt(str[i]);
}
findElement(arr,N);
}
public static void findElement(int arr[],int N){
int start=0;
int end=N;Arrays.sort(arr);
int mid=start+(end-start)/2;
System.out.print(arr[mid]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of size n containing distinct integers, where n is odd. Find the element which has the same number of lesser elements and the same number of greater elements in the array.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array.
Constraints
1 <= N <= 20
-10000 <= Arr[i] <= 10000Single integer n which has the same number of lesser elements and the same number of greater elements.Input:
3
3 1 2
Output:
2
Explanation:-
2 has one greater element 3 and one smaller element 1
Input:
5
2 3 4 9 1
Output:
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int m = 100001;
int main(){
int n,k;
cin>>n;
long long a[n],sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
cout<<a[n/2];;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
Reader sc = new Reader();
int t = sc.nextInt();
while(t!=0)
{
int n= sc.nextInt();
int k=sc.nextInt();
int arr[]= new int[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextInt();
System.out.println(findKthLargest(arr,k));
t--;
}
}
public static int largestEle(int []arr,int k)
{
PriorityQueue<Integer> pq= new PriorityQueue<>(Collections.reverseOrder ());
for(int i=0;i<arr.length;i++)
{
pq.offer(arr[i]);
}
for(int i=0;i<k-1;i++)
{
pq.poll();
}
return pq.peek();
}
public static int findKthLargest(int[] nums, int k) {
int start = 0, end = nums.length - 1, index = nums.length - k;
while (start < end) {
int pivot = partion(nums, start, end);
if (pivot < index) start = pivot + 1;
else if (pivot > index) end = pivot - 1;
else return nums[pivot];
}
return nums[start];
}
private static int partion(int[] nums, int start, int end) {
int pivot = start, temp;
while (start <= end) {
while (start <= end && nums[start] <= nums[pivot]) start++;
while (start <= end && nums[end] > nums[pivot]) end--;
if (start > end) break;
temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
}
temp = nums[end];
nums[end] = nums[pivot];
nums[pivot] = temp;
return end;
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code:
t=int(input())
while t>0:
t-=1
n,k=map(int,input().split())
arr=list(map(int,input().split()))
arr.sort(reverse=True)
print(arr[k-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N positive integers and a number K. The task is to find the kth largest element in the array.
Note: DO NOT USE sort() stl.First line of input contains number of testcases. For each testcase, there will be a single line of input containing number of elements in the array and K. Next line contains N elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^5
1 <= K <= NFor each testcase, print a single line of output containing the kth largest element in the array.Sample Input:
2
5 3
3 5 4 2 9
5 5
4 3 7 6 5
Sample Output:
4
3
Explanation:
Testcase 1: Third largest element in the array is 4.
Testcase 2: Fifth largest element in the array is 3., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
sort(a,a+n);
cout<<a[n-k]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: class Solution {
public static void printTriangle(){
System.out.println("*");
System.out.println("* *");
System.out.println("* * *");
System.out.println("* * * *");
System.out.println("* * * * *");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: j=1
for i in range(0,5):
for k in range(0,j):
print("*",end=" ")
if(j<=4):
print()
j=j+1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, we define the set A<sub>N</sub> as {1, 2, 3, ... N-1, N}.
Find the size of the largest subset of A<sub>N</sub>, such that if x belongs to the subset, then 2x does not belong to the subset.The first line of the input contains a single integer T β the number of test cases.
Each test case consists of one line containing a single integer N.
<b>Constraints:</b>
1 β€ T β€ 10<sup>5</sup>
1 β€ N β€ 10<sup>9</sup>For each test case, print one integer β the size of the largest subset of A<sub>N</sub> satisfying the condition in the problem statement.Sample Input:
4
1
2
3
4
Sample Output:
1
1
2
3
Explanation:
For test case 4, where N = 4, the subset {1, 3, 4} is the largest subset satisfying the conditions.
1 is in the subset whereas 2 is not.
3 is in the subset whereas 6 is not.
