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For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<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>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: static void Icecreams (int N, int D){
int x=N;
while(D-->0){
x-=x/2;
x*=3;
}
System.out.println(x);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<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>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
cout << x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<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>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
printf("%d", x);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<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>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: def Icecreams(N,D):
ans = N
while D > 0:
ans = ans - ans//2
ans = ans*3
D = D-1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Display the acceptance rate of requests defined as the ratio of accepted contacts vs all contacts. Multiply the ratio by 100 to get the rate.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'airbnb_contacts', 'columns': [{'name': 'id_guest', 'type': 'object'}, {'name': 'id_host', 'type': 'object'}, {'name': 'id_listing', 'type': 'object'}, {'name': 'ts_contact_at', 'type': 'datetime64[ns]'}, {'name': 'ts_reply_at', 'type': 'datetime64[ns]'}, {'name': 'ts_accepted_at', 'type': 'datetime64[ns]'}, {'name': 'ts_booking_at', 'type': 'datetime64[ns]'}, {'name': 'ds_checkin', 'type': 'datetime64[ns]'}, {'name': 'ds_checkout', 'type': 'datetime64[ns]'}, {'name': 'n_guests', 'type': 'int64'}, {'name': 'n_messages', 'type': 'int64'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: print(airbnb_contacts[airbnb_contacts['ts_accepted_at']=='NaT'].shape[0]/airbnb_contacts.shape[0]*100), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Display the acceptance rate of requests defined as the ratio of accepted contacts vs all contacts. Multiply the ratio by 100 to get the rate.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'airbnb_contacts', 'columns': [{'name': 'id_guest', 'type': 'object'}, {'name': 'id_host', 'type': 'object'}, {'name': 'id_listing', 'type': 'object'}, {'name': 'ts_contact_at', 'type': 'datetime64[ns]'}, {'name': 'ts_reply_at', 'type': 'datetime64[ns]'}, {'name': 'ts_accepted_at', 'type': 'datetime64[ns]'}, {'name': 'ts_booking_at', 'type': 'datetime64[ns]'}, {'name': 'ds_checkin', 'type': 'datetime64[ns]'}, {'name': 'ds_checkout', 'type': 'datetime64[ns]'}, {'name': 'n_guests', 'type': 'int64'}, {'name': 'n_messages', 'type': 'int64'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: SELECT 100.0*SUM(CASE
WHEN ts_accepted_at == 'NaT' THEN 1
ELSE 0
END)/COUNT(*) acceptance_rate
FROM airbnb_contacts, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 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());
int a[]=new int[100001];
String str[]=br.readLine().split(" ");
for(int i=0;i<str.length;i++){
a[Integer.parseInt(str[i])]++;
}
long cnt=0;
for(int i=0;i<100001;i++){
if(a[i]>0)
cnt++;
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 3, I have written this Solution Code: a = int(input())
arr = input().split()
print(len(set(arr))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
cout<<m.size();
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long fact=1;
for(int i=1; i<=n;i++){
fact*=i;
}
System.out.print(fact);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: def fact(n):
if( n==0 or n==1):
return 1
return n*fact(n-1);
n=int(input())
print(fact(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
int main(){
int t;
t=1;
while(t--){
int n;
cin>>n;
unsigned long long sum=1;
for(int i=1;i<=n;i++){
sum*=i;
}
cout<<sum<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
public static int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;break;}
ch*=(long)3;
}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
def Ifpossible(x,y) :
result = y-x
ans = 0
ch = 1
while ans<result :
ans+=ch
if ans==result:
return 1;
ch*=3
return 0;
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You need to make an order counter to keep track of the total number of orders received.
Complete the function <code> generateOrder() </code> which returns a <code>function func()</code>. This function <code>func</code> should maintain a <code> count (initially 0)</code>. Every time <code>func</code> is called, <code> count</code> must be incremented by 1 and the string <code>"Total orders = " + count</code> must be returned.
<b>Note:</b> The function generateOrder() will be called internally. You do not need to call it yourself. The generateOrder() takes no argument. It is called internally.The generateOrder() function returns a function that returns the string <code>"Total orders = " + count</code>, where <code>count</code> is the number of times the function is called.
const initC = generateOrder(starting);
console.log(initC()) //prints "Total orders = 1"
console.log(initC()) //prints "Total orders = 2"
console.log(initC()) //prints "Total orders = 3"
, I have written this Solution Code: let generateOrder = function() {
let prefix = "Total orders = ";
let count = 0;
let totalOrders = function(){
count++
return prefix + count;
}
return totalOrders;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let us define a function f as f(x)=x<sup>2</sup>+2x+3. Given an integer t, find f(f(f(t)+t)+f(f(t))).
Here, it is guaranteed that the answer is an integer not greater than 2×10<sup>9</sup>.The input consists of a single integer.
<b>Constraints</b>
t is an integer between 0 and 10 (inclusive).Print the answer as an integer.<b>Sample Input 1</b>
0
<b>Sample Output 1</b>
1371
<b>Sample Input 2</b>
3
<b>Sample Output 2</b>
722502, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int f(int x){return x*x+2*x+3;}
int main(){
int t;
cin >> t;
cout << f(f(f(t)+t)+f(f(t))) << '\n';
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
int n = io.nextInt(), k = io.nextInt(), j = io.nextInt() - 1;
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = io.nextInt();
}
int cost = arr[j];
arr[j] = Integer.MAX_VALUE;
Arrays.sort(arr);
for(int i = 0; i < k - 1; i++) {
cost += arr[i];
}
io.println(cost);
io.close();
}
static IO io = new IO();
static class IO {
private byte[] buf;
private InputStream in;
private PrintWriter pw;
private int total, index;
public IO() {
buf = new byte[1024];
in = System.in;
pw = new PrintWriter(System.out);
}
public int next() throws IOException {
if(total < 0)
throw new InputMismatchException();
if(index >= total) {
index = 0;
total = in.read(buf);
if(total <= 0)
return -1;
}
return buf[index++];
}
public int nextInt() throws IOException {
int n = next(), integer = 0;
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for(int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public long nextLong() throws IOException {
long integer = 0l;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public double nextDouble() throws IOException {
double doub = 0;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n) && n != '.') {
if(n >= '0' && n <= '9') {
doub *= 10;
doub += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
if(n == '.') {
n = next();
double temp = 1;
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
temp /= 10;
doub += (n - '0') * temp;
n = next();
}
else
throw new InputMismatchException();
}
}
return doub * neg;
}
public String nextString() throws IOException {
StringBuilder sb = new StringBuilder();
int n = next();
while(isWhiteSpace(n))
n = next();
while(!isWhiteSpace(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
public String nextLine() throws IOException {
int n = next();
while(isWhiteSpace(n))
n = next();
StringBuilder sb = new StringBuilder();
while(!isEndOfLine(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
private boolean isWhiteSpace(int n) {
return n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1;
}
private boolean isEndOfLine(int n) {
return n == '\n' || n == '\r' || n == -1;
}
public void print(Object obj) {
pw.print(obj);
}
public void println(Object... obj) {
if(obj.length == 1)
pw.println(obj[0]);
else {
for(Object o: obj)
pw.print(o + " ");
pw.println();
}
}
public void flush() throws IOException {
pw.flush();
}
public void close() throws IOException {
pw.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code:
a=input().split()
b=input().split()
for j in [a,b]:
for i in range(0,len(j)):
j[i]=int(j[i])
n,k,j=a[0],a[1],a[2]
c_j=b[j-1]
b.sort()
if b[k-1]<=c_j:
b[k-1]=c_j
sum=0
for i in range(0,k):
sum+=b[i]
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, k, j; cin>>n>>k>>j;
vector<int> vect;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i!=j)
vect.pb(a);
else
ans += a;
}
sort(all(vect));
for(int i=0; i<k-1; i++){
ans += vect[i];
}
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 think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays of size N which have the same values but in different orders, we need to make a second array the same as a first array using a minimum number of swaps.
