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;
}