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Arrange Even and Odd number such that Odd comes after Even


Reading Time - 2 mins

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)

  1. Simply initialize two variables named left and right at starting index and ending index respectively.
  2. 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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