4 is in the subset whereas 8 is not., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br;
static long compute(long n){
if(n==1||n==2) return 1;
if(n==3) return 2;
if(n%2==0){
long ans=(n/2);
return ans+compute(ans/2);
}
else {
long ans=(n+1)/2;
if(ans%2==1) return ans+compute(ans/2);
else return ans+compute((ans/2)-1);
}
}
public static void main(String[] args) throws Exception
{
br= new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int t=Integer.parseInt(br.readLine());
while(t-->0){
long n=Long.parseLong(br.readLine());
long ans=compute(n);
pw.println(ans);
}
pw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, we define the set A<sub>N</sub> as {1, 2, 3, ... N-1, N}.
Find the size of the largest subset of A<sub>N</sub>, such that if x belongs to the subset, then 2x does not belong to the subset.The first line of the input contains a single integer T β the number of test cases.
Each test case consists of one line containing a single integer N.
<b>Constraints:</b>
1 β€ T β€ 10<sup>5</sup>
1 β€ N β€ 10<sup>9</sup>For each test case, print one integer β the size of the largest subset of A<sub>N</sub> satisfying the condition in the problem statement.Sample Input:
4
1
2
3
4
Sample Output:
1
1
2
3
Explanation:
For test case 4, where N = 4, the subset {1, 3, 4} is the largest subset satisfying the conditions.
1 is in the subset whereas 2 is not.
3 is in the subset whereas 6 is not.
4 is in the subset whereas 8 is not., I have written this Solution Code: //Author: Xzirium
//Time and Date: 15:50:05 28 September 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(T);
while(T--)
{
READV(N);
ll ans=0;
while (N>0)
{
ans+=(N+1)/2;
N=N/4;
}
cout<<ans<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 β€ A, B β€ 4Print a single word β the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine();
String[] arr=s.split(" ");
int a=Integer.parseInt(arr[0]);
int b=Integer.parseInt(arr[1]);
long ab=(long)Math.pow(a,b);
long ba=(long)Math.pow(b,a);
if(ab>ba) System.out.print("First");
else if(ab<ba) System.out.print("Second");
else System.out.print("Equal");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 β€ A, B β€ 4Print a single word β the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , I have written this Solution Code: a,b=map(int,input().split())
if a**b > b**a:
print("First")
elif a**b <b**a:
print("Second")
else :
print("Equal"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two positive integers A and B.
Print "First", if A<sup>B</sup> > B<sup>A</sup>.
Print "Second", if A<sup>B</sup> < B<sup>A</sup>.
Otherwise, print "Equal".The only line contains two space-separated integers A and B.
<b>Constraints:</b>
1 β€ A, B β€ 4Print a single word β the answer to the problem.Sample Input 1:
1 3
Sample Output 1:
Second
Sample Explanation 1:
As 1<sup>3</sup> < 3<sup>1</sup>, we print "Second".
Sample Input 2:
2 2
Sample Output 2:
Equal
Sample Explanation 2:
As 2<sup>2</sup> = 2<sup>2</sup>, we print "Equal". , I have written this Solution Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
#define ll long long
#define ld long double
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long powm(long long a, long long b,long long mod) {
long long res = 1;
while (b > 0) {
if (b & 1)
res = res * a %mod;
a = a * a %mod;
b >>= 1;
}
return res;
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
#ifndef ONLINE_JUDGE
if(fopen("input.txt","r"))
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
}
#endif
int a,b;
cin>>a>>b;
int ans1=1;
for(int i=0;i<b;i++)
ans1*=a;
int ans2=1;
for(int i=0;i<a;i++)
ans2*=b;
if(ans1>ans2)
cout<<"First";
else if(ans1<ans2)
cout<<"Second";
else
cout<<"Equal";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long x;
BigInteger sum = new BigInteger("0");
for(int i=0;i<n;i++){
x=sc.nextLong();
sum= sum.add(BigInteger.valueOf(x));
}
sum=sum.divide(BigInteger.valueOf(n));
System.out.print(sum);
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, cur = 0, rem = 0;
cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
cur += (p + rem)/n;
rem = (p + rem)%n;
}
cout << cur;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: n = int(input())
a =list
a=list(map(int,input().split()))
sum=0
for i in range (0,n):
sum=sum+a[i]
print(int(sum//n))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) β the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines β the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],β¦, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between xβk and x+k.
