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Sort Array By Parity II - Array - Easy - LeetCode - MiniTV

Sort Array By Parity II - Array - Easy - LeetCode - मिनी टीवी

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

 

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
 

Note:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

public class Solution {
    public int[] SortArrayByParityII(int[] A) {
        if(A.Length<=1){
            return A;
        }
        
        int evenIndexWithOddValue = -1;
        int oddIndexWithEvenValue= -1;
        
        evenIndexWithOddValue= FindNextEvenIndexWithOddValue(A,0);
        oddIndexWithEvenValue= FindNextOddIndexWithEvenValue(A,0);
        
        while(evenIndexWithOddValue != -1 && oddIndexWithEvenValue != -1){
            int temp = A[evenIndexWithOddValue];
            A[evenIndexWithOddValue] = A[oddIndexWithEvenValue];
            A[oddIndexWithEvenValue]=temp;
            
            evenIndexWithOddValue= FindNextEvenIndexWithOddValue(A,evenIndexWithOddValue+1);
            oddIndexWithEvenValue= FindNextOddIndexWithEvenValue(A,oddIndexWithEvenValue+1);
        }
        
        return A;
    }
    
    private int FindNextEvenIndexWithOddValue(int[] A, int start){
        for(int i=start;i<A.Length;i++){
            if(A[i]%2==1 && i%2==0){
                return i;
            }
        }
        return -1;
    }
    
    private int FindNextOddIndexWithEvenValue(int[] A, int start){
        for(int i=start;i<A.Length;i++){
            if(A[i]%2==0 && i%2==1){
                return i;
            }
        }
        return -1;
    }
}

 


Time Complexity: O(n^2)

Space Complexity: O(1)