Note:- It is guaranteed that the elements of the array are uniqueFirst line of input contains the size of array N, second line of input contains N space separated integers depicting values of first array, third line of input contains N space separated integers depicting values of second array.
Constraints:-
1 < = N < = 10000
1 < = Arr[i] < = 1000000000Print the minimum number of swaps required to make the second array equal to first.Sample Input:-
5
1 2 3 4 5
3 1 2 5 4
Sample Output:-
3
Sample Input:-
4
3 6 4 8
4 6 8 3
Sample Output:-
2, 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 s[] = br.readLine().split(" ");
int a[] = new int[n];
int b[] = new int[n];
for(int i=0;i<n;i++)
a[i] = Integer.parseInt(s[i]);
String s1[] = br.readLine().split(" ");
for(int i=0;i<n;i++)
b[i] = Integer.parseInt(s1[i]);
int t = miniSwap(a,b,n);
System.out.print(t);
}
public static int miniSwap(int []a, int b[], int n){
int count=0;
for(int i=0;i<n;i++){
if(a[i]!=b[i]){
count++;
for(int j=i+1;j<n;j++){
if(a[i]==b[j])
swap(i,j,b);
}
}
}
return count;
}
public static void swap(int i,int j,int []b){
int temp = b[i];
b[i] = b[j];
b[j] = temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays of size N which have the same values but in different orders, we need to make a second array the same as a first array using a minimum number of swaps.
Note:- It is guaranteed that the elements of the array are uniqueFirst line of input contains the size of array N, second line of input contains N space separated integers depicting values of first array, third line of input contains N space separated integers depicting values of second array.
Constraints:-
1 < = N < = 10000
1 < = Arr[i] < = 1000000000Print the minimum number of swaps required to make the second array equal to first.Sample Input:-
5
1 2 3 4 5
3 1 2 5 4
Sample Output:-
3
Sample Input:-
4
3 6 4 8
4 6 8 3
Sample Output:-
2, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
// Function returns the minimum number of swaps
// required to sort the array
// This method is taken from below post
// https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/
int minSwapsToSort(int arr[], int n)
{
// Create an array of pairs where first
// element is array element and second element
// is position of first element
pair<int, int> arrPos[n];
for (int i = 0; i < n; i++)
{
arrPos[i].first = arr[i];
arrPos[i].second = i;
}
// Sort the array by array element values to
// get right position of every element as second
// element of pair.
sort(arrPos, arrPos + n);
// To keep track of visited elements. Initialize
// all elements as not visited or false.
vector<bool> vis(n, false);
// Initialize result
int ans = 0;
// Traverse array elements
for (int i = 0; i < n; i++)
{
// already swapped and corrected or
// already present at correct pos
if (vis[i] || arrPos[i].second == i)
continue;
// find out the number of node in
// this cycle and add in ans
int cycle_size = 0;
int j = i;
while (!vis[j])
{
vis[j] = 1;
// move to next node
j = arrPos[j].second;
cycle_size++;
}
// Update answer by adding current cycle.
ans += (cycle_size - 1);
}
// Return result
return ans;
}
// method returns minimum number of swap to make
// array B same as array A
int minSwapToMakeArraySame(int a[], int b[], int n)
{
// map to store position of elements in array B
// we basically store element to index mapping.
map<int, int> mp;
for (int i = 0; i < n; i++)
mp[b[i]] = i;
// now we're storing position of array A elements
// in array B.
for (int i = 0; i < n; i++)
b[i] = mp[a[i]];
/* We can uncomment this section to print modified
b array
for (int i = 0; i < N; i++)
cout << b[i] << " ";
cout << endl; */
// returing minimum swap for sorting in modified
// array B as final answer
return minSwapsToSort(b, n);
}
// Driver code to test above methods
signed main()
{
int n;
cin>>n;
int a[n];
int b[n];
for(int i=0;i<n;i++){
cin>>a[i];}
for(int i=0;i<n;i++){
cin>>b[i];
}
cout << minSwapToMakeArraySame(a, b, n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a Pyramid pattern of '*' of height N.
For N = 5, the following pattern is printed.
<center> * </center>
<center> *** </center>
<center> ***** </center>
<center> ******* </center>
<center>*********</center>The input contains N as an input.
</b>Constraint:</b>
1 <b>≤</b> N <b>≤</b> 50Print a Pyramid pattern of '*' of height N.Sample Input:
5
Sample Output:
<center> * </center>
<center> *** </center>
<center> ***** </center>
<center> ******* </center>
<center>*********</center>
Sample Input:
2
Sample Output:
<center> * </center>
<center> *** </center>, I have written this Solution Code: import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=1;i<=n;i++){
int cnt=2*i-1;
for(int j=0;j<n-i;j++)System.out.print(" ");
for(int j=0;j<cnt;j++)System.out.print("*");
for(int j=0;j<n-i;j++)System.out.print(" ");
System.out.println("");
}
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 Binary Tree, print Right view of it. <b>Right view</b> of a Binary Tree is set of nodes visible when tree is viewed from right side.
Right view of following tree is 1 3 7 8.
1
/ \
2 3
/ \ / \
4 5 6 7
\
8<b>Your Task:</b>
This is a function problem. You don't have to take input. Just complete the function <b>rightView()</b> that takes node as parameter and prints the right view. Also the functional part visible to you contains information about the Node class
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
1 <= node values <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>
For <b>Custom Input:</b>
First line of input should contains the number of test cases T. For each test case, there will be two lines of input.
First line contains number of nodes N. Second line will be a string representing the tree as described below:
The values in the string are in the order of level order traversal of the tree where, numbers denote node values, and a character “N” denotes NULL child.
<b>Note:</b> If a node has been declared Null using 'N', no information about its children will be given further in the array.Print the right view of the binary tree.Sample Input:
2
3
1 3 2
5
10 20 30 40 60
Sample Output:
1 2
10 30 60
Explanation:
Test case 1: Below is the given tree
1
/ \
3 2
For the above test case the right view is: 1 2
Test case 2: Below is the given tree
10
/ \
20 30
/ \
40 60
Right View is: 10 30 60., I have written this Solution Code: private static void printRightView(Node root)
{
if (root == null)
return;
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty())
{
// number of nodes at current level
int n = queue.size();
// Traverse all nodes of current level
for (int i = 1; i <= n; i++) {
Node temp = queue.poll();
// Print the right most element at
// the level
if (i == n)
System.out.print(temp.data + " ");
// Add left node to queue
if (temp.left != null)
queue.add(temp.left);
// Add right node to queue
if (temp.right != null)
queue.add(temp.right);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String StrInput[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(StrInput[0]);
int s = Integer.parseInt(StrInput[1]);
int arr[] = new int[n];
String StrInput2[] = br.readLine().trim().split(" ");
for(int i=0;i<n;i++)
{
arr[i] = Integer.parseInt(StrInput2[i]);
}
int sum = arr[0];
int startingindex = 0;
int endingindex = 1;
int j = 0;
int i;
for(i=1;i<=n;i++)
{
if(sum < s && arr[i] != 0)
{
sum += arr[i];
}
while(sum > s && startingindex < i-1)
{
sum -= arr[startingindex];
startingindex++;
}
if(sum == s)
{
endingindex = i+1;
if(arr[0] == 0)
{
System.out.print(startingindex+2 + " " + endingindex);
}
else
{
System.out.print(startingindex+1 + " "+ endingindex);
}
break;
}
if(i == n && sum < s)
{
System.out.print(-1);
break;
}
}
}
catch(Exception e)
{
System.out.print(-1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
int sum=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){cout<<1<<" "<<i+1;return 0;}
if(m.find(sum-k)!=m.end()){
cout<<m[sum-k]+2<<" "<<i+1;
return 0;
}
m[sum]=i;
}
cout<<-1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: def sumFinder(N,S,a):
currentSum = a[0]
start = 0
i = 1
while i <= N:
while currentSum > S and start < i-1:
currentSum = currentSum - a[start]
start += 1
if currentSum == S:
return (start+1,i)
if i < N:
currentSum = currentSum + a[i]
i += 1
return(-1)
N, S = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
ans = sumFinder(N,S,a)
if(ans==-1):
print(ans)
else:
print(ans[0],ans[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args)
throws IOException
{
BufferedReader bi = new BufferedReader(
new InputStreamReader(System.in));
int num[] = new int[3];
String[] strNums;
strNums = bi.readLine().split(" ");
for (int i = 0; i < strNums.length; i++) {
num[i] = Integer.parseInt(strNums[i]);
}
int base_1 = num[0];
int base_2 = num[1];
int height = num[2];
int area = ((base_1 + base_2) * height) / 2;
System.out.println(area);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: a,b,c = input("").split()
a,b,c = float(a),float(b),float(c)
area = ((a+b)/2)*c
print(int(area)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Since area of rectangle can be slightly easy, you need to calculate the area of the trapezium. You are given the lengths of the parallel sides of the trapezium and its height (distance between the parallel sides).The first and the only line of input contains three integers lengths of parallel sides followed by the height.