The last line contains m integers b[1], b[2],β¦, b[m]: the size of each toy.
<b>Constraints</b>
1 β€ n,m β€ 200000
0 β€ k β€ 10<sup>9</sup>
1 β€ a[i],b[i] β€ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
int m = sc.nextInt();
long k = sc.nextLong();
long[] boyArray = new long[n];
long[] toyArray = new long[m];
for (int i = 0; i < n; i++) {
boyArray[i] = sc.nextLong();
}
for (int i = 0; i < m; i++) {
toyArray[i] = sc.nextLong();
}
Arrays.sort(boyArray);
Arrays.sort(toyArray);
int cnt = 0;
int l = 0 , h = n-1;
for (int i = 0; i < m; i++) {
while( h >= l){
if(toyArray[i] < boyArray[l]-k || toyArray[i] > boyArray[h]+k){
break;
}
else if(toyArray[i] >= boyArray[l]-k && toyArray[i] <= boyArray[l]+k){
l++;
cnt++;
break;
}
else if(toyArray[i] >= boyArray[h]-k && toyArray[i] <= boyArray[h]+k){
h--;
cnt++;
break;
}
else{
l++;
}
}
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],β¦, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between xβk and x+k.
The last line contains m integers b[1], b[2],β¦, b[m]: the size of each toy.
<b>Constraints</b>
1 β€ n,m β€ 200000
0 β€ k β€ 10<sup>9</sup>
1 β€ a[i],b[i] β€ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: n,m,k=[int(x) for x in input().split()[:3]]
a=[int(x) for x in input().split()[:n]]
b=[int(x) for x in input().split()[:m]]
a.sort()
b.sort()
j = 0
sum1 = 0
for i in a:
while(j<len(b)):
if(i-k<=b[j] and b[j]<=i+k):
j = j+1
sum1 +=1
break
elif i+k<b[j]:
break
else:
j +=1
if(j>=len(b)):
break
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],β¦, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between xβk and x+k.
The last line contains m integers b[1], b[2],β¦, b[m]: the size of each toy.
<b>Constraints</b>
1 β€ n,m β€ 200000
0 β€ k β€ 10<sup>9</sup>
1 β€ a[i],b[i] β€ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 200010
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
int c[max1];
int main()
{
int n,m;
cin>>n>>m;
ll k;
cin>>k;
ll a[n],b[m];
FOR(i,n){
cin>>a[i];
}
FOR(i,m){
cin>>b[i];}
sort(a,a+n);
sort(b,b+m);
int i=0,j=0;
int cnt=0;
while(i!=n && j!=m){
if(abs(b[j]-a[i])<=k){cnt++;i++;j++;}
else{
if(b[j]>a[i]){i++;}
else{j++;}
}
}
out(cnt);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String arr[][]=new String[n][n];
String transpose[][]=new String[n][n];
int row;
int cols;
for(row=0;row<n;row++)
{
String rowNum=br.readLine();
String rowVals[]=rowNum.split(" ");
for(cols=0; cols<n;cols++)
{
arr[row][cols]=rowVals[cols];
}
}
for(row=0;row<n;row++)
{
for(cols=0; cols<n;cols++)
{
transpose[row][cols]=arr[cols][row];
System.out.print(transpose[row][cols]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
l4=[]
for j in range(x):
l3=[]
for i in range(x):
l3.append(l1[i][j])
l4.append(l3)
for i in range(x):
for j in range(x):
print(l4[i][j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code:
def countOddSum(a, n):
# 'odd' stores number of
# odd numbers upto ith index
# 'c_odd' stores number of
# odd sum subarrays starting
# at ith index
# 'Result' stores the number
# of odd sum subarrays
c_odd = 0;
result = 0;
odd = False;
# First find number of odd
# sum subarrays starting at
# 0th index
for i in range(n):
if (a[i] % 2 == 1):
if(odd == True):
odd = False;
else:
odd = True;
if (odd):
c_odd += 1;
# Find number of odd sum
# subarrays starting at ith
# index add to result
for i in range(n):
result += c_odd;
if (a[i] % 2 == 1):
c_odd = (n - i - c_odd);
return result;
n=int(input())
arr=list(map(int,input().strip().split()))[:n]
print(countOddSum(arr,n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String args[]) throws IOException
{
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long a[] = new long[n];
String line = br.readLine(); // to read multiple integers line
String[] strs = line.trim().split("\\s+");
for (int i = 0; i < n; i++) {
a[i] = Long.parseLong(strs[i]);
}
long[] prefix=new long[n];
prefix[0]=a[0];
for(int i=1;i<n;i++){
prefix[i]=prefix[i-1]+a[i];
}
long e=1;
long o=0;
for(int i=0;i<n;i++){
if(prefix[i]%2==0)e++;
else o++;
}
e*=o;
System.out.print(e);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Peter has the task to write a program that processes N strings. Some of them have Russian and European format "day.month.year", while others use American format β "month/day/year".
