Instruction
stringlengths 261
35k
| Response
stringclasses 1
value |
|---|---|
For this Question: Pairs and vectors can be used together to achieve some amazing results. Here we will learn to use a vector that holds pairs.
You are given a vector V of size N. The vector hold pair of integers. Example V={(1,2),(3,4)...}. Now, you need to sum the second elements.First line contains N denoting the size of the array. The second line contains 2*N elements where the (2i - 1)'th and (2i)'th element represent the i'th pair.
Constraints:
1 <= N <= 10^5
0 <= Vi <= 10^5For each testcase, in a new line, print the required output.Input:
5
1 2 3 4 5 6 7 8 9 10
Output:
30
Explanation:
Sum = 10+8+6+4+2 = 30, I have written this Solution Code: input()
print(sum([int(i) for (j, i) in enumerate(input().split()) if j % 2 == 1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pairs and vectors can be used together to achieve some amazing results. Here we will learn to use a vector that holds pairs.
You are given a vector V of size N. The vector hold pair of integers. Example V={(1,2),(3,4)...}. Now, you need to sum the second elements.First line contains N denoting the size of the array. The second line contains 2*N elements where the (2i - 1)'th and (2i)'th element represent the i'th pair.
Constraints:
1 <= N <= 10^5
0 <= Vi <= 10^5For each testcase, in a new line, print the required output.Input:
5
1 2 3 4 5 6 7 8 9 10
Output:
30
Explanation:
Sum = 10+8+6+4+2 = 30, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader reader =new BufferedReader(new InputStreamReader(System.in));
int a=Integer.parseInt(reader.readLine())*2;
long sum=0;
String arr[]=reader.readLine().split(" ");
reader.close();
for(int i=0;i<a;i++){
if(i%2==1)
{
sum=sum+Integer.parseInt(arr[i]);
}
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pairs and vectors can be used together to achieve some amazing results. Here we will learn to use a vector that holds pairs.
You are given a vector V of size N. The vector hold pair of integers. Example V={(1,2),(3,4)...}. Now, you need to sum the second elements.First line contains N denoting the size of the array. The second line contains 2*N elements where the (2i - 1)'th and (2i)'th element represent the i'th pair.
Constraints:
1 <= N <= 10^5
0 <= Vi <= 10^5For each testcase, in a new line, print the required output.Input:
5
1 2 3 4 5 6 7 8 9 10
Output:
30
Explanation:
Sum = 10+8+6+4+2 = 30, 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 = 1e3 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n, ans = 0; cin >> n;
for(int i = 1; i <= 2*n; i++){
int p; cin >> p;
if(i%2 == 0)
ans += p;
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
mat[i][j] = sc.nextInt();
}
}
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
System.out.print(mat[i][j]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code: n = int(input())
for _ in range(n):
print(input()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int arr[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<arr[i][j]<<" ";
}
cout<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, 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 integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long cnt=0;
long ans=0;
for(int i=0;i<n;i++){
if(a[i]&1){cnt++;}
else{
ans+=(cnt*(cnt+1))/2;
cnt=0;
}
}
ans+=(cnt*(cnt+1))/2;
cout<<ans;
}
, 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 integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
c=0
result=0
for i in arr:
if i % 2 != 0:
c += 1
else:
result += (c*(c+1)) / 2
c=0
result += (c*(c+1)) / 2
print(int(result)), 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 integers. Find the number of subarrays of this array that are powerful.
A subarray [L, R] (1 <= L <= R <= N) is said to be powerful if the product A<sub>L</sub> * A<sub>L+1</sub> * ... * A<sub>R-1</sub> * A<sub>R</sub> is odd.The first line of input contains a single integer N
The second line of input contains N integers representing the elements of the array Arr
<b>Constraints </b>
1 <= N <= 100000
1 <= Arr[i] <= 100000Output the number of powerful subarrays of array Arr.Sample Input 1
5
2 4 4 5 3
Sample output 1
3
Sample Input 2
3
1 5 1
Sample Output 2
6
<b>Explanation:</b>
(3), (5), (3, 5) are the required subarrays.
(1), (2), (1), (1, 5), (5, 1]) (1, 5, 1) are the required subarrays., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader inputStreamReader=new InputStreamReader(System.in);
BufferedReader reader=new BufferedReader(inputStreamReader);
int n =Integer.parseInt(reader.readLine());
String str=reader.readLine();
String[] strarr=str.split(" ");
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(strarr[i]);
}
long noOfsubArrays=0;
int start=0;
boolean sFlag=false;
for(int i=0;i<n;i++){
if(arr[i]%2!=0){
if(!sFlag){
start=i;
noOfsubArrays++;
sFlag=true;
continue;
}
noOfsubArrays+=2;
int temp=start;
temp++;
while(temp<i){
temp++;
noOfsubArrays++;
}
}else if(arr[i]%2==0 && sFlag){
sFlag=false;
}
}
System.out.println(noOfsubArrays);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N Power Stones in Morag represented in an array PowerStones, where PowerStones[i] is the power of ith stone.
You are one of the guardians, your task is to choose two stones and smash them together. Suppose you have stones with power x and y with x <= y then the result of the smash is :
a. If x == y, then both stones get destroyed.
b. If x != y, then the power of stone x is destroyed and the
stone y has a new power y - x.
At the end of the task, there is at most one stone left. Print the smallest possible power of the left stone.First Line contains an integer N, length of array PowerStones
Second- line contains N space- separated integers of the array
Constraints :
1 <= N <= 30
1 <= PowerStones[i] <= 100Print the smallest possible power of the left stone
Note:-
If the size of the array becomes zero then 0 will be considered as the smallest possible power.Sample Input:-
6
2 7 4 1 8 1
Sample Output:-
1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to
[2, 7, 1, 8, 1] then,
we can combine 7 and 8 to get 1, so the array converts to
[2, 1, 1, 1] then,
we can combine 2 and 1 to get 1, so the array converts to [1, 1, 1]
then,
we can combine 1 and 1 to get 0, so the array converts to [1],
then that's the optimal value
Sample Input:-
5
2 1 1 1 1
Sample Output:-
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int laststonepower(vector<int>& stones) {
int sum=0, n=stones.size();
for(int i=0; i<n; i++){
sum += stones[i];
}
bool dp[n+1][sum+1];
for(int i=0; i<n+1; i++){
for(int j=0; j<sum+1; j++){
if(i==0)
dp[i][j]=false;
if(j==0)
dp[i][j]=true;
}
}
for(int i=1; i<n+1; i++){
for(int j=1; j<sum+1; j++){
if(stones[i-1]<=j){
dp[i][j]=dp[i-1][j]+dp[i-1][j-stones[i-1]];
}
else
dp[i][j]=dp[i-1][j];
}
}
vector<int> temp;
for(int j=0; j<=(sum/2); j++){
if(dp[n][j]==true){
temp.push_back(j);
}
}
int max=-1;
for(int i=0; i<temp.size(); i++){
if(temp[i]>max)
max=temp[i];
}
int res = sum - (2*max);
return res;
}
int main()
{
int N;
cin >> N;
int c;
vector<int> powerstones;
for(int i =0; i < N ; i++){
cin >> c;
powerstones.push_back(c);
}
cout << laststonepower(powerstones) << endl ;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a sequence P, which is represented by the following recurrence
P(N) = P(N-2) + P(N-3)
P(0) = P(1) = P(2) = 1
Now given a number N your task is to find the Nth number in this sequence.
Note: Since the output could be very long, take mod 1000000007. And try to give the most optimum solution as you can.The input line contains T, denoting the number of testcases. Each testcase contains single line containing an integer N.
