Print the lexicographically smallest array by swapping elements whose sum is odd

Given an array of N integers. The task is to find the lexicographically smallest array possible by applying the given operation any number of times. The operation is to pick two elements a_i and a_j (1<=i, j<=N) such that a_i + a_j is odd, and then swap a_i and a_j. 
Examples: 
 

Input : a[] = {1, 5, 4, 3, 2} 
Output : 1 2 3 4 5 
Explanation : First swap (5, 2) and then (4, 3). This is the 
lexicographically smallest possible array that can be obtained by 
swapping array elements satisfies given condition
Input : a[] = {4, 2} 
Output : 4 2 
Explanation : Not possible to swap any elements.

Approach: Observe that swapping of 2 elements is possible if they have different parity. If all elements in the array have the same parity (odd + odd and even + even is not odd), no swaps are possible. Hence the answer will be the input array only. Otherwise, you can actually swap any pair of elements. Assume you want to swap 2 elements, a and b, and they have the same parity. There must be a third element c that has a different parity. Without loss of generality, assume the array is [a, b, c]. Let's do the following swaps:
 

• Swap a and c: [c language="," 1="a"][/c]
• Swap b and c: [b, c, a]
• Swap a and c: [b, a, c]

In other words, use c as an intermediate element to swap a and b, and it'll always return to its original position afterward. Since swapping is possible between any pair of elements, we can always sort the array, which will be the lexicographically smallest array. 
Below is the implementation of the above approach: 
 

// CPP program to find possible 
// lexicographically smaller array
// by swapping only elements whose sum is odd 
#include <bits/stdc++.h>
using namespace std;

// Function to find possible lexicographically smaller array
void lexicographically_smaller(int arr[], int n)
{
    // To store number of even and odd integers
    int odd = 0, even = 0;

    // Find number of even and odd integers
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2)
            odd++;
        else
            even++;
    }

    // If it possible to make
    // lexicographically smaller
    if (odd && even)
        sort(arr, arr + n);

    // Print the array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Driver code
int main()
{
    int arr[] = { 1, 5, 4, 3, 2 };

    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    lexicographically_smaller(arr, n);

    return 0;
}

// Java program to find possible 
// lexicographically smaller array
// by swapping only elements whose sum is odd 
import java.util.*;

class GFG 
{

// Function to find possible lexicographically smaller array
static void lexicographically_smaller(int arr[], int n)
{
    // To store number of even and odd integers
    int odd = 0, even = 0;

    // Find number of even and odd integers
    for (int i = 0; i < n; i++) 
    {
        if (arr[i] % 2 == 1)
            odd++;
        else
            even++;
    }

    // If it possible to make
    // lexicographically smaller
    if (odd > 0 && even > 0)
        Arrays.sort(arr);

    // Print the array
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 5, 4, 3, 2 };

    int n = arr.length;

    // Function call
    lexicographically_smaller(arr, n);
}
}

// This code is contributed by Rajput-Ji

# Python3 program to find possible 
# lexicographically smaller array
# by swapping only elements whose sum is odd 

# Function to find possible 
# lexicographically smaller array
def lexicographically_smaller(arr, n):
    
    # To store number of even and odd integers
    odd, even = 0, 0;

    # Find number of even and odd integers
    for i in range(n):
        if (arr[i] % 2 == 1):
            odd += 1;
        else:
            even += 1;

    # If it possible to make
    # lexicographically smaller
    if (odd > 0 and even > 0):
        arr.sort();

    # Print the array
    for i in range(n):
        print(arr[i], end = " ");

# Driver code
if __name__ == '__main__':

    arr = [ 1, 5, 4, 3, 2 ];

    n = len(arr);

    # Function call
    lexicographically_smaller(arr, n);

# This code contributed by Rajput-Ji

// C# program to find possible 
// lexicographically smaller array by
// swapping only elements whose sum is odd 
using System;
    
class GFG 
{

// Function to find possible 
// lexicographically smaller array
static void lexicographically_smaller(int []arr, int n)
{
    // To store number of even and odd integers
    int odd = 0, even = 0;

    // Find number of even and odd integers
    for (int i = 0; i < n; i++) 
    {
        if (arr[i] % 2 == 1)
            odd++;
        else
            even++;
    }

    // If it possible to make
    // lexicographically smaller
    if (odd > 0 && even > 0)
        Array.Sort(arr);

    // Print the array
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 5, 4, 3, 2 };

    int n = arr.Length;

    // Function call
    lexicographically_smaller(arr, n);
}
}

// This code is contributed by 29AjayKumar

<script>

// javascript program to find possible 
// lexicographically smaller array
// by swapping only elements whose sum is odd 

// Function to find possible lexicographically smaller array
function lexicographically_smaller(arr , n)
{
    // To store number of even and odd integers
    var odd = 0, even = 0;

    // Find number of even and odd integers
    for (var i = 0; i < n; i++) 
    {
        if (arr[i] % 2 == 1)
            odd++;
        else
            even++;
    }

    // If it possible to make
    // lexicographically smaller
    if (odd > 0 && even > 0)
        arr.sort((a,b)=>a-b);

    // Print the array
    for (i = 0; i < n; i++)
        document.write(arr[i] + " ");
}

// Driver code
var arr = [ 1, 5, 4, 3, 2 ];

var n = arr.length;

// Function call
lexicographically_smaller(arr, n);


// This code is contributed by 29AjayKumar 
</script>

1 2 3 4 5

Time Complexity: O(N log N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.