Constraints
1 <= side1, side2 <= 100
2 <= height <= 100
height is always divisible by 2Output a single integer, the area of the trapezium.
Sample Input
3 4 2
Sample Output
7
Explanation: The area of the given trapezium is 7., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c;
cin>>a>>b>>c;
int x=(a+b)*c;
x=x/2;
cout<<x;}, 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: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., 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));
int t = Integer.parseInt(in.readLine());
while(t-- > 0){
String str[] = in.readLine().trim().split(" ");
int k = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int z = Integer.parseInt(str[2]);
int cnt = Math.abs(x-z);
int rem = k - cnt;
if(cnt < rem){
if(cnt != 0){
System.out.println(cnt-1);
}
else{
System.out.println(0);
}
}
else if(rem < cnt){
if(rem != 0){
System.out.println(rem-1);
}
else{
System.out.println(0);
}
}
else{
System.out.println(0);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: k = int(input())
for i in range(k):
z = 0
a = input().split()
a = [int(i) for i in a]
rad = 360/a[0]
dia = abs(a[2]-a[1])
ang = rad*dia
if(ang/2 > 90):
z = a[0]- dia - 1
elif(ang/2 < 90):
z = dia - 1
else:
z = 0
print(z)
a.clear(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int t,k,r,s,a,b;
cin>>t;
while(t--)
{
cin>>k>>r>>s;
b=max(r,s);
a=min(r,s);
if(k%2==0&&(b-a)==k/2)
cout<<0;
else
if((k%2!=0&&(b-a)<=k/2)||(k%2==0&&(b-a)<k/2))
cout<<b-a-1;
else if((b-a)>k/2)
cout<<k-(b-a)-1;
cout<<endl;
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 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: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String exp = br.readLine();
if(exp.charAt(0)=='+'|| exp.charAt(0)=='-'|| exp.charAt(0)=='*' || exp.charAt(0)=='/'||exp.charAt(exp.length()-1)=='+'|| exp.charAt(exp.length()-1)=='-'|| exp.charAt(exp.length()-1)=='*' || exp.charAt(exp.length()-1)=='/'){
System.out.println("String Invalid");
return;
}
Stack<Integer> operands = new Stack<>();
Stack<Character> operators = new Stack<>();
for (int i = 0; i < exp.length(); i++) {
char ch = exp.charAt(i);
if (Character.isDigit(ch)) {
operands.push(ch - '0');
} else if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
while (operators.size() > 0 && precedence(ch) <= precedence(operators.peek())) {
int val2 = operands.pop();
int val1 = operands.pop();
char op = operators.pop();
int opval = operation(val1, val2, op);
operands.push(opval);
}
operators.push(ch);
}
}
while (!operators.isEmpty()) {
int val2 = operands.pop();
int val1 = operands.pop();
char op = operators.pop();
int opval = operation(val1, val2, op);
operands.push(opval);
}
int val = operands.pop();
System.out.println(val);
}
public static int precedence(char op) {
if (op == '+') {
return 1;
} else if (op == '-') {
return 1;
} else if (op == '*') {
return 2;
} else {
return 2;
}
}
public static int operation(int val1, int val2, char op) {
if (op == '+') {
return val1 + val2;
} else if (op == '-') {
return val1 - val2;
} else if (op == '*') {
return val1 * val2;
} else {
return val1 / val2;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: string=input()
try:
if string[0] in ["+","-","/","*"]:
print("String Invalid")
else:
print(int(eval(string)))
except:
print("String Invalid"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: Array.prototype.peek = function () {
return this[this.length - 1];
};
function Main(input)
{
console.log(infix2postfix(input))
}
/**
*This function accepts a string input for calculation, turns it into
*an array of token in infix notation, futher into postfix notation and
*returns an answer to the inputStr passed in or false if an error was
*found in the inputStr
**/
function infix2postfix(inputStr) {
//check if input length is empty
if((inputStr.length === 0) || (inputStr === null) || inputStr.charAt(0) === '+' || inputStr.charAt(0) === '-' || inputStr.charAt(0) === '*'
|| inputStr.charAt(0) === '/'||
inputStr.charAt(inputStr.length-1) === '+' || inputStr.charAt(inputStr.length-1) === '-' || inputStr.charAt(inputStr.length-1) === '*' ||
inputStr.charAt(inputStr.length-1) === '/' ){
return "String Invalid";
}
else
{var infix = new infixClass(inputStr.replace(/ /g, ''));
infix.convert2Tokens();
infix.toPosfix();
return infix.getAns();}
}
/**
Class that performs series of operations required for perfoming infix to postfix
**/
var infixClass = function(inputStr){
var output = [];//array that holds the output in postfix notation
var stack = [];//stack for holding the operators during conversion
var infixAsToken = [];//array of inputStr as tokens
var operators = ["^", "*", "/", "+", "-"];//holds array of operators
var braces = ["(", ")"];//host array of braces
var temp = "";
var error = false;//if there is an error this variable is set to true
var precedence = {"^":3, "*":2, "/":2, "+":1, "-":1, "(":0, ")":0};//holds an array of operators and their precedence value
/**
converts the inputString to postfix
**/
this.convert2Tokens = function(){
//check if the first character is + or -
if((inputStr.charAt(0) == "+") || (inputStr.charAt(0) == "-")){
inputStr = "0" + inputStr;
}
//loop through all the values
for(var i=0; i<inputStr.length; i++){
//check if its an operator or a braces
//Also checks if the temp buffer is empty. If its not, it pushes it unto the stack and emptys the temp
if (this.isAnOperator(inputStr.charAt(i)) || this.isBraces(inputStr.charAt(i))){
if(temp.length !== 0){
infixAsToken.push(temp);
temp = "";
}
infixAsToken.push(inputStr.charAt(i));
continue;
}
//if the character is a decimal no add it to the temp
if(inputStr.charAt(i) == "."){
temp = temp + ".";
continue;
}
//if the character is a no, then add it to temp
if(!isNaN(inputStr.charAt(i))){
temp = temp + inputStr.charAt(i);
continue;
}
}//loop ends
//check if there are still values in the temp after running through the array
if(temp.length !== 0){
infixAsToken.push(temp);
temp = "";
}
};
/**
Checks if there are any error in the input
**/
// this.check4Error = function(){
// if(infixAsToken[0] === '+' || infixAsToken[0] === '-' || infixAsToken[0] === '*'|| infixAsToken[0] === '/')
// return true;
// return false;//no error found
// };
/**
performs infix2postfix conversion
**/
this.toPosfix = function(){
//loop through all the infixtokens
for(var i = 0; i<infixAsToken.length; i++){
if(this.isAnumber(infixAsToken[i])){//if its a no, push to the output stack
output.push(infixAsToken[i]);
}else if(this.isAnOperator(infixAsToken[i])){//if its an operator
if(stack.length === 0){//if the stack is empty, push the new operator to the stack
stack.push(infixAsToken[i]);//push to stack
}else{
//while whats at the top of the stack has greater precedence, pop it off to the output
while(precedence[stack.peek()] >= precedence[infixAsToken[i]]){
//if both the new entry and previous entry top of the stack are ^. Just push to output
if((infixAsToken[i] === "^") && (precedence[stack.peek()] === precedence[infixAsToken[i]])){
break;
}
output.push(stack.pop());
if(stack.length === 0){//after poping of, if the stack is now empty exit loop