Constraints:
1 <= T <= 15
1 <= N <=100For each testcase, print the nth number of the P sequence in a newline.Sample Input:
2
12
10
Sample Output:
21
12
Explanation:
Testcase 1:
As per the recurrence relation given in the question, we are basically adding last second and last third numbers of the sequence in bottom up manner. Let's see this case
P(3) = P(1) + P(0) = 2
P(4) = P(2) + P(1) = 2
P(5) = P(3) + P(2) = 3
P(6) = P(4) + P(3) = 4
P(7) = P(5) + P(4) = 5
P(8) = P(6) + P(5) = 7
P(9) = P(7) + P(6) = 9
P(10) = P(8) + P(7) = 12
P(11) = P(9) + P(8) = 16
P(12) = P(10) + P(9) = 21
, 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());
long[] dp = new long[101];
Arrays.fill(dp,-1);
while(t-- > 0) {
int n = Integer.parseInt(br.readLine());
if(dp[n] != -1) {
System.out.println(dp[n]);
} else {
System.out.println(solve(n,dp));
}
}
}
static long solve(int n,long[] dp) {
int mod = 1000000007;
if(n < 3) return 1;
if(dp[n] != -1) return dp[n];
return dp[n] = (solve(n-3,dp)%mod+solve(n-2,dp)%mod)%mod;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a sequence P, which is represented by the following recurrence
P(N) = P(N-2) + P(N-3)
P(0) = P(1) = P(2) = 1
Now given a number N your task is to find the Nth number in this sequence.
Note: Since the output could be very long, take mod 1000000007. And try to give the most optimum solution as you can.The input line contains T, denoting the number of testcases. Each testcase contains single line containing an integer N.
Constraints:
1 <= T <= 15
1 <= N <=100For each testcase, print the nth number of the P sequence in a newline.Sample Input:
2
12
10
Sample Output:
21
12
Explanation:
Testcase 1:
As per the recurrence relation given in the question, we are basically adding last second and last third numbers of the sequence in bottom up manner. Let's see this case
P(3) = P(1) + P(0) = 2
P(4) = P(2) + P(1) = 2
P(5) = P(3) + P(2) = 3
P(6) = P(4) + P(3) = 4
P(7) = P(5) + P(4) = 5
P(8) = P(6) + P(5) = 7
P(9) = P(7) + P(6) = 9
P(10) = P(8) + P(7) = 12
P(11) = P(9) + P(8) = 16
P(12) = P(10) + P(9) = 21
, I have written this Solution Code:
// author-Shivam gupta
#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 100001
#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;
using vl = vector<ll>;
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;
}
int Tot;
long long a[100010];
map <long long,int> mp;
void dfs(int now,int ee,int les)
{
if (les==0||now+1==ee)
{
Tot++;
return;
}
for (int i=0;i<=les;i++)
dfs(now+1,ee,les-i);
}
void dfs2(int now,long long num,int les)
{
if (now==0) num+=les*a[0];
if (les==0||now==0)
{
mp[num]--;
if (mp[num]==0) mp.erase(num);
return;
}
for (int i=0;i<=les;i++)
dfs2(now-1,num+i*a[now],les-i);
}
ll solve(ll x, ll y){
int a[50],b[50];
FOR(i,50){
a[i]=b[i]=0;}
int i=0;
while(x>0){
a[i]=x%2;
x/=2;
i++;
}
i=0;
while(y>0){
b[i]=y%2;
y=y/2;
i++;
}
int ans=0;
FOR(i,50){
if(a[i]!=b[i]){ans++;}
}
return ans;
}
int main()
{
ll a[101];
a[0]=a[1]=a[2]=1;
for(int i =3;i<=100;i++){
a[i]=a[i-2]+a[i-3];
a[i]=a[i]%MOD;
}
int t;
cin>>t;
while(t--){
int n;
cin>>n;
out(a[n]);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.The first line of the input contains n ( length of num )
The second line contains the array num.
<b>Constraints</b>
1 ≤ nums. length ≤ 100
-100 ≤ nums[i] ≤ 100Print the sorted arraySample Input
6
1 1 2 2 2 3
Sample Output
3 1 1 2 2 2
Explanation: '
3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 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));
int n = Integer.parseInt(br.readLine());
String str = br.readLine();
String[] str1 = str.split(" ");
int[] arr = new int[n];
HashMap<Integer,Integer> map = new HashMap<>();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < n; ++i) {
arr[i] = Integer.parseInt(str1[i]);
}
arr = frequencySort(arr);
for(int i : arr) {
System.out.print(i+" ");
}
}
static Map<Integer,Integer>map;
public static int[] frequencySort(int[] nums)
{
map=new HashMap<Integer,Integer>();
for(int i:nums){
if(map.containsKey(i)){
map.put(i,1+map.get(i));
}else{
map.put(i,1);
}
}
Integer[]arr=new Integer[nums.length];
int k=0;
for(int i:nums){
arr[k++]=i;
}
Arrays.sort(arr,new Comp());
k=0;
for(int i:arr){
nums[k++]=i;
}
return nums;
}
}
class Comp implements Comparator<Integer>{
Map<Integer,Integer>map=Main.map;
public int compare(Integer a,Integer b){
if(map.get(a)>map.get(b))return 1;
else if(map.get(b)>map.get(a))return -1;
else{
if(a>b)return -1;
else 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: Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.The first line of the input contains n ( length of num )
The second line contains the array num.
<b>Constraints</b>
1 ≤ nums. length ≤ 100
-100 ≤ nums[i] ≤ 100Print the sorted arraySample Input
6
1 1 2 2 2 3
Sample Output
3 1 1 2 2 2
Explanation: '
3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3., I have written this Solution Code: import numpy as np
from collections import defaultdict
d=defaultdict(list)
d_f=defaultdict (int)
n=int(input())
a=np.array([input().strip().split()],int).flatten()
for i in a:
d_f[i]+=1
for i in d_f:
d[d_f[i]].append(i)
d=sorted(d.items())
for i in d:
i[1].sort(reverse=True)
for i in d:
for j in i[1]:
for _ in range(i[0]):
print(j,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.The first line of the input contains n ( length of num )
The second line contains the array num.
<b>Constraints</b>
1 ≤ nums. length ≤ 100
-100 ≤ nums[i] ≤ 100Print the sorted arraySample Input
6
1 1 2 2 2 3
Sample Output
3 1 1 2 2 2
Explanation: '
3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3., I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-10 12:51:16
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
bool static comparator(pair<int, int> m, pair<int, int> n) {
if (m.second == n.second)
return m.first > n.first; // m>n can also be written it will return the same
else
return m.second < n.second;
}
vector<int> frequencySort(vector<int>& nums) {
unordered_map<int, int> mp;
for (auto k : nums)
mp[k]++;
vector<pair<int, int>> v1;
for (auto k : mp)
v1.push_back(k);
sort(v1.begin(), v1.end(), comparator);
vector<int> v;
for (auto k : v1) {
while (k.second != 0) {
v.push_back(k.first);
k.second--;
}
}
return v;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
vector<int> res = frequencySort(a);
for (auto& it : res) {
cout << it << " ";
}
cout << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code:
#include <iostream>
using namespace std;
int Dishes(int N, int T){
return T-N;
}
int main(){
int n,k;
scanf("%d%d",&n,&k);
printf("%d",Dishes(n,k));
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, 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));
BufferedWriter bo=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
try{
t=Integer.parseInt(br.readLine());
}
catch(Exception e)
{
return;
}
while(t-->0)
{
String[] g=br.readLine().split(" ");
int n=Integer.parseInt(g[0]);
int k=Integer.parseInt(g[1]);
if(k>n || (k==1) || (k>26))
{
if(n==1 && k==1)
bo.write("a\n");
else
bo.write(-1+"\n");
}
else
{
int extra=k-2;
boolean check=true;
while(n>extra)
{
if(check==true)
bo.write("a");
else
bo.write("b");
if(check==true)
check=false;
else
check=true;
n--;
}
for(int i=0;i<extra;i++)
bo.write((char)(i+99));
bo.write("\n");
}
}
bo.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: t=int(input())
for tt in range(t):
n,k=map(int,input().split())
if (k==1 and n>1) or (k>n):
print(-1)
continue
s="abcdefghijklmnopqrstuvwxyz"
ss="ab"
if (n-k)%2==0:
a=ss*((n-k)//2)+s[:k]
else:
a=ss*((n-k)//2)+s[:2]+"a"+s[2:k]
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef unsigned long long int ull;
const long double PI = acos(-1);
const ll mod=1e9+7;
const ll mod1=998244353;
const int inf = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,k;
cin >> n >> k;
if(k==1){
if(n>1){
cout << -1 << endl;
}else{
cout << 'a' << endl;
}
}else if(n<k){
cout << -1 << endl;
}else if(n==k){
string s="";
for(int i=0 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}else{
string s="";
for(int i=0 ; i<(n-k+2) ; i++){
if(i%2){
s+="b";
}else{
s+="a";
}
}
for(int i=2 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
cin >> test;
while(test--){
TEST_CASE();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a queue of integers and N queries. Your the task is to perform these operations:-
enqueue:-this operation will add an element to your current queue.