Here, year is a number from 1 to 9999, which may contain leading zeroes to get 2, 3 or 4 digits, month β number from 1 to 12, which may contain leading zero to get 2 digits, day β number from 1 to 31, which may contain leading zeros to get 2 digits.
Print each date in two formats: first as "DD.MM.YYYY', second as "MM/DD/YYYY".
For better understanding see sample.The first line of input contains one integer N (1 <= N <= 100).
Each of the following N lines contain one date in format "day.month.year" or "month/day/year".Print π lines. Each line should contain the date in two formats: first as "DD.MM.YYYY", second as "MM/DD/YYYY".Sample Input
2
11.12.2000
1.2.1
Sample Output
11.12.2000 12/11/2000
01.02.0001 02/01/0001
Sample Input
2
20.10.2100
1/29/3000
Sample Output
20.10.2100 10/20/2100
29.01.3000 01/29/3000, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int tc= scanner.nextInt();
for(int i=0;i<tc;i++){
String s=scanner.next();
if(s.indexOf(".")==-1){
String[] str=s.split("/");
printOut1(str);
}else{
String[] str=s.split("[.]");
printOut(str);
}
System.out.println();
}
}
public static void printOut(String str[]){
if(str[0].length()!=2){
System.out.print("0");
}
System.out.print(str[0]);
System.out.print(".");
if(str[1].length()!=2){
System.out.print("0");
}
System.out.print(str[1]);
System.out.print(".");
for(int i=0;i<4-str[2].length();i++){
System.out.print("0");
}
System.out.print(str[2]);
System.out.print(" ");
if(str[1].length()!=2){
System.out.print("0");
}
System.out.print(str[1]);
System.out.print("/");
if(str[0].length()!=2){
System.out.print("0");
}
System.out.print(str[0]);
System.out.print("/");
for(int i=0;i<4-str[2].length();i++){
System.out.print("0");
}
System.out.print(str[2]);
}
public static void printOut1(String str[]){
if(str[1].length()!=2){
System.out.print("0");
}
System.out.print(str[1]);
System.out.print(".");
if(str[0].length()!=2){
System.out.print("0");
}
System.out.print(str[0]);
System.out.print(".");
for(int i=0;i<4-str[2].length();i++){
System.out.print("0");
}
System.out.print(str[2]);
System.out.print(" ");
if(str[0].length()!=2){
System.out.print("0");
}
System.out.print(str[0]);
System.out.print("/");
if(str[1].length()!=2){
System.out.print("0");
}
System.out.print(str[1]);
System.out.print("/");
for(int i=0;i<4-str[2].length();i++){
System.out.print("0");
}
System.out.print(str[2]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Peter has the task to write a program that processes N strings. Some of them have Russian and European format "day.month.year", while others use American format β "month/day/year".
Here, year is a number from 1 to 9999, which may contain leading zeroes to get 2, 3 or 4 digits, month β number from 1 to 12, which may contain leading zero to get 2 digits, day β number from 1 to 31, which may contain leading zeros to get 2 digits.
Print each date in two formats: first as "DD.MM.YYYY', second as "MM/DD/YYYY".