break;
}
}
stack.push(infixAsToken[i]);
}//console.log(stack + output);
}else if(this.isBraces(infixAsToken[i])){//if its a braces
if(infixAsToken[i] === "("){//if its a left brace, push into the stack
stack.push("(");
}else{
while(stack.peek() != "("){//if its a right, pop off
output.push(stack.pop());
}
stack.pop();//get the ( out of the stack
}
}
}
if(stack.length !== 0){
while(stack.length !== 0){
output.push(stack.pop());
}
}
return output;//return the answer in postfix
};
/**
This method is for performing the final calculation on the postfix
and it returns the answer as a float
**/
this.getAns = function(){
var no = 0;
var tempAns = 0;
//count the no of operators and store it
for(var i = 0; i<output.length; i++){
if(this.isAnOperator(output[i])){
++no;
}
}
//first loop for the no of operators found
for(i = 0; i<no; i++){
//run through the no of available values left
for(var j = 0; j<output.length; j++){
if(this.isAnOperator(output[j])){
tempAns = this.calculate(output[j - 2], output[j - 1], output[j]);
output[j - 2] = tempAns;
output.splice((j - 1),2);
break;
}
}
}
return output[0];
};
/**
This method do the calculation
**/
this.calculate = function(a, b, op){
a = parseInt(a);
b = parseInt(b);
var ans = 0;
switch(op){
case "^" : ans = Math.pow(a, b) ;
break;
case "*" : ans = a * b;
break;
case "/" : ans = a / b;
break;
case "+" : ans = a + b;
break;
case "-" : ans = a - b;
break;
}
return ans;
};
/**
checks if the character passed is a numer
**/
this.isAnumber= function(char){
return !isNaN(char);
};
/**
checks if the character passed is an operator
**/
this.isAnOperator = function(char){
if(operators.indexOf(char) != -1){
return true;
}
return false;
};
/**
checks if character passed is a braces
**/
this.isBraces = function(char){
if(braces.indexOf(char) != -1){
return true;
}
return false;
};
};
Main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String exp = br.readLine();
if(exp.charAt(0)=='+'|| exp.charAt(0)=='-'|| exp.charAt(0)=='*' || exp.charAt(0)=='/'||exp.charAt(exp.length()-1)=='+'|| exp.charAt(exp.length()-1)=='-'|| exp.charAt(exp.length()-1)=='*' || exp.charAt(exp.length()-1)=='/'){
System.out.println("String Invalid");
return;
}
Stack<Integer> operands = new Stack<>();
Stack<Character> operators = new Stack<>();
for (int i = 0; i < exp.length(); i++) {
char ch = exp.charAt(i);
if (Character.isDigit(ch)) {
operands.push(ch - '0');
} else if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
while (operators.size() > 0 && precedence(ch) <= precedence(operators.peek())) {
int val2 = operands.pop();
int val1 = operands.pop();
char op = operators.pop();
int opval = operation(val1, val2, op);
operands.push(opval);
}
operators.push(ch);
}
}
while (!operators.isEmpty()) {
int val2 = operands.pop();
int val1 = operands.pop();
char op = operators.pop();
int opval = operation(val1, val2, op);
operands.push(opval);
}
int val = operands.pop();
System.out.println(val);
}
public static int precedence(char op) {
if (op == '+') {
return 1;
} else if (op == '-') {
return 1;
} else if (op == '*') {
return 2;
} else {
return 2;
}
}
public static int operation(int val1, int val2, char op) {
if (op == '+') {
return val1 + val2;
} else if (op == '-') {
return val1 - val2;
} else if (op == '*') {
return val1 * val2;
} else {
return val1 / val2;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: string=input()
try:
if string[0] in ["+","-","/","*"]:
print("String Invalid")
else:
print(int(eval(string)))
except:
print("String Invalid"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String which contains Operators and Numbers.
Operators can be: ['+', '- ', '*', '/']
Numbers: Positive integers.
The task is to evaluate the given expression based on the <b>BODMAS</b> rule.
<b>Note</b>: If the String starts or ends with any of the operators then that String would be considered invalid.Input contains a single line containing the string S.
Constraints:
1 <= numbers <= 9Return the result computed if the string is valid otherwise, it should return "String Invalid"Sample Input:
2 + 5 + 5 - 6 / 3 * 4
Sample Output:
4
Sample Input:
+ 2 + 6 + 3
Sample Output:
String Invalid, I have written this Solution Code: Array.prototype.peek = function () {
return this[this.length - 1];
};
function Main(input)
{
console.log(infix2postfix(input))
}
/**
*This function accepts a string input for calculation, turns it into
*an array of token in infix notation, futher into postfix notation and
*returns an answer to the inputStr passed in or false if an error was
*found in the inputStr
**/
function infix2postfix(inputStr) {
//check if input length is empty
if((inputStr.length === 0) || (inputStr === null) || inputStr.charAt(0) === '+' || inputStr.charAt(0) === '-' || inputStr.charAt(0) === '*'
|| inputStr.charAt(0) === '/'||
inputStr.charAt(inputStr.length-1) === '+' || inputStr.charAt(inputStr.length-1) === '-' || inputStr.charAt(inputStr.length-1) === '*' ||
inputStr.charAt(inputStr.length-1) === '/' ){
return "String Invalid";
}
else
{var infix = new infixClass(inputStr.replace(/ /g, ''));
infix.convert2Tokens();
infix.toPosfix();
return infix.getAns();}
}
/**
Class that performs series of operations required for perfoming infix to postfix
**/
var infixClass = function(inputStr){
var output = [];//array that holds the output in postfix notation
var stack = [];//stack for holding the operators during conversion
var infixAsToken = [];//array of inputStr as tokens
var operators = ["^", "*", "/", "+", "-"];//holds array of operators
var braces = ["(", ")"];//host array of braces
var temp = "";
var error = false;//if there is an error this variable is set to true
var precedence = {"^":3, "*":2, "/":2, "+":1, "-":1, "(":0, ")":0};//holds an array of operators and their precedence value
/**
converts the inputString to postfix
**/
this.convert2Tokens = function(){
//check if the first character is + or -
if((inputStr.charAt(0) == "+") || (inputStr.charAt(0) == "-")){
inputStr = "0" + inputStr;
}
//loop through all the values
for(var i=0; i<inputStr.length; i++){
//check if its an operator or a braces
//Also checks if the temp buffer is empty. If its not, it pushes it unto the stack and emptys the temp
if (this.isAnOperator(inputStr.charAt(i)) || this.isBraces(inputStr.charAt(i))){
if(temp.length !== 0){
infixAsToken.push(temp);
temp = "";
}
infixAsToken.push(inputStr.charAt(i));
continue;
}
//if the character is a decimal no add it to the temp
if(inputStr.charAt(i) == "."){
temp = temp + ".";
continue;
}
//if the character is a no, then add it to temp
if(!isNaN(inputStr.charAt(i))){
temp = temp + inputStr.charAt(i);
continue;
}
}//loop ends
//check if there are still values in the temp after running through the array
if(temp.length !== 0){
infixAsToken.push(temp);
temp = "";
}
};
/**
Checks if there are any error in the input
**/
// this.check4Error = function(){
// if(infixAsToken[0] === '+' || infixAsToken[0] === '-' || infixAsToken[0] === '*'|| infixAsToken[0] === '/')
// return true;
// return false;//no error found
// };
/**