dequeue:-this operation will delete the element from the starting of the queue
displayfront:-this operation will print the element presented at front
Note:-if queue is empty than dequeue operation will do nothing, and 0 will be printed as a front element of queue if it is empty.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>enqueue()</b>:- that takes integer to be added as a parameter.
<b>dequeue()</b>:- that takes no parameter.
<b>displayfront()</b> :- that takes no parameter.
Constraints:
1 <= N(Number of queries) <= 10<sup>3</sup>You don't need to print anything else other than in displayfront() in which you require to print the element at front of your queue in a new line, if the queue is empty you just need to print 0.Sample Input:-
7
displayfront
enqueue 2
displayfront
enqueue 4
displayfront
dequeue
displayfront
Sample Output:-
0
2
2
4
Sample input:
5
enqueue 4
enqueue 5
displayfront
dequeue
displayfront
Sample output:-
4
5, I have written this Solution Code: class Queue
{
private Node front, rear;
private int currentSize;
class Node {
Node next;
int val;
Node(int val) {
this.val = val;
next = null;
}
}
public Queue()
{
front = null;
rear = null;
currentSize = 0;
}
public boolean isEmpty()
{
return (currentSize <= 0);
}
public void dequeue()
{
if (isEmpty())
{
}
else{
front = front.next;
currentSize--;
}
}
//Add data to the end of the list.
public void enqueue(int data)
{
Node oldRear = rear;
rear = new Node(data);
if (isEmpty())
{
front = rear;
}
else
{
oldRear.next = rear;
}
currentSize++;
}
public void displayfront(){
if(isEmpty()){
System.out.println("0");
}
else{
System.out.println(front.val);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called amicable if there is no character in the string which appears exactly once. For example, "abcccba" is amicable, whereas "abcba" is not amicable as the character 'c' appears only once. Your task is to find the total number of amicable substrings of a given string S.
Note:
Two substrings are considered different if either their starting position or their ending position are different, even if their contents may be equal.The first line consists of a single integer T – the number of test cases.
Then T lines follow, the i<sup>th</sup> line containing a string S for the i<sup>th</sup> test case.
<b>Constraints:</b>
1 ≤ T ≤ 10<sup>5</sup>
The sum of lengths of strings across all test cases is less than or equal to 10<sup>5</sup>.For each test case, print a single line containing a single integer – the number of amicable substrings of string S.Sample Input 1:
3
abbac
bvb
sasa
Sample Output 1:
2
0
1
Explanation:
For the first test case, the amicable substrings are "bb" and "abba".
For the second test case, there are no amicable substrings.
For the third test case, there is only one amicable substring "sasa".
Sample Input 2:
3
meeoowww
idqaislac
eommrejv
Sample Output 2:
10
0
1, 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;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(t);
while (t--)
{
reads(s);
int n = sz(s), ans = 0, valid = 0;
vi l(26), sl(26), cnt(n + 1);
FOR (i, 1, n)
{
int ch = s[i - 1] - 'a';
valid++;
FOR (j, sl[ch] + 1, l[ch])
{
cnt[j]--;
if (j > 0 && cnt[j] == 0) valid++;
}
FOR (j, l[ch] + 1, i)
{
cnt[j]++;
if (j > 0 && cnt[j] == 1) valid--;
}
sl[ch] = l[ch];
l[ch] = i;
ans += valid;
}
print(ans);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called amicable if there is no character in the string which appears exactly once. For example, "abcccba" is amicable, whereas "abcba" is not amicable as the character 'c' appears only once. Your task is to find the total number of amicable substrings of a given string S.
Note:
Two substrings are considered different if either their starting position or their ending position are different, even if their contents may be equal.The first line consists of a single integer T – the number of test cases.
Then T lines follow, the i<sup>th</sup> line containing a string S for the i<sup>th</sup> test case.
<b>Constraints:</b>
1 ≤ T ≤ 10<sup>5</sup>
The sum of lengths of strings across all test cases is less than or equal to 10<sup>5</sup>.For each test case, print a single line containing a single integer – the number of amicable substrings of string S.Sample Input 1:
3
abbac
bvb
sasa
Sample Output 1:
2
0
1
Explanation:
For the first test case, the amicable substrings are "bb" and "abba".
For the second test case, there are no amicable substrings.
For the third test case, there is only one amicable substring "sasa".
Sample Input 2:
3
meeoowww
idqaislac
eommrejv
Sample Output 2:
10
0
1, I have written this Solution Code: import java.util.*;
import java.io.*;
public class Main {
static long mod = 1000000007;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
int t = sc.nextInt();
while( t-- > 0) {
char arr[] = sc.next().toCharArray();
int n = arr.length;
long total = ((1L)*(n)*(n+1))/2;
long ans = 0;
for( int i = 0 ;i < n ;i++) {
int a[] = new int[26];
int two = 0;
int one = 0;
int zero = 26;
for( int j = i ; j < n && two != 26 ;j++) {
int idx = arr[j]-'a';
if( a[idx] == 1) {
a[idx]++;
one--;
two++;
}
else if( a[idx] == 0) {
one++;
zero--;
a[idx]++;
}
if( one != 0) {
ans++;
}
}
for( int j = 0 ;j < 26 ; j++) {
a[j] = 0;
}
}
out.println(total - ans);
}
out.flush();
}
public static boolean ifpowof2(long n ) {
return ((n&(n-1)) == 0);
}
static boolean isprime(long x ) {
if( x== 2) {
return true;
}
if( x%2 == 0) {
return false;
}
for( long i = 3 ;i*i <= x ;i+=2) {
if( x%i == 0) {
return false;
}
}
return true;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int)n + 1];
for (int i = 0; i <= n; i++) {
prime[i] = true;
}
for (long p = 2; p * p <= n; p++) {
if (prime[(int)p] == true) {
for (long i = p * p; i <= n; i += p)
prime[(int)i] = false;
}
}
return prime;
}
public static int[] nextLargerElement(int[] arr, int n) {
Stack<Integer> stack = new Stack<>();
int rtrn[] = new int[n];
rtrn[n-1] = -1;
stack.push( n-1);
for( int i = n-2 ;i >= 0 ; i--){
int temp = arr[i];
int lol = -1;
while( !stack.isEmpty() && arr[stack.peek()] <= temp){
if(arr[stack.peek()] == temp ) {
lol = stack.peek();
}
stack.pop();
}
if( stack.isEmpty()){
if( lol != -1) {
rtrn[i] = lol;
}
else {
rtrn[i] = -1;
}
}
else{
rtrn[i] = stack.peek();
}
stack.push( i);
}
return rtrn;
}
static void mysort(int[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
int loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static void mySort(long[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
long loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
static long rightmostsetbit(long n) {
return n&-n;
}
static long leftmostsetbit(long n)
{
long k = (long)(Math.log(n) / Math.log(2));
return k;
}
static HashMap<Long,Long> primefactor( long n){
HashMap<Long ,Long> hm = new HashMap<>();
long temp = 0;
while( n%2 == 0) {
temp++;
n/=2;
}
if( temp!= 0) {
hm.put( 2L, temp);
}
long c = (long)Math.sqrt(n);
for( long i = 3 ; i <= c ; i+=2) {
temp = 0;
while( n% i == 0) {
temp++;
n/=i;
}
if( temp!= 0) {
hm.put( i, temp);
}
}
if( n!= 1) {
hm.put( n , 1L);
}
return hm;
}
static ArrayList<Long> allfactors(long abs) {
HashMap<Long,Integer> hm = new HashMap<>();
ArrayList<Long> rtrn = new ArrayList<>();
for( long i = 2 ;i*i <= abs; i++) {
if( abs% i == 0) {
hm.put( i , 0);
hm.put(abs/i, 0);
}
}
for( long x : hm.keySet()) {
rtrn.add(x);
}
if( abs != 0) {
rtrn.add(abs);
}
return rtrn;
}
public static int[][] prefixsum( int n , int m , int arr[][] ){
int prefixsum[][] = new int[n+1][m+1];
for( int i = 1 ;i <= n ;i++) {