For better understanding see sample.The first line of input contains one integer N (1 <= N <= 100).
Each of the following N lines contain one date in format "day.month.year" or "month/day/year".Print π lines. Each line should contain the date in two formats: first as "DD.MM.YYYY", second as "MM/DD/YYYY".Sample Input
2
11.12.2000
1.2.1
Sample Output
11.12.2000 12/11/2000
01.02.0001 02/01/0001
Sample Input
2
20.10.2100
1/29/3000
Sample Output
20.10.2100 10/20/2100
29.01.3000 01/29/3000, I have written this Solution Code: n = int(input())
while(n > 0):
s = input()
d = ""
m = ""
y = ""
if('.' in s):
s = s.split('.')
for i in range(0, 2-len(s[0])):
d += '0'
d += s[0]
for i in range(0, 2-len(s[1])):
m += '0'
m += s[1]
for i in range(0, 4-len(s[2])):
y += '0'
y += s[2]
else:
s = s.split('/')
for i in range(0, 2-len(s[0])):
m += '0'
m += s[0]
for i in range(0, 2-len(s[1])):
d += '0'
d += s[1]
for i in range(0, 4-len(s[2])):
y += '0'
y += s[2]
print(d+'.'+m+'.'+y, end = ' ')
print(m+'/'+d+'/'+y)
n = n-1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: static void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
System.out.print(x+4*j+" ");
}
System.out.println();
x+=6;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: def Pattern(N):
x=0
for i in range (0,N):
for j in range (0,N):
print(x+4*j,end=' ')
print()
x = x+6
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create an n*n matrix with values [1,. , n) just below the diagonal2<=n<=99A 2D matrix of dimension n with values just below the diagonalSample Input:
2
Output:
[[0 0]
[1 0]]
Sample Input:
10
Output:
[[0 0 0 0 0 0 0 0 0 0]
[1 0 0 0 0 0 0 0 0 0]
[0 2 0 0 0 0 0 0 0 0]
[0 0 3 0 0 0 0 0 0 0]
[0 0 0 4 0 0 0 0 0 0]
[0 0 0 0 5 0 0 0 0 0]
[0 0 0 0 0 6 0 0 0 0]
[0 0 0 0 0 0 7 0 0 0]
[0 0 0 0 0 0 0 8 0 0]
[0 0 0 0 0 0 0 0 9 0]], I have written this Solution Code: import numpy as np
n=int(input())
d=np.diag(np.arange(1,n),k=-1)
print(d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: function PokemonMaster(a,b) {
// write code here
// do no console.log the answer
// return the output using return keyword
const ans = (a - b >= 0) ? 1 : 0
return ans
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: def PokemonMaster(A,B):
if(A>=B):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: static int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-- > 0) {
long n = Long.parseLong(br.readLine());
System.out.println(greyCode(n));
}
}
static long greyCode(long n) {
return n ^ (n>>1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int x; cin >> x;
cout << (x ^ (x >> 1)) << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: t=int(input())
for _ in range(t):
a=int(input())
x=list(map(int,(list(bin(a)[2:]))))
y=[x[0]]
for i in range(1,len(x)):
y.append(x[i-1]^x[i])
c=''
p=0
s=0
#print(y)
for i in range(len(y)-1,-1,-1):
s+=y[i]*2**p
p+=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.StringTokenizer;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
};
public static void main(String[] args) {
FastReader sc = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextLong();
}
Arrays.sort(arr);
long min=Long.MAX_VALUE;
long max=Long.MIN_VALUE;
for(int i=0;i<n/2;i++)
{
min=Math.min(min,arr[n-i-1]+arr[i]);
max=Math.max(max,arr[n-i-1]+arr[i]);
}
System.out.println(max-min);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: n = int(input())
bruh = input().strip().split(" ")
for i in range(len(bruh)):
bruh[i] = int(bruh[i])
bruh.sort()
maxSum = 0
minSum = 100000000000000000000
for i in range(len(bruh)//2):
add = bruh[i] + bruh[-(i+1)]
if add > maxSum:
maxSum = add
if add < minSum:
minSum = add
print(maxSum - minSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> v(n);
For(i, 0, n){
cin>>v[i];
}
sort(all(v));
vector<int> cur;
For(i, 0, n/2){
cur.pb(v[i]+v[n-i-1]);
}
sort(all(cur));
cout<<cur[n/2-1]-cur[0];
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>numOfWords</code>, which should take a string