performs infix2postfix conversion
**/
this.toPosfix = function(){
//loop through all the infixtokens
for(var i = 0; i<infixAsToken.length; i++){
if(this.isAnumber(infixAsToken[i])){//if its a no, push to the output stack
output.push(infixAsToken[i]);
}else if(this.isAnOperator(infixAsToken[i])){//if its an operator
if(stack.length === 0){//if the stack is empty, push the new operator to the stack
stack.push(infixAsToken[i]);//push to stack
}else{
//while whats at the top of the stack has greater precedence, pop it off to the output
while(precedence[stack.peek()] >= precedence[infixAsToken[i]]){
//if both the new entry and previous entry top of the stack are ^. Just push to output
if((infixAsToken[i] === "^") && (precedence[stack.peek()] === precedence[infixAsToken[i]])){
break;
}
output.push(stack.pop());
if(stack.length === 0){//after poping of, if the stack is now empty exit loop
break;
}
}
stack.push(infixAsToken[i]);
}//console.log(stack + output);
}else if(this.isBraces(infixAsToken[i])){//if its a braces
if(infixAsToken[i] === "("){//if its a left brace, push into the stack
stack.push("(");
}else{
while(stack.peek() != "("){//if its a right, pop off
output.push(stack.pop());
}
stack.pop();//get the ( out of the stack
}
}
}
if(stack.length !== 0){
while(stack.length !== 0){
output.push(stack.pop());
}
}
return output;//return the answer in postfix
};
/**
This method is for performing the final calculation on the postfix
and it returns the answer as a float
**/
this.getAns = function(){
var no = 0;
var tempAns = 0;
//count the no of operators and store it
for(var i = 0; i<output.length; i++){
if(this.isAnOperator(output[i])){
++no;
}
}
//first loop for the no of operators found
for(i = 0; i<no; i++){
//run through the no of available values left
for(var j = 0; j<output.length; j++){
if(this.isAnOperator(output[j])){
tempAns = this.calculate(output[j - 2], output[j - 1], output[j]);
output[j - 2] = tempAns;
output.splice((j - 1),2);
break;
}
}
}
return output[0];
};
/**
This method do the calculation
**/
this.calculate = function(a, b, op){
a = parseInt(a);
b = parseInt(b);
var ans = 0;
switch(op){
case "^" : ans = Math.pow(a, b) ;
break;
case "*" : ans = a * b;
break;
case "/" : ans = a / b;
break;
case "+" : ans = a + b;
break;
case "-" : ans = a - b;
break;
}
return ans;
};
/**
checks if the character passed is a numer
**/
this.isAnumber= function(char){
return !isNaN(char);
};
/**
checks if the character passed is an operator
**/
this.isAnOperator = function(char){
if(operators.indexOf(char) != -1){
return true;
}
return false;
};
/**
checks if character passed is a braces
**/
this.isBraces = function(char){
if(braces.indexOf(char) != -1){
return true;
}
return false;
};
};
Main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an undirected graph, the task is to detect if there is a cycle in the undirected graph or not.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.
Print 'Yes' if there is a cycle in the graph otherwise print 'No'Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 3
0 1
1 2
2 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static ArrayList<Integer> ar[];
static boolean st[];
static boolean status=false;
public static void main (String[] args) throws Exception {
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
try{
String sa[]=rd.readLine().split(" ");
int n=Integer.parseInt(sa[0]);
int e=Integer.parseInt(sa[1]);
st=new boolean[n+1];
ar=new ArrayList[n+1];
for(int i=0;i<n;i++)
ar[i]=new ArrayList<Integer>();
for(int i=0;i<e;i++)
{
sa=rd.readLine().split(" ");
int u=Integer.parseInt(sa[0]);;
int v=Integer.parseInt(sa[1]);;
ar[u].add(v);
ar[v].add(u);
}
dfs(0);
if(status)
System.out.println("Yes");
else
System.out.println("No");
}
catch(Exception e)
{
if(status)
System.out.println("Yes");
else
System.out.println("Yes");
}
}
static int p=-1;
static void dfs(int k)
{
try{
Stack <Integer> st1=new Stack<>();
int l=0;
st1.add(k);
while(!st1.empty())
{
if(!st1.empty())
l=st1.pop();
if(st[l])
{
status=true;
return;
}
st[l]=true;
for(int i=0;i<ar[l].size();i++)
{
if(!st[ar[l].get(i)])
st1.push(ar[l].get(i));
}
}
}
catch(Exception e)
{
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an undirected graph, the task is to detect if there is a cycle in the undirected graph or not.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.
Print 'Yes' if there is a cycle in the graph otherwise print 'No'Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 3
0 1
1 2
2 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, I have written this Solution Code:
def dfs(adjList,vis,start,prev):
vis[start]=True
for i in adjList[start]:
if i == prev:
continue
if vis[i]:
return True
else:
dfs(adjList,vis,i,start)
return False
np=list(map(int,input().rstrip().split()))
n=np[0]
e=np[1]
adjList = [[] for i in range(n+1)]
for i in range(e):
edge = list(map(int,input().rstrip().split()))
adjList[edge[0]].append(edge[1])
adjList[edge[1]].append(edge[0])
vis = [False for i in range(n+1)]
flag=0
for i in range(n):
if not vis[i]:
if dfs(adjList,vis,i,0):
flag=1
break
if flag:
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 undirected graph, the task is to detect if there is a cycle in the undirected graph or not.The first line of input contains two integers N and M which denotes the no of vertices and no of edges in the graph respectively.
Next M lines contain space-separated integers u and v denoting that there is a directed edge from u to v.
Constraints:
1 <= N, M <= 1000
0 <= u, v <= N-1, u != v
There are no self loops or multiple edges.
Print 'Yes' if there is a cycle in the graph otherwise print 'No'Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
Yes
Explanation:
There is a cycle with nodes 0, 1, 2, 3
Sample Input 2:
4 3
0 1
1 2
2 3
Sample Output 2:
No
Explanation:
There is no cycle in this graph, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
vector<int> g[N];
int vis[N];
bool flag = 0;
void dfs(int u, int p){
vis[u] = 1;
for(auto i: g[u]){
if(i == p) continue;
if(vis[i] == 1)
flag = 1;
if(vis[i] == 0) dfs(i, u);
}
vis[u] = 2;
}
signed main() {
IOS;
int n, m;
cin >> n >> m;
for(int i = 1; i <= m; i++){
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
for(int i = 0; i < n; i++){
if(vis[i]) continue;
dfs(i, n);
}
if(flag)
cout << "Yes";
else
cout << "No";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: the user would input a number n, print the mean median mode respectively, for the list of natural numbers till n (included)
note - let mode remain the smallest element in case of non uniqueness.
print mean and median up to 2 decimal places, but let mode remain integer.32
2
1-, I have written this Solution Code: import statistics as st
n = int(input())
lst = [i for i in range(1,n+1)]
print(round(st.mean(lst),2))
print(round(st.median(lst),2))
print(st.mode(lst)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: the user would input a number n, print the mean median mode respectively, for the list of natural numbers till n (included)
note - let mode remain the smallest element in case of non uniqueness.
