Make all array elements even by replacing any pair of array elements with their sum

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Given an array arr[] consisting of N positive integers, the task is to make all array elements even by replacing any pair of array elements with their sum.

Examples:

Input: arr[] = {5, 6, 3, 7, 20}
Output: 3
Explanation:
Operation 1: Replace arr[0] and arr[2] by their sum ( = 5 + 3 = 8) modifies arr[] to {8, 6, 8, 7, 20}.
Operation 2: Replace arr[2] and arr[3] by their sum ( = 7 + 8 = 15) modifies arr[] to {8, 6, 15, 15, 20}.
Operation 3: Replace arr[2] and arr[3] by their sum ( = 15 + 15 = 30) modifies arr[] to {8, 6, 30, 30, 20}.

Input: arr[] = {2, 4, 16, 8, 7, 9, 3, 1}
Output: 2

Approach: The idea is to keep replacing two odd array elements by their sum until all array elements are even. Follow the steps below to solve the problem:

Initialize a variable, say moves, to store the minimum number of replacements required.
Calculate the total number of odd elements present in the given array and store it in a variable, say cnt.
If the value of cnt is odd, then print (cnt / 2 + 2) as the result. Otherwise, print cnt / 2 as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of replacements required to make
// all array elements even
void minMoves(int arr[], int N)
{
    // Stores the count of odd elements
    int odd_element_cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Increase count of odd elements
        if (arr[i] % 2 != 0) {
            odd_element_cnt++;
        }
    }
 
    // Store number of replacements required
    int moves = (odd_element_cnt) / 2;
 
    // Two extra moves will be required
    // to make the last odd element even
    if (odd_element_cnt % 2 != 0)
        moves += 2;
 
    // Print the minimum replacements
    cout << moves;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 6, 3, 7, 20 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    minMoves(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the minimum number
// of replacements required to make
// all array elements even
static void minMoves(int arr[], int N)
{
   
    // Stores the count of odd elements
    int odd_element_cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
        // Increase count of odd elements
        if (arr[i] % 2 != 0) 
        {
            odd_element_cnt++;
        }
    }
 
    // Store number of replacements required
    int moves = (odd_element_cnt) / 2;
 
    // Two extra moves will be required
    // to make the last odd element even
    if (odd_element_cnt % 2 != 0)
        moves += 2;
 
    // Print the minimum replacements
    System.out.print(moves);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 6, 3, 7, 20 };
    int N = arr.length;
 
    // Function call
    minMoves(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;
public class GFG
{
 
  // Function to find the minimum number
  // of replacements required to make
  // all array elements even
  static void minMoves(int []arr, int N)
  {
 
    // Stores the count of odd elements
    int odd_element_cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
      // Increase count of odd elements
      if (arr[i] % 2 != 0) 
      {
        odd_element_cnt++;
      }
    }
 
    // Store number of replacements required
    int moves = (odd_element_cnt) / 2;
 
    // Two extra moves will be required
    // to make the last odd element even
    if (odd_element_cnt % 2 != 0)
      moves += 2;
 
    // Print the minimum replacements
    Console.Write(moves);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 5, 6, 3, 7, 20 };
    int N = arr.Length;
 
    // Function call
    minMoves(arr, N);
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
# Function to find the minimum number
# of replacements required to make
# all array elements even
def minMoves(arr, N):
   
    # Stores the count of odd elements
    odd_element_cnt = 0;
 
    # Traverse the array
    for i in range(N):
 
        # Increase count of odd elements
        if (arr[i] % 2 != 0):
            odd_element_cnt += 1;
 
    # Store number of replacements required
    moves = (odd_element_cnt) // 2;
 
    # Two extra moves will be required
    # to make the last odd element even
    if (odd_element_cnt % 2 != 0):
        moves += 2;
 
    # Print the minimum replacements
    print(moves);
 
# Driver Code
if __name__ == '__main__':
    arr = [5, 6, 3, 7, 20];
    N = len(arr);
 
    # Function call
    minMoves(arr, N);
 
    # This code is contributed by 29AjayKumar

Javascript

<script>
 
// javascript program for the above approach
 
// Function to find the minimum number
// of replacements required to make
// all array elements even
function minMoves(arr, N)
{
    // Stores the count of odd elements
    var odd_element_cnt = 0;
     
    var i;
    // Traverse the array
    for(i = 0; i < N; i++) {
 
        // Increase count of odd elements
        if (arr[i] % 2 != 0) {
            odd_element_cnt++;
        }
    }
 
    // Store number of replacements required
    var moves = Math.floor((odd_element_cnt)/2);
 
    // Two extra moves will be required
    // to make the last odd element even
    if (odd_element_cnt % 2 != 0)
        moves += 2;
 
    // Print the minimum replacements
    document.write(moves);
}
 
// Driver Code
    var arr = [5, 6, 3, 7, 20];
    N =  arr.length;
 
    // Function call
    minMoves(arr, N);
 
</script>

Output:

Time complexity: O(N)
Auxiliary Space: O(1)

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