for( int j = 1 ; j<= m ; j++) {
int toadd = 0;
if( arr[i-1][j-1] == 1) {
toadd = 1;
}
prefixsum[i][j] = toadd + prefixsum[i][j-1] + prefixsum[i-1][j] - prefixsum[i-1][j-1];
}
}
return prefixsum;
}
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;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: If the total income in Wonderland is strictly more than 100 dollars, a tax of 10 dollars is deducted. Alice's total income is X dollars. She was wondering how much money will she get. Help Alice to find out.The first line of input will contain a single integer T, denoting the number of test cases. The first and only line of each test case contains a single integer X — Alice's total income.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ X ≤ 1000For each test case, output on a new line, the amount of money Alice gets.Sample Input :
4
5
105
101
100
Sample Output :
5
95
91
100, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
while(x-->0){
int n = sc.nextInt();
if(n > 100){
System.out.println(n-10);
}
else{
System.out.println(n);
}
}
// your code goes here
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: If the total income in Wonderland is strictly more than 100 dollars, a tax of 10 dollars is deducted. Alice's total income is X dollars. She was wondering how much money will she get. Help Alice to find out.The first line of input will contain a single integer T, denoting the number of test cases. The first and only line of each test case contains a single integer X — Alice's total income.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ X ≤ 1000For each test case, output on a new line, the amount of money Alice gets.Sample Input :
4
5
105
101
100
Sample Output :
5
95
91
100, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
// operator overload of << for vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
for (const auto &x : v)
os << x << " ";
return os;
}
int32_t main() {
ios_base::sync_with_stdio(true);
cin.tie(nullptr);
cout.tie(nullptr);
auto start = std::chrono::high_resolution_clock::now();
int tt;
cin >> tt;
while (tt--) {
int x;
cin >> x;
if (x > 100) {
cout << x - 10 << "\n";
} else {
cout << x << "\n";
}
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two integer N and M. You have to construct an array A having N non- negative integers such that the sum of its elements is M. As there are many such arrays possible, find the maximum <a href = "https://en.wikipedia.org/wiki/Median">median</a> of all such possible arrays.The first line of the input contains two integers N and M.
Constraints:
1 <= N, M <= 10<sup>9</sup>Print the maximum median possible.Sample Input:
3 5
Sample Output:
2
Explanation:
One possible array: [1, 2, 2], I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
int n, s;
cin >> n >> s;
int l = 0, r = s + 1;
while (r - l > 1) {
int M = (l + r) / 2;
int m = n / 2 + 1;
if (m * M <= s) {
l = M;
} else {
r = M;
}
}
cout << l;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
bool EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
int main(){
int n1,n2,v1,v2;
cin>>n1>>n2>>v1>>v2;
if(EqualOrNot(n1,n2,v1,v2)){
cout<<"Yes";}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: def EqualOrNot(h1,h2,v1,v2):
if (v2>v1 and (h1-h2)%(v2-v1)==0):
return True
else:
return False
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<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>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -1, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
t = int(input())
def findFloor(arr, l, h, x, res):
if(l<=h):
m = l+(h-l)//2
if(arr[m] == x):
return m
if(arr[m] > x):
return findFloor(arr, l, m-1, x, res)
if(arr[m] < x):
res = m
return findFloor(arr, m+1, h, x, res)
else:
return res
def findCeil(arr, l, h, x, res):
if(l<=h):
m = l+(h-l)//2
if(arr[m] == x):
return m
if(arr[m] < x):
return findCeil(arr, m+1, h, x, res)
res = m
return findCeil(arr, l, m-1, x, res)
else:
return res
for _ in range(t):
x = int(input())
f = findFloor(arr, 0, n-1, x, -1)
c = findCeil(arr, 0, n-1, x, -1)
floor = -1
ceil = -1
if(f!=-1):
floor = arr[f]
if(c!=-1):
ceil = arr[c]
print(floor,ceil), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<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>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -1, I have written this Solution Code: static void floorAndCeil(int a[], int n, int x){
int it = lower(a,n,x);
if(it==0){
if(a[it]==x){
System.out.println(x+" "+x);
}
else{
System.out.println("-1 "+a[it]);
}
}
else if (it==n){
it--;
System.out.println(a[it]+" -1");
}
else{
if(a[it]==x){
System.out.println(x+" "+x);
}
else{
it--;
System.out.println(a[it]+" "+a[it+1]);
}
}
}
static int lower(int a[], int n,int k){
int l=0;
int h=n-1;
int m;
int ans=n;
while(l<=h){
m=l+h;
m/=2;
if(a[m]<k){
l=m+1;
}
else{
h=m-1;
ans=m;
}
}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of N integers a[], and Q queries. For each query, you will be given one element K your task is to print the maximum element from the array which is less than or equal to the given element(Floor), and the minimum element from the array which is greater than or equal to the given element(Ceil).<b>In case of Java only</b>
<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>floorAndCeil()</b> that takes the array a[], integer N and integer k as arguments.
<b>Custom Input</b>
The first line of input contains a single integer N. The Second line of input contains N space-separated integers depicting the values of the array. The third line of input contains a single integer Q. The next Q line of input contains a single integer the value of K.
Constraints:-
1 <= N <= 100000
1 <= K, Arr[i] <= 1000000000000
1 <= Q <= 10000In a new line Print two space-separated integers depicting the values of Floor and Ceil of the given number. If the floor or ceil of the element does not exist print -1.Sample Input:-
5
2 5 6 11 15
5
2
4
8
1
16
Sample Output:-
2 2
2 5
6 11
-1 2
15 -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 max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n;
cin>>n;
vector<int> v;
int x;
FOR(i,n){
cin>>x;
v.EB(x);}
int q;
cin>>q;
while(q--){
cin>>x;
auto it = lower_bound(v.begin(),v.end(),x);
if(it==v.begin()){
if(*it==x){
cout<<x<<" "<<x;
}
else{
cout<<-1<<" "<<*it;
}
}
else if (it==v.end()){
it--;
cout<<*it<<" -1";
}
else{
if(*it==x){
cout<<x<<" "<<x;
}
else{
it--;
cout<<*it<<" ";
it++;
cout<<*it;}
}
END;
}
}
, 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 check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
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();
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++){
if(n%i==0){System.out.print("NO");return;}
}
System.out.print("YES");
}
}
, 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 check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: import math
def isprime(A):
if A == 1:
return False
sqrt = int(math.sqrt(A))
for i in range(2,sqrt+1):
if A%i == 0:
return False
return True
inp = int(input())
if isprime(inp):
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, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Monk is standing at the door of his classroom. There are currently N students in the class, i'th student got Ai candies.
There are still M more students to come. At every instant, a student enters the class and wishes to be seated with a student who has exactly the same number of candies. For each student, Monk shouts YES if such a student is found, NO otherwise.First line contains two space- separated integers N and M.