print mean and median up to 2 decimal places, but let mode remain integer.32
2
1-, I have written this Solution Code: import statistics
n = int(input())
inputs = []
for i in range(n):
inputs.append(int(i+1))
mode = statistics.mode(inputs)
median = statistics.median(inputs)
mean = statistics.mean(inputs)
print(round(mean,2))
print(round(median,2))
print(mode), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void sieve(boolean prime[], int n) {
int i,j;
for(i = 0; i <= n; i++)
prime[i] = true;
for(i = 2; i*i <= n; i++)
if(prime[i])
for(j = i*i; j<=n; j+=i)
prime[j] = false;
}
public static void main (String[] args) throws IOException {
int num = 10000005;
boolean prime[] = new boolean[num+1];
sieve(prime, num);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
while(T --> 0) {
int n = Integer.parseInt(br.readLine().trim());
int count = 0;
for(int i=(n/2)+1; i<=n; i++)
if(prime[i])
count++;
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
bool a[max1];
long b[max1];
void pre(){
b[0]=0;b[1]=0;
for(int i=0;i<max1;i++){
a[i]=false;
}
long cnt=0;
for(int i=2;i<max1;i++){
if(a[i]==false){
cnt++;
for(int j=i+i;j<=max1;j=j+i){a[j]=true;}
}
b[i]=cnt;
}
}
int main(){
pre();
int t;
cin>>t;
while(t--){
long n;
cin>>n;
cout<<(b[n]-b[(n)/2])<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., 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();
//////////////
typedef unsigned long long ull;
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(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2 = br.readLine();
countCommonCharacters(str1, str2);
}
static void countCommonCharacters(String s1, String s2) {
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for (int i = 0; i < s1.length(); i++)
arr1[s1.codePointAt(i) - 97]++;
for (int i = 0; i < s2.length(); i++)
arr2[s2.codePointAt(i) - 97]++;
int lenToCut = 0;
for (int i = 0; i < 26; i++) {
int leastOccurrence = Math.min(arr1[i], arr2[i]);
lenToCut += (2 * leastOccurrence);
}
System.out.println(findRelation((s1.length() + s2.length() - lenToCut) % 6));
}
static String findRelation(int value) {
switch (value) {
case 1:
return "Friends";
case 2:
return "Love";
case 3:
return "Affection";
case 4:
return "Marriage";
case 5:
return "Enemy";
default:
return "Siblings";
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: name1 = input().strip().lower()
name2 = input().strip().lower()
listA = [0]*26
listB = [0]*26
for i in name1:
listA[ord(i)-ord('a')] = listA[ord(i)-ord('a')] + 1
for i in name2:
listB[ord(i)-ord('a')] = listB[ord(i)-ord('a')] + 1
count = 0
for i in range(0,26):
count = count+abs(listA[i]-listB[i])
res = ["Siblings","Friends","Love","Affection", "Marriage", "Enemy"]
print(res[count%6]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
string s1,s2;
cin>>s1>>s2;
string a[6];
a[1]= "Friends";
a[2]= "Love";
a[3]="Affection";
a[4]= "Marriage";
a[5]= "Enemy";
a[0]= "Siblings";
int b[26],c[26];
for(int i=0;i<26;i++){b[i]=0;c[i]=0;}
for(int i=0;i<s1.length();i++){
b[s1[i]-'a']++;
}
for(int i=0;i<s2.length();i++){
c[s2[i]-'a']++;
}
int sum=0;
for(int i=0;i<26;i++){
sum+=abs(b[i]-c[i]);
}
cout<<a[sum%6];
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, 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));
br.readLine();
String[] line = br.readLine().split(" ");
int happyBalloons = 0;
for(int i=1;i<=line.length;++i){
int num = Integer.parseInt(line[i-1]);
if(num%2 == i%2 ){
happyBalloons++;
}
}
System.out.println(happyBalloons);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: x=int(input())
arr=input().split()
for i in range(0,x):
arr[i]=int(arr[i])
count=0
for i in range(0,x):
if(arr[i]%2==0 and (i+1)%2==0):
count+=1
elif (arr[i]%2!=0 and (i+1)%2!=0):
count+=1
print (count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, 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(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i%2 == a%2)
ans++;
}
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: Given a singly linked list and an element K, your task is to insert the element at the tail of the 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>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: a,b=[int(x) for x in input().split()]
c=[int(x) for x in input().split()]
for i in c:
print(i,end=" ")
print(b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a singly linked list and an element K, your task is to insert the element at the tail of the 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>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: public static Node addElement(Node head,int k) {
Node temp=head;
while(temp.next!=null){
temp=temp.next;}
Node x= new Node(k);
temp.next = x;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: n=1
index=0
li=[]
while n!=0:
n=int(input())
li.append(n)
index=index+1
#li = list(map(int,input().strip().split()))
for i in range(0,len(li)-1):
print(li[i],end=" ")
print(li[len(li)-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=100001;
int a;
for(int i=0;i<n;i++){
a=sc.nextInt();
System.out.print(a+" ");
if(a==0){break;}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n;
while(cin >> n){
cout << n << " ";
if(n == 0) break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., 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 s1 = br.readLine();
String s2 = br.readLine();
boolean flag = true;
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for(int i=0; i<s1.length(); i++){
arr1[s1.charAt(i)-97]++;
}
for(int i=0; i<s2.length(); i++){
arr2[s2.charAt(i)-97]++;
}
for(int i=0; i<25; i++){
if(arr1[i]!=arr2[i]){
flag = false;
break;
}
}
if(flag==true)
System.out.print("YES");
else System.out.print("NO");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: s1 = input().strip()
s2 = input().strip()
dict1 = dict()
dict2 = dict()
for i in s1:
dict1[i] = dict1.get(i, 0) + 1
for j in s2:
dict2[j] = dict2.get(j, 0) + 1
print(("NO", "YES")[dict1 == dict2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[26],B[26];
signed main()
{
string s,p;
cin>>s>>p;
for(int i=0;i<s.size();i++)
{
int y=s[i]-'a';
A[y]++;
}
for(int i=0;i<p.size();i++)
{
int y=p[i]-'a';
B[y]++;
}int ch=1;
for(int i=0;i<26;i++)
{
if(B[i]!=A[i])ch=0;
}
if(ch==1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: // str1 and str2 are the two input strings
function isAnagram(str1,str2){
// Get lengths of both strings
let n1 = str1.length;
let n2 = str2.length;
// If length of both strings is not same,
// then they cannot be anagram
if (n1 != n2)
return "NO";
str1 = str1.split('')
str2 = str2.split('')
// Sort both strings
str1.sort();
str2.sort()
// Compare sorted strings
for (let i = 0; i < n1; i++)
if (str1[i] != str2[i])
return "NO";
return "YES";
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
if(t%2==0)
System.out.println("Even");
else
System.out.println("Odd");
}
catch (Exception e){
return;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: n = int(input())
if n % 2 == 0:
print("Even")
else:
print("Odd"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch(num % 2)
{
case 0:
printf("Even");
break;
/* Else if n%2 == 1 */
case 1:
printf("Odd");
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(n*(n+1)/2);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: for t in range(int(input())):
n = int(input())
print(n*(n+1)//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, 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;
cout<<(n*(n+1))/2<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given 2 arrays A and B, each of the size N. Each element of these arrays is either a positive integer or -1. The total number of -1's that can appear over these 2 arrays are ≥1 and ≤ 2.
Now, you need to find the number of ways in which we can replace each -1 with a non- negative integer, such that the sum of both of these arrays is equal.First line: An integer N.