and return its result as an integer which is the number of space seperated words in it (Use JS In built functions)Function will take a string as argumentFunction will return the number of space separated wordsconsole. log(numOfWords("Hi World")) // prints 2
console. log(fnumOfWords("hi")) // prints 1
console. log(numOfWords("My name is Newton")) // prints 4, I have written this Solution Code: import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
String s=sc.nextLine();
int count=1;
for(int i=0;i<s.length();i++){
if(s.charAt(i)==' '){
count++;
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>numOfWords</code>, which should take a string
and return its result as an integer which is the number of space seperated words in it (Use JS In built functions)Function will take a string as argumentFunction will return the number of space separated wordsconsole. log(numOfWords("Hi World")) // prints 2
console. log(fnumOfWords("hi")) // prints 1
console. log(numOfWords("My name is Newton")) // prints 4, I have written this Solution Code:
function numOfWords(line){
// write code here
// return the output , do not use console.log here
return line.split(" ").length
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(i,end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is an integer q is given as input.You are asked q queries each of which can be of type 1 or type 2.
Type 1 : Report the area of square having side length S.
Type 2 : Report the area of rectangle of having side lengths L and R.The first line of the input contains single integer Q, number of queries.
Next Q line contains the queries.
Each query first contains type of query
If query is of type one then you are given a single integer S as input.
If query is of type two then you are given two integers L and R as input.
Constraints
1 <= Q <= 25
1 <= T <= 2
1 <= S, L, R <= 10<sup>3</sup>Output Q line each containing answer to the asked query.Sample Input
2
2 5 4
1 5
Sample Output
20
25, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader inputReader = new BufferedReader(new InputStreamReader(System.in));
int queries = Integer.parseInt(inputReader.readLine());
for(int q = 0; q < queries; q++) {
String[] inputNums = inputReader.readLine().trim().split(" ");
if(Integer.parseInt(inputNums[0]) == 1) {
System.out.println(Integer.parseInt(inputNums[1]) * Integer.parseInt(inputNums[1]));
} else if (Integer.parseInt(inputNums[0]) == 2) {
System.out.println(Integer.parseInt(inputNums[1]) * Integer.parseInt(inputNums[2]));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is an integer q is given as input.You are asked q queries each of which can be of type 1 or type 2.
Type 1 : Report the area of square having side length S.
Type 2 : Report the area of rectangle of having side lengths L and R.The first line of the input contains single integer Q, number of queries.
Next Q line contains the queries.
Each query first contains type of query
If query is of type one then you are given a single integer S as input.
If query is of type two then you are given two integers L and R as input.
Constraints
1 <= Q <= 25
1 <= T <= 2
1 <= S, L, R <= 10<sup>3</sup>Output Q line each containing answer to the asked query.Sample Input
2
2 5 4
1 5
Sample Output
20
25, I have written this Solution Code: q = int(input())
while q:
arr = list(map(int, input().split()))
if arr[0] == 1:
print(arr[1]*arr[1])
else:
print(arr[1]*arr[2])
q-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is an integer q is given as input.You are asked q queries each of which can be of type 1 or type 2.
Type 1 : Report the area of square having side length S.
Type 2 : Report the area of rectangle of having side lengths L and R.The first line of the input contains single integer Q, number of queries.
Next Q line contains the queries.
Each query first contains type of query
If query is of type one then you are given a single integer S as input.
If query is of type two then you are given two integers L and R as input.