Second line contains N + M space- separated integers, the candies of the students.output M new line, Monk's answer to the M students.
Print "YES" (without the quotes) or "NO" (without the quotes) pertaining to the Monk's answer.
Constraints:
1 ≤ N, M ≤ 100000
0 ≤ Ai ≤ 1000000000Sample Input 1:
2 3
3 2 9 11 2
Sample Output 1:
NO
NO
YES
Explanations:
Initially students with 3 and 2 candies are in the class.
A student with 9 candies enters, No student with 9 candies in class. Hence, "NO"
A student with 11 candies enters, No student with 11 candies in class. Hence, "NO"
A student with 2 candies enters, Student with 2 candies found in class. Hence, "YES", 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));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
String s;
StringTokenizer st;
if ((s = br.readLine().trim()) != null) {
st = new StringTokenizer(s);
int N = Integer.parseInt(st.nextToken());
int M = Integer.parseInt(st.nextToken());
Set<Long> arr = new HashSet<Long>();
StringBuilder ans = new StringBuilder();
if ((s = br.readLine().trim()) != null) {
st = new StringTokenizer(s);
for (int i = 0; i < N; i++) {
long num = Long.parseLong(st.nextToken());
arr.add(num);
}
for (int i = 0; i < M; i++) {
long num = Long.parseLong(st.nextToken());
if (arr.contains(num)) {
ans.append("YES\n");
} else {
arr.add(num);
ans.append("NO\n");
}
}
out.print(ans);
}
}
out.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Monk is standing at the door of his classroom. There are currently N students in the class, i'th student got Ai candies.
There are still M more students to come. At every instant, a student enters the class and wishes to be seated with a student who has exactly the same number of candies. For each student, Monk shouts YES if such a student is found, NO otherwise.First line contains two space- separated integers N and M.
Second line contains N + M space- separated integers, the candies of the students.output M new line, Monk's answer to the M students.
Print "YES" (without the quotes) or "NO" (without the quotes) pertaining to the Monk's answer.
Constraints:
1 ≤ N, M ≤ 100000
0 ≤ Ai ≤ 1000000000Sample Input 1:
2 3
3 2 9 11 2
Sample Output 1:
NO
NO
YES
Explanations:
Initially students with 3 and 2 candies are in the class.
A student with 9 candies enters, No student with 9 candies in class. Hence, "NO"
A student with 11 candies enters, No student with 11 candies in class. Hence, "NO"
A student with 2 candies enters, Student with 2 candies found in class. Hence, "YES", I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
struct node
{
long long int data;
struct node* left;
struct node* right;
};
struct node* newnode(long long int req)
{
struct node* temp = (struct node*)malloc(sizeof(struct node));
temp->data=req;
temp->left=NULL;
temp->right=NULL;
return temp;
}
struct node* insert(struct node* item,long long int val)
{
if(item==NULL)
return newnode(val);
if(val < item->data)
item->left=insert(item->left,val);
else if(val > item->data)
item->right=insert(item->right,val);
return item;
}
bool check(struct node* res,long long int p)
{
while(res!=NULL)
{
if(p < res->data)
res= res->left;
else if(p > res->data)
res= res->right;
else
{
return true;
}
}
return false;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t=1;
//cin>>t;
while(t--)
{
struct node *root=NULL;
long long int n,m,a;
cin>>n>>m;
unordered_map<long long int,int> hash;
for(int i=0;i<n;i++)
{
cin>>a;
root=insert(root,a);
}
for(int i=0;i<m;i++)
{
cin>>a;
if(check(root,a))
{
cout<<"YES\n";
}
else
{
if(hash[a]==0)
{
hash[a]=1;
cout<<"NO\n";
}
else
cout<<"YES\n";
}
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Monk is standing at the door of his classroom. There are currently N students in the class, i'th student got Ai candies.
There are still M more students to come. At every instant, a student enters the class and wishes to be seated with a student who has exactly the same number of candies. For each student, Monk shouts YES if such a student is found, NO otherwise.First line contains two space- separated integers N and M.
Second line contains N + M space- separated integers, the candies of the students.output M new line, Monk's answer to the M students.
Print "YES" (without the quotes) or "NO" (without the quotes) pertaining to the Monk's answer.
Constraints:
1 ≤ N, M ≤ 100000
0 ≤ Ai ≤ 1000000000Sample Input 1:
2 3
3 2 9 11 2
Sample Output 1:
NO
NO
YES
Explanations:
Initially students with 3 and 2 candies are in the class.
A student with 9 candies enters, No student with 9 candies in class. Hence, "NO"
A student with 11 candies enters, No student with 11 candies in class. Hence, "NO"
A student with 2 candies enters, Student with 2 candies found in class. Hence, "YES", I have written this Solution Code: import numpy as np
from collections import defaultdict
n,m=input().strip().split()
n=int(n)
m=int(m)
a=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in range(n):
d[a[i]]+=1
for i in range(m):
if(d[a[n+i]]):
print("YES")
else:
print("NO")
d[a[n+i]]+=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String str = read.readLine();
System.out.println(maxProduct(str));
}
public static long maxProduct(String str) {
StringBuilder sb = new StringBuilder(str);
int x = sb.length();
int[] dpl = new int[x];
int[] dpr = new int[x];
modifiedOddManacher(sb.toString(), dpl);
modifiedOddManacher(sb.reverse().toString(), dpr);
long max=1;
for(int i=0;i<x-1;i++)
max=Math.max(max, (1+(dpl[i]-1)*2L)*(1+(dpr[x-(i+1)-1]-1)*2L));
return max;
}
private static void modifiedOddManacher(String str, int[] dp){
int x = str.length();
int[] center = new int[x];
for(int l=0,r=-1,i=0;i<x;i++){
int radius = (i > r) ? 1 : Math.min(center[l+(r-i)], r-i+1);
while(i-radius>=0 && i+radius<x && str.charAt(i-radius)==str.charAt(i+radius)) {
dp[i+radius] = radius+1;
radius++;
}
center[i] = radius--;
if(i+radius>r){
l = i-radius;
r = i+radius;
}
}
for(int i=0, max=1;i<x;i++){
max = Math.max(max, dp[i]);
dp[i] = max;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: s=input()
n = len(s)
hlen = [0]*n
center = right = 0
for i in range(n):
if i < right:
hlen[i] = min(right - i, hlen[2*center - i])
while 0 <= i-1-hlen[i] and i+1+hlen[i] < len(s) and s[i-1-hlen[i]] == s[i+1+hlen[i]]:
hlen[i] += 1
if right < i+hlen[i]:
center, right = i, i+hlen[i]
left = [0]*n
right = [0]*n
for i in range(n):
left[i+hlen[i]] = max(left[i+hlen[i]], 2*hlen[i]+1)
right[i-hlen[i]] = max(right[i-hlen[i]], 2*hlen[i]+1)
for i in range(1, n):
left[~i] = max(left[~i], left[~i+1]-2)
right[i] = max(right[i], right[i-1]-2)
for i in range(1, n):
left[i] = max(left[i-1], left[i])
right[~i] = max(right[~i], right[~i+1])
print(max(left[i-1]*right[i] for i in range(1, n))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountl
#define m_p make_pair
#define inf 200000000000000
#define MAXN 1000001
#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);
}
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
#define S second
#define F first
#define int long long
/////////////
int v1[100001]={};
int v2[100001]={};
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
string s;
cin>>s;
n=s.length();
vector<int> d1(n);
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && s[i - k] == s[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
int c=0;
for(int i=0;i<n;++i)
{
int x=2*d1[i]-1;
int j=i+d1[i]-1;
while(v1[j]<x&&j>=i)
{
v1[j]=x;
x-=2;
++c;
--j;
}
}
for(int i=1;i<n;++i)
{
v1[i]=max(v1[i],v1[i-1]);
}
for(int i=n-1;i>=0;--i)
{
int x=2*d1[i]-1;
int j=i-d1[i]+1;
while(v2[j]<x&&j<=i)
{
v2[j]=x;
x-=2;
++j;
++c;
}
}
for(int i=n-2;i>=0;--i)
{
v2[i]=max(v2[i],v2[i+1]);
}
int ans=0;
for(int i=1;i<n;++i)
{
ans=max(ans,v1[i-1]*v2[i]);
}
cout<<ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find out the shortest distance between two even positive integers in an array <b>A</b> and print the distance. If there is one or zero even positive integers in the array then return -1.