Second line: N space- separated integers, where the of these denotes A[i].
Third line: N space- separated integers, where the i<sup>th</sup> of these denotes B[i].
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
-1 ≤ A[i], B[i] ≤ 10<sup>9</sup>
The -1's may spread out among both arrays, and their quantity is between 1 and 2 (both inclusive).If there exists a finite number X, then print it. If the answer is not a finite integer, then print 'Infinite'.Sample Input
4
1 2 -1 4
3 3 3 1
Sample Output
1
Explanation
We can replace the only -1 by 3, so that the sum of both of these arrays become 10., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
typedef complex<double> base;
typedef long double ld;
typedef long long ll;
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
const int maxn=(int)(1e5+5);
const ll mod=(ll)(998244353);
int a[maxn],b[maxn];
int main()
{
ios_base::sync_with_stdio(0);
int n,ctr=0;cin>>n;
ll sum1=0,sum2=0;
bool a_q=false,b_q=false;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]==-1)
{
a_q=true;
ctr++;
}
else
{
sum1+=a[i];
}
}
for(int i=0;i<n;i++)
{
cin>>b[i];
if(b[i]==-1)
{
ctr++;
b_q=true;
}
else
{
sum2+=b[i];
}
}
// cout<<a_q<<" "<<b_q<<" "<<ctr<<endl;
if(ctr==1)
{
ll now=(a_q?(sum2-sum1):(sum1-sum2));
if(now>=0)
{
cout<<1<<endl;
}
else
{
cout<<0<<endl;
}
}
else
{
if(a_q && b_q)
{
cout<<"Infinite"<<endl;
}
else
{
ll now=(a_q?sum2-sum1:sum1-sum2);
if(now>=0)
{
cout<<(now+1)<<endl;
}
else
{
cout<<0<<endl;
}
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given 2 arrays A and B, each of the size N. Each element of these arrays is either a positive integer or -1. The total number of -1's that can appear over these 2 arrays are ≥1 and ≤ 2.
Now, you need to find the number of ways in which we can replace each -1 with a non- negative integer, such that the sum of both of these arrays is equal.First line: An integer N.
Second line: N space- separated integers, where the of these denotes A[i].
Third line: N space- separated integers, where the i<sup>th</sup> of these denotes B[i].
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
-1 ≤ A[i], B[i] ≤ 10<sup>9</sup>
The -1's may spread out among both arrays, and their quantity is between 1 and 2 (both inclusive).If there exists a finite number X, then print it. If the answer is not a finite integer, then print 'Infinite'.Sample Input
4
1 2 -1 4
3 3 3 1
Sample Output
1
Explanation
We can replace the only -1 by 3, so that the sum of both of these arrays become 10., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
class Main {
public static void main(String args[]) throws
Exception {
BufferedReader br = new BufferedReader(new
InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int arr1[] = new int[n];
int arr2[] = new int[n];
int sum_a = 0, sum_b = 0, a = 0, b = 0;
String str1[] = br.readLine().trim().split(" ");
String str2[] = br.readLine().trim().split(" ");
for (int i = 0; i < n; i++) {
arr1[i] = Integer.parseInt(str1[i]);
arr2[i] = Integer.parseInt(str2[i]);
if(arr1[i] >= 0) {
sum_a = sum_a + arr1[i];
}
else if (arr1[i] == - 1)
a++;
if(arr2[i] >= 0) {
sum_b = sum_b + arr2[i];
}
else if(arr2[i] == -1) {
b++;
}
}
if(a == 1 && b == 0) {
if(sum_a <= sum_b) {
System.out.println(1);
}
else {
System.out.println(0);
}
}
else if(a == 0 && b == 1) {
if(sum_a >= sum_b) {
System.out.println(1);
}
else {
System.out.println(0);
}
}
else if(a == 2 && b == 0) {
if(sum_a <= sum_b) {
System.out.println(java.lang.Math.abs(sum_a-sum_b) + 1);
} else {
System.out.println(0);
}
} else if(a == 0 && b == 2) {
if(sum_a >= sum_b) {
System.out.println(java.lang.Math.abs(sum_a-sum_b)+1);
}
else {
System.out.println(0);
}
} else {
System.out.println("Infinite");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
if(s==null){
System.exit(0);
}
StringTokenizer st = new StringTokenizer(s, " ");
int power = Integer.parseInt(st.nextToken());
int multiple = Integer.parseInt(st.nextToken());
int res = power;
for(int i = 1;i<=multiple;i++){
res = res*2;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int x, n;
cin >> x >> n;
cout << x*(1 << n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: x,n = map(int,input().split())
print(x*(2**n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int noofterm=Integer.parseInt(br.readLine());
int arr[] = new int[noofterm];
String s[] = br.readLine().split(" ");
for(int i=0; i<noofterm;i++){
arr[i]= Integer.parseInt(s[i]);
}
System.out.println(unique(arr));
}
public static int unique(int[] inputArray)
{
int result = 0;
for(int i=0;i<inputArray.length;i++)
{
result ^= inputArray[i];
}
return (result>0 ? result : -1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: n = int(input())
a = [int (x) for x in input().split()]
mapp={}
for index,val in enumerate(a):
if val in mapp:
del mapp[val]
else:
mapp[val]=1
for key, value in mapp.items():
print(key), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main()
{
int n;
cin>>n;
int p=0;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
p^=a;
}
cout<<p<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
int n = io.nextInt(), k = io.nextInt(), j = io.nextInt() - 1;
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = io.nextInt();
}
int cost = arr[j];
arr[j] = Integer.MAX_VALUE;
Arrays.sort(arr);
for(int i = 0; i < k - 1; i++) {
cost += arr[i];
}
io.println(cost);
io.close();
}
static IO io = new IO();
static class IO {
private byte[] buf;
private InputStream in;
private PrintWriter pw;
private int total, index;
public IO() {
buf = new byte[1024];
in = System.in;
pw = new PrintWriter(System.out);
}
public int next() throws IOException {
if(total < 0)
throw new InputMismatchException();
if(index >= total) {
index = 0;
total = in.read(buf);
if(total <= 0)
return -1;
}
return buf[index++];
}
public int nextInt() throws IOException {
int n = next(), integer = 0;
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for(int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public long nextLong() throws IOException {
long integer = 0l;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public double nextDouble() throws IOException {
double doub = 0;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n) && n != '.') {
if(n >= '0' && n <= '9') {
doub *= 10;
doub += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
if(n == '.') {
n = next();
double temp = 1;
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
temp /= 10;
doub += (n - '0') * temp;
n = next();
}
else
throw new InputMismatchException();
}
}
return doub * neg;
}
public String nextString() throws IOException {
StringBuilder sb = new StringBuilder();
int n = next();
while(isWhiteSpace(n))
n = next();
while(!isWhiteSpace(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
public String nextLine() throws IOException {
int n = next();
while(isWhiteSpace(n))
n = next();
StringBuilder sb = new StringBuilder();
while(!isEndOfLine(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
private boolean isWhiteSpace(int n) {
return n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1;
}
private boolean isEndOfLine(int n) {
return n == '\n' || n == '\r' || n == -1;
}
public void print(Object obj) {
pw.print(obj);
}
public void println(Object... obj) {
if(obj.length == 1)
pw.println(obj[0]);
else {
for(Object o: obj)
pw.print(o + " ");
pw.println();
}
}
public void flush() throws IOException {
pw.flush();
}
public void close() throws IOException {
pw.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code:
a=input().split()
b=input().split()
for j in [a,b]:
for i in range(0,len(j)):
j[i]=int(j[i])
n,k,j=a[0],a[1],a[2]
c_j=b[j-1]
b.sort()
if b[k-1]<=c_j:
b[k-1]=c_j
sum=0
for i in range(0,k):
sum+=b[i]
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, k, j; cin>>n>>k>>j;
vector<int> vect;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i!=j)
vect.pb(a);
else
ans += a;
}
sort(all(vect));
for(int i=0; i<k-1; i++){
ans += vect[i];
}
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: Actually, a pair of chopsticks need not have two sticks that are the same length. You can eat with a pair of sticks as long as the length difference is no greater than D. Samar is appointed as a head chef of a restaurant. He is provided with N sticks, the i<sup>th</sup>stick of which is L[i] units long. More than one set of chopsticks cannot share the same stick. Assist him in matching the sticks to create as many functional pairs of chopsticks as possible.The first line contains two space-separated integers N and D. The next N lines contain one integer each, the i<sup>th</sup>line giving the value of L[i].
