Push Zeros to end
Push Zeros to end
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Push Zeros to end
Change in the input array/list itself. You don't need to return or print the elements.
You need to do this in one scan of array only. Don't use extra space.
The first line contains an Integer 't' which denotes the number of test cases or queries to be run. Then the test cases follow.
First line of each test case or query contains an integer 'N' representing the size of the array/list.
Second line contains 'N' single space separated integers representing the elements in the array/list.
For each test case, print the elements of the array/list in the desired order separated by a single space.
Output for every test case will be printed in a separate line.
1 <= t <= 10^2
0 <= N <= 10^5
Time Limit: 1 sec
1
7
2 0 0 1 3 0 0
2 1 3 0 0 0 0
All the zeros have been pushed towards the end of the array/list. Another important fact is that the order of the non-zero elements have been maintained as they appear in the input array/list.
2
5
0 3 0 2 0
5
9 0 0 8 2
3 2 0 0 0
9 8 2 0 0
void pushZeroesEnd(int *input, int size){ //Write your code here
int count=0,k=0; for(int i=0;i<size;i++){ if(input[i]==0)count++; else input[k++]=input[i]; } for(int i=0;i<count;i++) input[k++]=0;}
#include <iostream>using namespace std;
void pushZeroesEnd(int *input, int size){ //Write your code here
int count=0,k=0; for(int i=0;i<size;i++){ if(input[i]==0)count++; else input[k++]=input[i]; } for(int i=0;i<count;i++) input[k++]=0;}
int main(){
int t; cin >> t; while (t--) {
int size;
cin >> size; int *input = new int[size];
for (int i = 0; i < size; i++) { cin >> input[i]; }
pushZeroesEnd(input, size);
for (int i = 0; i < size; i++) { cout << input[i] << " "; }
cout << endl; delete[] input; }
return 0;}
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