Constraints
1 <= Q <= 25
1 <= T <= 2
1 <= S, L, R <= 10<sup>3</sup>Output Q line each containing answer to the asked query.Sample Input
2
2 5 4
1 5
Sample Output
20
25, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int x;
for(int i=0;i<n;i++){
cin>>x;
if(x==1){
cin>>x;
cout<<x*x<<endl;
}
else{
int a,b;
cin>>a>>b;
cout<<a*b<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.util.Arrays;
class Main {
public static double getMedia(long[]a,long[]b){
if(a.length>b.length){
long[]temp;
temp=a;
a=b;b=temp;
}
int lo=0;
int hi=a.length;
int tl=a.length+b.length;
while(lo<=hi){
int al=(lo+hi)/2;
int bl=((tl+1)/2)-al;
long alm1=(al==0)? Long.MIN_VALUE:a[al-1];
long alf=(al==a.length)? Long.MAX_VALUE:a[al];
long blm1=(bl==0)? Long.MIN_VALUE:b[bl-1];
long blf=(bl==b.length)? Long.MAX_VALUE:b[bl];
if(alm1<=blf && blm1<=alf){
double median = 0.0;
if(tl%2==0){
long lmax=Math.max(alm1,blm1);
long rmin=Math.min(alf,blf);
median=(lmax+rmin)/2.0;
return median;
}else{
long lmax=Math.max(alm1,blm1);
median=lmax;
return median;
}
}
else if(alm1>blf){
hi=al-1;
}else if(blm1>alf){
lo=al+1;
}
}
return 0;
}
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int m=Integer.parseInt(str[1]);
long[] a=new long[n];
long[] b=new long[m];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
a[i]=Long.parseLong(str[i]);
}
str=br.readLine().split(" ");
for(int i=0;i<m;i++){
b[i]=Long.parseLong(str[i]);
}
System.out.format("%.2f",getMedia(a,b));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: def getMedian(arr1, arr2, n, m):
i=j=0
sort=[]
while i<n and j<m:
if arr1[i] < arr2[j]:
sort.append(arr1[i])
i+=1
else:
sort.append(arr2[j])
j+=1
while i<n:
sort.append(arr1[i])
i+=1
while j<m:
sort.append(arr2[j])
j+=1
if (n+m)%2==1:
ind=(len(sort)//2)+1
num=sort[ind-1]
else:
mid=len(sort)//2
num=(sort[mid-1]+sort[mid])/2
return num
n, m= input().split()
n, m= int(n),int(m)
ar1=list(map(int, input().split()))
ar2=list(map(int, input().split()))
print("%.2f" %getMedian(ar1, ar2, n, m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted arrays A and B of size n and m respectively, return the median of the two sorted arrays.The first line of input contains n, m the length of arrays A and B.
The next two lines contain the input of arrays A and B.
<b>Constraints</b>
1 ≤ n, m ≤ 1000
-10<sup>6</sup> ≤ A[i], B[i] ≤ 10<sup>6</sup>Print the median of two sorted arrays upto two decimal places.Sample Input :
2 1
1 3
2
Sample Output :
2.00, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-02 14:05:28
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
if (nums2.size() < nums1.size())
return findMedianSortedArrays(nums2, nums1);
int n1 = nums1.size();
int n2 = nums2.size();
int lo = 0, hi = n1;
while (lo <= hi) {
int cut1 = (lo + hi) / 2;
int cut2 = (n1 + n2 + 1) / 2 - cut1;
int left1 = cut1 == 0 ? INT_MIN : nums1[cut1 - 1];
int left2 = cut2 == 0 ? INT_MIN : nums2[cut2 - 1];
int right1 = cut1 == n1 ? INT_MAX : nums1[cut1];
int right2 = cut2 == n2 ? INT_MAX : nums2[cut2];
if (left1 <= right2 && left2 <= right1) {
if ((n1 + n2) % 2 == 0)
return (max(left1, left2) + min(right1, right2)) / 2.0;
else
return max(left1, left2);
} else if (left1 > right2) {
hi = cut1 - 1;
} else
lo = cut1 + 1;
}
return 0.0;
}
int main() {
int n, m;
cin >> n >> m;
vector<int> a(n), b(m);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < m; i++) {
cin >> b[i];
}
printf("%0.2f", findMedianSortedArrays(a, b));
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==0):
print("Yes")
else:
print("No");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
Long m = sc.nextLong();
if(n%m==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
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