<b>Note</b>: If the index of one even positive integer is i and the index of another even positive integer is j then their distance will be |i- j|.The first line of the input contains N denoting the size of the array.
The second line of the input contains N space- separated integers denoting an array A.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>9</sup> ≤ A[i] ≤ 10<sup>9</sup>Print the shortest distance if more than one even positive integer is present or return -1 if only one or no even positive integer is present<b>Sample Input</b>
2
1 2
<b>Sample Output</b>
-1
<b>Explanation</b>
There is only 1 even positive integer in the array. So the output will be -1., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int shortEven(int n, vector<int> &arr) {
// write code here
vector<int> even;
for (int i = 0; i < n; i++) {
if (!(arr[i] & 1) and arr[i] > 0) {
even.push_back(i);
}
}
if (even.size() <= 1) {
return -1;
} else {
int ans = INT_MAX;
for (int i = 1; i < even.size(); i++) {
ans = min(abs(even[i] - even[i - 1]), ans);
}
return ans;
}
}
int main() {
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; ++i) {
cin >> arr[i];
}
int ans = shortEven(n, arr);
cout << ans << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Negi is fascinated with the binary representation of the number. Tell him the number of set bits (ones) in the binary representation of an integer N.The first line of the input contains single integer N.
Constraints
1 <= N <= 1000000000000The output should contain a single integer, the number of set bits (ones) in the binary representation of an integer N.Sample Input
7
Sample Output
3
Sample Input
16
Sample Output
1, 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));
long n = Long.parseLong(br.readLine());
int count = 0;
try{
while (n > 0) {
count += n & 1;
n >>= 1;
}
}catch(Exception e){
return ;
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Negi is fascinated with the binary representation of the number. Tell him the number of set bits (ones) in the binary representation of an integer N.The first line of the input contains single integer N.
Constraints
1 <= N <= 1000000000000The output should contain a single integer, the number of set bits (ones) in the binary representation of an integer N.Sample Input
7
Sample Output
3
Sample Input
16
Sample Output
1, I have written this Solution Code: a=int(input())
l=bin(a)
print(l.count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Negi is fascinated with the binary representation of the number. Tell him the number of set bits (ones) in the binary representation of an integer N.The first line of the input contains single integer N.
Constraints
1 <= N <= 1000000000000The output should contain a single integer, the number of set bits (ones) in the binary representation of an integer N.Sample Input
7
Sample Output
3
Sample Input
16
Sample Output
1, 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
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
signed main()
{
int n;
cin>>n;
int cnt=0;
while(n>0)
{
int p=n%2LL;
cnt+=p;
n/=2LL;
}
cout<<cnt<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code: def DiceProblem(N):
return (7-N)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
static int diceProblem(int N){
return (7-N);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How would you add your own method to the Array object so
the following code would work?
const arr = [1, 2, 3]
console. log(arr.average())
// 2input will be an array, run like this
const anyArray = [5,6...]
anyArray.average should return average of the arraysAverage of the given arrayconst myArray = [1,2,3,4,5]
console.log(myArray.average())
// 3, I have written this Solution Code: Array.prototype.average = function() {
// calculate sum
var sum = this.reduce(function(prev, cur) { return prev + cur; });
// return sum divided by number of elements
return sum / this.length;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in =new BufferedReader(new InputStreamReader(System.in));
StringBuilder s = new StringBuilder();
String text=null;
while ((text = in.readLine ()) != null)
{
s.append(text);
}
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)==s.charAt(i+1)){
int flag=0;
s.delete(i,i+2);
int left=i-1;
len=len-2;
i=i-2;
if(i<0){
i=-1;
}
}
}
System.out.println(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: s=input()
l=["aa","bb","cc","dd","ee","ff","gg","hh","ii","jj","kk","ll","mm","nn","oo","pp","qq","rr","ss","tt","uu","vv","ww","xx","yy","zz"]
while True:
do=False
for i in range(len(l)):
if l[i] in s:
do=True
while l[i] in s:
s=s.replace(l[i],"")
if do==False:
break
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 you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, 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
string s;
cin>>s;
int len=s.length();
char stk[410000];
int k = 0;
for (int i = 0; i < len; i++)
{
stk[k++] = s[i];
while (k > 1 && stk[k - 1] == stk[k - 2])
k -= 2;
}
for (int i = 0; i < k; i++)
cout << stk[i];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b><em>Array List In Java</em></b>
There is a list, having n integers that may have duplicates, is given. Print all unique elements in sorted order.There is an integer n is given in first line of input.
In Second line, n space separated integers are given.
<b>Constraints</b>
1 <= n <= 10<sup>4</sup>Print all unique elements in sorted order.Sample Input:
7
1 2 4 3 5 4 3
Sample Output:
1 2 3 4 5, I have written this Solution Code: // Java program for the above approach
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
public class Main {
public static void main(String[] args)
{
TreeSet<Integer> st=new TreeSet<>();
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++){
int a=sc.nextInt();
st.add(a);
}
for(Integer x:st){
System.out.print(x);
System.out.print(" ");
}
System.out.println("");
return;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
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));
String st = br.readLine();
int len = 0;
int c=0;
for(int i=0;i<st.length();i++){
if(st.charAt(i)=='A'){
c++;
len = Math.max(len,c);
}else{
c=0;
}
}
System.out.println(len);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: S=input()
max=0
flag=0
for i in range(0,len(S)):
if(S[i]=='A' or S[i]=='B'):
if(S[i]=='A'):
flag+=1
if(flag>max):
max=flag
else:
flag=0
print(max), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, 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(){
string s; cin>>s;
int ct = 0;
int ans = 0;
for(char c: s){
if(c == 'A')
ct++;
else
ct=0;
ans = max(ans, ct);
}
cout<<ans;
}
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: You are given an array a of n integers a<sub>1</sub>, a<sub>2</sub>,... , a<sub>n</sub>. Your task is to response to the queries like : How many number's values are between l and r ?The first line of the input contains a single space separated integer n denoting the length of the array.
The second line of the input contains n space separated integers denoting an array a.
The third line of the input contains a single space separated integer k denoting the number of queries.
