**INPUT:** 7 2 4 9 10 11 13 27

**OUTPUT:** 10 2 4 9 7 11 13 27

**ALGORITHM**

**TIME COMPLEXITY:** O(N)

**SPACE COMPLEXITY:** O(1)

- Simply initialize two variables named left and right at starting index and ending index respectively.
- Loop till left is less than right and do

- If element at index left is odd then check if element at index right is odd or even.
- If the element at index right is odd then simply decrement right variable till we get any even number and right index being greater than left index.
- Else swap the element at left and right index.
- Loop terminates when left equals right.

#include <bits/stdc++.h> using namespace std; int main() { int arr[] = {4,1,8,9,11,3,77,2}; int N = sizeof(arr)/sizeof(arr[0]); int left = 0, right =N-1; //left at start index and right at end index while(left < right) //till left index is less than right index { if(arr[left]%2) //if array at left index is odd { while((arr[right]%2 ==1) and right > left) //then loop backwards if element at right index is odd right --; swap(arr[left++],arr[right--]);//swap the even and odd elements to bring even element at front and odd at back. } else left++; } for(int i=0;i<N;i++) cout<<arr[i]<<" "; return 0; }

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