<b>Constraints</b>
1 ≤ N ≤ 10,000 (10<sup>5</sup>)
0 ≤ D ≤ 1,000,000,000 (10<sup>9</sup>)
1 ≤ L[i] ≤ 1,000,000,000 (10<sup>9</sup>) for all integers i from 1 to NOutput a single line containing the maximum number of pairs of chopsticks Samar can form.Sample Input :
5 2
1
3
3
9
4
Sample Output :
2
Explanation :
The 5 sticks have lengths 1, 3, 3, 9, and 4 respectively. The maximum allowed difference in the lengths of two sticks forming a pair is at most 2. It is clear that the 4th stick (length 9) cannot be used with any other stick. The remaining 4 sticks can be paired as (1st and 3rd) and (2nd and 5th) to form 2 pairs of usable chopsticks., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int d=sc.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
int count=0;
Arrays.sort(arr);
for(int i=0;i<n-1;i++){
if(arr[i+1]-arr[i]<=d){
count++;
i++;
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Actually, a pair of chopsticks need not have two sticks that are the same length. You can eat with a pair of sticks as long as the length difference is no greater than D. Samar is appointed as a head chef of a restaurant. He is provided with N sticks, the i<sup>th</sup>stick of which is L[i] units long. More than one set of chopsticks cannot share the same stick. Assist him in matching the sticks to create as many functional pairs of chopsticks as possible.The first line contains two space-separated integers N and D. The next N lines contain one integer each, the i<sup>th</sup>line giving the value of L[i].
<b>Constraints</b>
1 ≤ N ≤ 10,000 (10<sup>5</sup>)
0 ≤ D ≤ 1,000,000,000 (10<sup>9</sup>)
1 ≤ L[i] ≤ 1,000,000,000 (10<sup>9</sup>) for all integers i from 1 to NOutput a single line containing the maximum number of pairs of chopsticks Samar can form.Sample Input :
5 2
1
3
3
9
4
Sample Output :
2
Explanation :
The 5 sticks have lengths 1, 3, 3, 9, and 4 respectively. The maximum allowed difference in the lengths of two sticks forming a pair is at most 2. It is clear that the 4th stick (length 9) cannot be used with any other stick. The remaining 4 sticks can be paired as (1st and 3rd) and (2nd and 5th) to form 2 pairs of usable chopsticks., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
long int n, d;
cin >> n >> d;
long int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
int cnt = 0;
for (int i = 1; i < n; i++) {
if (arr[i] - arr[i - 1] <= d) {
cnt++;
i++;
}
}
cout << cnt << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, 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 in = br.readLine().trim();
boolean printed = false;
for(int i = 1; i <= (in.length() / 2); i++) {
ArrayList<Long> valueArray = new ArrayList<>();
boolean incremented = false;
int len = i;
for(int j = 0; j < in.length(); ) {
String temp = in.substring(j, Math.min((len+j), in.length()));
valueArray.add(Long.parseLong(temp));
j+=len;
if(Long.parseLong(temp) % 10 == 9) {
int digits = findLength(Long.parseLong(temp));
if(Long.parseLong(temp) % ((long)Math.pow(10, digits) - 1) == 0)
++len;
incremented = true;
}
}
long[] arrli = new long[valueArray.size()];
for(int k = 0; k < valueArray.size(); k++) {
arrli[k] = valueArray.get(k);
}
if(incremented) {
--len;
}
if(isBeauty(arrli)) {
System.out.println("YES " + arrli[0]);
printed = true;
break;
}
}
if(!printed)
System.out.println("NO");
}
public static int findLength(long n) {
int count = 0;
while(n > 0) {
n /= 10;
++count;
}
return count;
}
public static boolean isBeauty(long[] tempOne) {
for(int i = 1; i < tempOne.length; i++) {
if(tempOne[i-1] + 1 != tempOne[i])
return false;
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, I have written this Solution Code: def isSequence(string):
start = 0
n = len(string)
for i in range(0,n//2):
new_str = string[0:i+1]
num = int(new_str)
start = num
while len(new_str) < n:
num +=1
new_str += str(num)
if(new_str == string):
return start
return -1
string = input()
start = isSequence(string)
if start != -1:
print("YES",end=" ")
print(start)
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, I have written this Solution Code: #include <sstream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <string>
#include <cassert>
#include <ctime>
#include <map>
#include <math.h>
#include <cstdio>
#include <set>
#include <deque>
#include <memory.h>
#include <queue>
#pragma comment(linker, "/STACK:64000000")
typedef long long ll;
using namespace std;
const int MAXK = -1;
const int MAXN = -1;
const int MOD = 0; // 1000 * 1000 * 1000 + 7;
int main() {
#ifdef _MSC_VER
freopen("input.txt", "r", stdin);
#endif
int T;
T=1;
while (T--) {
string s;
cin >> s;
int n = s.length();
ll ans = -1;
for (int len = 1; len * 2 <= n; len++) {
ll x = 0;
for (int i = 0; i < len; i++) x = 10 * x + (s[i] - '0');
string t = "";
ll x0 = x;
while (t.length() < s.length()) t += to_string(x++);
if (s == t) {
ans = x0;
break;
}
}
if (ans == -1) cout << "NO" << endl;
else cout << "YES " << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Anya owns N triplets of integers (A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub>), for each i from 1 to N. She asks you to find a sequence of integers which satisfies the following conditions:
1. The sequence contains at least C<sub>i</sub> distinct integers from the closed interval [A<sub>i</sub>, B<sub>i</sub>], for each i from 1 to N.
2. Out of all sequences satisfying the first condition, choose a sequence with the minimum possible number of elements.
For simplicity, she asks you to just print the length of such a sequence.The first line of the input contains a single integer N denoting the number of triplets.
Then N lines follow, where the i<sup>th</sup> line contains three integers A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub> for each i from 1 to N.
<b> Constraints: </b>
1 ≤ N ≤ 2000
1 ≤ A<sub>i</sub> ≤ B<sub>i</sub> ≤ 2000
0 ≤ C<sub>i</sub> ≤ B<sub>i</sub> - A<sub>i</sub> + 1Print a single integer — the minimum possible sequence length.Sample Input 1:
1
1 3 3
Sample Output 1:
3
Sample Explanation 1:
Since there are only 3 elements in the closed interval [1,3], and we need to take 3 of them, clearly the smallest possible length is 3.
Sample Input 2:
2
1 3 1
3 5 1
Sample Output 2:
1
Sample Explanation 2:
We can take the sequence consisting of a single element {3}., I have written this Solution Code: l = [0]*2001
k = []
n = int(input())
for i in range(n):
a,b,c = map(int,input().split())
k.append([b,(a,c)])
k.sort()
for b,aa in k:
a = aa[0]
c = aa[1]
cnt = 0
for i in range(b,a-1,-1):
if l[i]:
cnt+=1
if cnt>=c:
continue
else:
for i in range(b,a-1,-1):
if not l[i]:
l[i]=1
cnt+=1
if cnt==c:
break
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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