The following k lines contain a pair of integers l and r - query, described above.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>9</sup> ≤ a[i] ≤ 10<sup>9</sup>
1 ≤ k ≤ 10<sup>5</sup>
-10<sup>9</sup> ≤ l ≤ r ≤ 10<sup>9</sup>The output must consist of k integers - responses for the queries.Sample Input
5
10 1 10 3 4
4
1 10
2 9
3 4
2 2
Sample Output
5 2 2 0, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(int l = LLONG_MIN, int r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
vector<int> a;
int searchLeft(int x, int n) {
int l = 0, r = n - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] <= x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return l;
}
int searchRight(int x, int n) {
int l = 0, r = n - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (a[mid] < x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return l + 1;
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, k;
cin >> n;
a.assign(n, 0);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
cin >> k;
while (k--) {
int x, y;
cin >> x >> y;
x -= 1, y += 1;
int left = searchLeft(x, n);
int right = searchRight(y, n);
cout << right - left - 1 << " ";
}
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 as an input, print a pattern as shown in sample.Input contains a single integer N
Constraints:-
1<=N<=15Print the patter for N.Sample Input:
3
Sample Output:
3 3 3 3 3
3 2 2 2 3
3 2 1 2 3
3 2 2 2 3
3 3 3 3 3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max 100
void print(int a[][max], int size)
{
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
cout << a[i][j]<<" ";
}
cout <<endl;
}
}
void innerPattern(int n) {
int size = 2 * n - 1;
int front = 0;
int back = size - 1;
int a[max][max];
while (n != 0)
{
for (int i = front; i <= back; i++) {
for (int j = front; j <= back; j++) {
if (i == front || i == back ||
j == front || j == back)
a[i][j] = n;
}
}
++front;
--back;
--n;
}
print(a, size);
}
int main()
{
int n ;
cin>>n;
innerPattern(n);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of several arrays that each contain integers and your goal is to write a function that
will sum up all the numbers in all the arrays. For example, if the input is [[3, 2], [1], [4, 12]] then your
program should output 22 because 3 + 2 + 1 + 4 + 12 = 22An array containing arrays which can contain any number of elements.Sum of all the elements in all of the arrays.Sample input:-
[[3, 2], [1], [4, 12]]
Sample output:-
22
Explanation:-
3 + 2 + 1 + 4 + 12 = 22, I have written this Solution Code: function sum_array(arr) {
// store our final answer
var sum = 0;
// loop through entire array
for (var i = 0; i < arr.length; i++) {
// loop through each inner array
for (var j = 0; j < arr[i].length; j++) {
// add this number to the current final sum
sum += arr[i][j];
}
}
console.log(sum);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String s = in.readLine().trim();
char[] express = s.toCharArray();
StringBuilder ans = new StringBuilder("");
Stack<Character> stack = new Stack<>();
for(int i=0; i<express.length; i++){
if(express[i]>=65 && express[i]<=90){
ans.append(express[i]);
}else if(express[i]==')'){
while(stack.peek()!='('){
ans.append(stack.peek());
stack.pop();
}
stack.pop();
}else if(stack.empty() || express[i]=='(' || precedence(express[i])>=precedence(stack.peek())){
stack.push(express[i]);
}else{
while(!stack.empty() && stack.peek()!='(' && precedence(stack.peek())<=express[i]){
ans.append(stack.peek());
stack.pop();
}
stack.push(express[i]);
}
}
while(!stack.empty()){
ans.append(stack.peek());
stack.pop();
}
System.out.println(ans);
}
static int precedence(char operator){
int precedenceValue;
if(operator=='+' || operator=='-')
precedenceValue = 1;
else if(operator=='*' || operator=='/')
precedenceValue = 2;
else if(operator == '^')
precedenceValue = 3;
else
precedenceValue = -1;
return precedenceValue;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: string=input()
stack=list()
preced={'+':1,'-':1,'*':2,'/':2,'^':3}
for i in string:
if i!=')':
if i=='(':
stack.append(i)
elif i in ['*','+','/','-','^']:
while(stack and stack[-1]!='(' and preced[i]<=preced[stack[-1]]):
print(stack.pop(),end="")
stack.append(i)
else:
print(i,end="")
else:
while(stack and stack[-1]!='('):
print(stack.pop(),end="")
stack.pop()
while(stack):
print(stack.pop(),end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
//Function to return precedence of operators
int prec(char c)
{
if(c == '^')
return 3;
else if(c == '*' || c == '/')
return 2;
else if(c == '+' || c == '-')
return 1;
else
return -1;
}
// The main function to convert infix expression
//to postfix expression
void infixToPostfix(string s)
{
std::stack<char> st;
st.push('N');
int l = s.length();
string ns;
for(int i = 0; i < l; i++)
{
// If the scanned character is an operand, add it to output string.
if((s[i] >= 'a' && s[i] <= 'z')||(s[i] >= 'A' && s[i] <= 'Z'))
ns+=s[i];
// If the scanned character is an ‘(‘, push it to the stack.
else if(s[i] == '(')
st.push('(');
// If the scanned character is an ‘)’, pop and to output string from the stack
// until an ‘(‘ is encountered.
else if(s[i] == ')')
{
while(st.top() != 'N' && st.top() != '(')
{
char c = st.top();
st.pop();
ns += c;
}
if(st.top() == '(')
{
char c = st.top();
st.pop();
}
}
//If an operator is scanned
else{
while(st.top() != 'N' && prec(s[i]) <= prec(st.top()))
{
char c = st.top();
st.pop();
ns += c;
}
st.push(s[i]);
}
}
//Pop all the remaining elements from the stack
while(st.top() != 'N')
{
char c = st.top();
st.pop();
ns += c;
}
cout << ns << endl;
}
//Driver program to test above functions
int main()
{
string exp;
cin>>exp;
infixToPostfix(exp);
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 Doubly linked list and an integer K . Your task is to insert the integer K at the head of the given linked 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>insertnew()</b> that takes the head node and the integer K as a parameter.
<b>Constraints:</b>
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
2 1 2 3 4 5
, I have written this Solution Code: public static Node insertnew(Node head, int k) {
Node temp = new Node(k);
temp.next = head;
head.prev=temp;
return temp;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
import java.math.BigInteger;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
BigInteger sum;
String ip1 = sc.next();
String ip2 = sc.next();
BigInteger a = new BigInteger(ip1);
BigInteger b = new BigInteger(ip2);
sum = a.add(b);
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-03 02:46:30
**/
#include <bits/stdc++.h>
#define NX 105
#define MX 3350
using namespace std;
const int mod = 998244353;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
typedef long long INT;
const int pb = 10;
const int base_digits = 9;
const int base = 1000000000;
const int DIV = 100000;
struct bigint {
vector<int> a;
int sign;
bigint() : sign(1) {}
bigint(INT v) { *this = v; }
bigint(const string &s) { read(s); }
void operator=(const bigint &v) { sign = v.sign, a = v.a; }
void operator=(INT v) {
sign = 1;
if (v < 0) sign = -1, v = -v;
for (; v > 0; v = v / base) a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) ||
carry;
i++) {
if (i == (int)res.a.size()) res.a.push_back(0);
res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry) res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int)v.a.size() || carry; i++) {
res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry) res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int)a.size() || carry; i++) {
if (i == (int)a.size()) a.push_back(0);
INT cur = a[i] * (INT)v + carry;
carry = (int)(cur / base);
a[i] = (int)(cur % base);
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> DIVmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((INT)base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0) r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const { return DIVmod(*this, v).first; }
bigint operator%(const bigint &v) const { return DIVmod(*this, v).second; }
void operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)a.size() - 1, rem = 0; i >= 0; i--) {
INT cur = a[i] + rem * (INT)base;
a[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (INT)base) % v;
return m * sign;
}
void operator+=(const bigint &v) { *this = *this + v; }
void operator-=(const bigint &v) { *this = *this - v; }
void operator*=(const bigint &v) { *this = *this * v; }
void operator/=(const bigint &v) { *this = *this / v; }
bool operator<(const bigint &v) const {
if (sign != v.sign) return sign < v.sign;
if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const { return v < *this; }
bool operator<=(const bigint &v) const { return !(v < *this); }
bool operator>=(const bigint &v) const { return !(*this < v); }
bool operator==(const bigint &v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint &v) const { return *this < v || v < *this; }
void trim() {
while (!a.empty() && !a.back()) a.pop_back();
if (a.empty()) sign = 1;
}
bool isZero() const { return a.empty() || (a.size() == 1 && !a[0]); }
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
INT longValue() const {
INT res = 0;
for (int i = a.size() - 1; i >= 0; i--) res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) { return b.isZero() ? a : gcd(b, a % b); }
friend bigint lcm(const bigint &a, const bigint &b) { return a / gcd(a, b) * b; }
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
pos++;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x *
pb +
s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream &operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream &operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1) stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int)v.a.size() - 2; i >= 0; --i) stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<INT> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * pb;
vector<int> res;
INT cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int)a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && !res.back()) res.pop_back();
return res;
}
typedef vector<INT> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a5 = convert_base(this->a, base_digits, 5);
vector<int> b5 = convert_base(v.a, base_digits, 5);
vll a(a5.begin(), a5.end());
vll b(b5.begin(), b5.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int)c.size(); i++) {
INT cur = c[i] + carry;
res.a.push_back((int)(cur % DIV));
carry = (int)(cur / DIV);
}
res.a = convert_base(res.a, 5, base_digits);
res.trim();
return res;
}
inline bool isOdd() { return a[0] & 1; }
};
int main() {
bigint n, m;
cin >> n >> m;
cout << n + m << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: n,m = map(int,input().split())
print(n+m) , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <b>arr</b> and a target value <b>K</b>. Print the minimum integer value <b>V</b> such that when we change all values greater than V in the array to V, <b>the absolute difference (|sum - K|) between the sum of the array elements and K is minimized.</b>
Note: the answer is not necessarily a number from arr.The input line containing T, denoting the number of testcases. Each testcase contains 2 lines. First line contains N, size of array and target K separated by space. Second line contains elements of array.
