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Minimize increments required to make count of even and odd array elements equal
  • Last Updated : 04 Feb, 2021

Given an array arr[] of size N, the task is to find the minimum increments by 1 required to be performed on the array elements such that the count of even and odd integers in the given array becomes equal. If it is not possible, then print “-1”.

Examples:

Input: arr[] = {1, 3, 4, 9}
Output: 1
Explanation: 
Count of even and odd integers in the array are 1 and 3 respectively.
Increment arr[3] ( = 9) by 1 to make it 10(even).
So, as the count of even and odd integer are the same after the above steps. Hence, the minimum increment operations is 1.

Input: arr[] = {2, 2, 2, 2}
Output: 2

Approach: The idea to solve the given problem is as follows:



  • If N is even, then traverse the array and keep a count of odd and even integers. The absolute difference of the count of even and odd integers divided by 2 gives the minimum increment operations required to make even and odd numbers equal.
  • If N is odd, then it is not possible to make even and odd numbers equal, hence print “-1”.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find min operations
// to make even and odd count equal
int minimumIncrement(int arr[], int N)
{
    // Odd size will never make odd
    // and even counts equal
    if (N % 2 != 0) {
        cout << "-1";
        exit(0);
    }
 
    // Stores the count of even
    // numbers in the array arr[]
    int cntEven = 0;
 
    // Stores count of odd numbers
    // in the array arr[]
    int cntOdd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is an
        // even number
        if (arr[i] % 2 == 0) {
 
            // Update cntEven
            cntEven += 1;
        }
    }
 
    // Odd numbers in arr[]
    cntOdd = N - cntEven;
 
    // Return absolute difference
    // divided by 2
    return abs(cntEven - cntOdd) / 2;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 4, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << minimumIncrement(arr, N);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG
{
     
// Function to find min operations
// to make even and odd count equal
static int minimumIncrement(int arr[], int N)
{
   
    // Odd size will never make odd
    // and even counts equal
    if (N % 2 != 0)
    {
        System.out.println( "-1");
        System.exit(0);
    }
 
    // Stores the count of even
    // numbers in the array arr[]
    int cntEven = 0;
 
    // Stores count of odd numbers
    // in the array arr[]
    int cntOdd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
 
        // If arr[i] is an
        // even number
        if (arr[i] % 2 == 0)
        {
 
            // Update cntEven
            cntEven += 1;
        }
    }
 
    // Odd numbers in arr[]
    cntOdd = N - cntEven;
 
    // Return absolute difference
    // divided by 2
    return Math.abs(cntEven - cntOdd) / 2;
}
   
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 3, 4, 9 };
    int N = arr.length;
 
    // Function call
    System.out.println(minimumIncrement(arr, N));
}
}
 
// This code is contributed by code_hunt.

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Python3

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# Python3 program for the above approach
 
# Function to find min operations
# to make even and odd count equal
def minimumIncrement(arr, N):
     
    # Odd size will never make odd
    # and even counts equal
    if (N % 2 != 0):
        print("-1")
        return
 
    # Stores the count of even
    # numbers in the array arr[]
    cntEven = 0
 
    # Stores count of odd numbers
    # in the array arr[]
    cntOdd = 0
 
    # Traverse the array arr[]
    for i in range(N):
 
        # If arr[i] is an
        # even number
        if (arr[i] % 2 == 0):
 
            # Update cntEven
            cntEven += 1
 
    # Odd numbers in arr[]
    cntOdd = N - cntEven
 
    # Return absolute difference
    # divided by 2
    return abs(cntEven - cntOdd) // 2
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 3, 4, 9]
    N = len(arr)
 
    # Function call
    print (minimumIncrement(arr, N))
 
    # Thiss code is contributed by mohit kumar 29.

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C#

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// C# program to implement
// the above approach
using System;
class GFG
{
   
// Function to find min operations
// to make even and odd count equal
static int minimumIncrement(int[] arr, int N)
{
   
    // Odd size will never make odd
    // and even counts equal
    if (N % 2 != 0)
    {
        Console.WriteLine( "-1");
        Environment.Exit(0);
    }
 
    // Stores the count of even
    // numbers in the array arr[]
    int cntEven = 0;
 
    // Stores count of odd numbers
    // in the array arr[]
    int cntOdd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
 
        // If arr[i] is an
        // even number
        if (arr[i] % 2 == 0)
        {
 
            // Update cntEven
            cntEven += 1;
        }
    }
 
    // Odd numbers in arr[]
    cntOdd = N - cntEven;
 
    // Return absolute difference
    // divided by 2
    return Math.Abs(cntEven - cntOdd) / 2;
}
 
  // Driver Code
  public static void  Main()
  {
    int[] arr = { 1, 3, 4, 9 };
    int N = arr.Length;
 
    // Function call
    Console.WriteLine(minimumIncrement(arr, N));
  }
}
 
// This code is contributed by susmitakundugoaldanga.

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Output: 

1

 

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

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