Constraints:
1 <= T <= 50
1 <= N <= 10^4
1 <= arr[i], K <= 10^5For each testcase you need to print the minimum such integer possible in a new line.Input:
2
3 10
4 9 3
3 10
2 3 5
Output:
3
5
Explanation:
Testcase 1: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer., I have written this Solution Code: def findBestValue(arr,target):
org_sum=sum(arr)
if org_sum<target:
return max(arr)
l=1
r=max(arr)
while l<r:
mid=l+(r-l)//2;
tmp=0
for val in arr:
if val>mid:
tmp+=mid
else:
tmp+=val
if tmp<target:
l=mid+1
else:
r=mid
sum1=0
sum2=0
for val in arr:
if val>l:
sum1+=l
else:
sum1+=val
if val>l-1:
sum2+=l-1
else:
sum2+=val
if abs(sum2-target)<= abs(sum1-target):
return l-1
else:
return l
t=int(input())
while t:
n,target=map(int,input().split())
list1=list(map(int,input().split()))
print(findBestValue(list1,target))
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <b>arr</b> and a target value <b>K</b>. Print the minimum integer value <b>V</b> such that when we change all values greater than V in the array to V, <b>the absolute difference (|sum - K|) between the sum of the array elements and K is minimized.</b>
Note: the answer is not necessarily a number from arr.The input line containing T, denoting the number of testcases. Each testcase contains 2 lines. First line contains N, size of array and target K separated by space. Second line contains elements of array.
Constraints:
1 <= T <= 50
1 <= N <= 10^4
1 <= arr[i], K <= 10^5For each testcase you need to print the minimum such integer possible in a new line.Input:
2
3 10
4 9 3
3 10
2 3 5
Output:
3
5
Explanation:
Testcase 1: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer., 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);
long sum=0;
int j=0;
long ans=0;
long res;
long total=INT_MAX;
for(int i=0;i<=100000;i++){
if(i>a[j] && j!=n){sum+=a[j];j++;}
ans=sum+i*(n-j);
if(abs(ans-k)<total){res=i;total=abs(ans-k);}
}
cout<<res<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <b>arr</b> and a target value <b>K</b>. Print the minimum integer value <b>V</b> such that when we change all values greater than V in the array to V, <b>the absolute difference (|sum - K|) between the sum of the array elements and K is minimized.</b>
Note: the answer is not necessarily a number from arr.The input line containing T, denoting the number of testcases. Each testcase contains 2 lines. First line contains N, size of array and target K separated by space. Second line contains elements of array.
Constraints:
1 <= T <= 50
1 <= N <= 10^4
1 <= arr[i], K <= 10^5For each testcase you need to print the minimum such integer possible in a new line.Input:
2
3 10
4 9 3
3 10
2 3 5
Output:
3
5
Explanation:
Testcase 1: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main(String args[])throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
int N = Integer.parseInt(str[0]);
int K = Integer.parseInt(str[1]);
//`int D = Integer.parseInt(str[1]);
int arr[] = new int[N];
str = read.readLine().trim().split(" ");
for(int i = 0; i < N; i++)
arr[i] = Integer.parseInt(str[i]);
//int res[] = moveZeroes(arr);
//print(res);
System.out.println(findBestValue(arr, K));
}
}
static void print(int list[])
{
for(int i = 0; i < list.length; i++)
System.out.print(list[i] + " ");
}
static int findBestValue(int arr[], int target) {
int sum = 0;
int mx = Integer.MIN_VALUE;
int sz = arr.length;
int remaining_target = target;
int remaining_items = sz;
int one_part = target / sz;
for(int i = 0; i < sz; ++i) {
mx = Math.max(mx, arr[i]);
sum += arr[i];
if(arr[i] < one_part) {
remaining_items--;
remaining_target -= arr[i];
}
}
if(sum <= target) {
return mx;
}
int val1 = remaining_target / remaining_items;
int min_val = val1 - 1;
int min_diff = Math.abs(remaining_target - (min_val * remaining_items));
int diff1 = Math.abs(remaining_target - (val1 * remaining_items));
if(diff1 < min_diff) {
min_val = val1;
min_diff = diff1;
}
int val2 = val1 + 1;
int diff2 = Math.abs(remaining_target - (val2 * remaining_items));
return (diff2 < min_diff) ? val2 : min_val;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void printMax(int arr[], int n, int k) {
int j, max;
for(int i = 0; i <= n - k; i++) {
max = arr[i];
for(j = 1; j < k; j++) {
if(arr[i + j] > max) {
max = arr[i + j];
}
}
System.out.print(max + " ");
}
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(str1[0]);
int k = Integer.parseInt(str1[1]);
String str2[] = br.readLine().trim().split(" ");
int arr[] = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(str2[i]);
}
printMax(arr, n ,k);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: n,k=input().split()
n=int(n)
k=int(k)
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
m=max(arr[0:k])
for i in range(k-1,n):
if(arr[i] > m):
m=arr[i]
if(arr[i-k]==m):
m=max(arr[i-k+1:i+1])
print (m, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// A Dequeue (Double ended queue) based method for printing maximum element of
// all subarrays of size k
void printKMax(int arr[], int n, int k)
{
// Create a Double Ended Queue, Qi that will store indexes of array elements
// The queue will store indexes of useful elements in every window and it will
// maintain decreasing order of values from front to rear in Qi, i.e.,
// arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order
std::deque<int> Qi(k);
/* Process first k (or first window) elements of array */
int i;
for (i = 0; i < k; ++i) {
// For every element, the previous smaller elements are useless so
// remove them from Qi
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back(); // Remove from rear
// Add new element at rear of queue
Qi.push_back(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (; i < n; ++i) {
// The element at the front of the queue is the largest element of
// previous window, so print it
cout << arr[Qi.front()] << " ";
// Remove the elements which are out of this window
while ((!Qi.empty()) && Qi.front() <= i - k)
Qi.pop_front(); // Remove from front of queue
// Remove all elements smaller than the currently
// being added element (remove useless elements)
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back();
// Add current element at the rear of Qi
Qi.push_back(i);
}
// Print the maximum element of last window
cout << arr[Qi.front()];
}
// Driver program to test above functions
int main()
{
int n,k;
cin>>n>>k;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
printKMax(arr, n, k);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
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));
String st = br.readLine();
int len = 0;
int c=0;
for(int i=0;i<st.length();i++){
if(st.charAt(i)=='A'){
c++;
len = Math.max(len,c);
}else{
c=0;
}
}
System.out.println(len);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: S=input()
max=0
flag=0
for i in range(0,len(S)):
if(S[i]=='A' or S[i]=='B'):
if(S[i]=='A'):
flag+=1
if(flag>max):
max=flag
else:
flag=0
print(max), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, 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(){
string s; cin>>s;
int ct = 0;
int ans = 0;
for(char c: s){
if(c == 'A')
ct++;
else
ct=0;
ans = max(ans, ct);
}
cout<<ans;
}
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 |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.