Minimize adding odd and subtracting even numbers to make all array elements equal to K

Namachila November 18, 2025

0 0 5 minutes read

Given an array, arr[] of size N and an integer K, the task is to find the minimum number of operations required to make all array elements equal to K by performing the following operations any number of times:

Convert arr[i] to arr[i] + X, where X is an odd number.
Convert arr[i] to arr[i] – Y, where Y is an even number.

Examples:

Input: arr[] = {8, 7, 2, 1, 3}, K = 5
Output: 8
Explanation: To make all elements of the given array equal to K(= 5), following operations are required:
arr[0] = arr[0] + X, X = 1
arr[0] = arr[0] – Y, Y = 4
arr[1] = arr[1] – Y, Y = 2
arr[2] = arr[2] + X, X = 3
arr[3] = arr[3] + X, X = 3
arr[3] = arr[3] + X, X = 1
arr[4] = arr[4] + X, X = 1
arr[4] = arr[4] + X, X = 1

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, K = 3
Output: 9

Approach: The problem can be solved using the Greedy technique. Following are the observations:

Even + Even = Even
Even + Odd = Odd
Odd + Odd = Even
Odd + Even = Odd

Follow the steps below to solve the problem:

Traverse the given array and check the following conditions.
- If K > arr[i] and (K – arr[i]) % 2 == 0 then add two odd numbers(X) into arr[i]. Therefore, total 2 operations required.
- If K > arr[i] and (K – arr[i]) % 2 != 0 then add one odd numbers(X) into arr[i]. Therefore, total 1 operations required.
- If K < arr[i] and (arr[i] – arr[i]) % 2 == 0 then subtract one even numbers(Y) into arr[i]. Therefore, total 1 operations required.
- If K < arr[i] and (K – arr[i]) % 2 != 0 then add an odd numbers(X) into arr[i] and subtract an even numbers(Y) from arr[i]. Therefore, total 2 operations required.
Finally, print the total number of operations required to make all the array elements equal to K.

Below is the implementation of the above approach

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum operations
// required to make array elements equal to K
int MinOperation(int arr[], int N, int K)
{
    // Stores minimum count of operations
    int cntOpe = 0;
 
    // Traverse the given array
    for (int i = 0; i < N; i++) {
 
        // If K is greater than arr[i]
        if (K > arr[i]) {
 
            // If (K - arr[i]) is even
            if ((K - arr[i]) % 2 == 0) {
 
                // Update cntOpe
                cntOpe += 2;
            }
            else {
 
                // Update cntOpe
                cntOpe += 1;
            }
        }
 
        // If K is less than arr[i]
        else if (K < arr[i]) {
 
            // If (arr[i] - K) is even
            if ((K - arr[i]) % 2 == 0) {
 
                // Update cntOpe
                cntOpe += 1;
            }
            else {
 
                // Update cntOpe
                cntOpe += 2;
            }
        }
    }
 
    return cntOpe;
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 7, 2, 1, 3 };
    int K = 5;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << MinOperation(arr, N, K);
 
    return 0;
}

Java

// Java program to implement 
// the above approach 
class GFG{
     
// Function to find the minimum 
// operations required to make 
// array elements equal to K 
public static int MinOperation(int arr[], 
                               int N, int K) 
{ 
  // Stores minimum count of 
  // operations 
  int cntOpe = 0; 
 
  // Traverse the given array 
  for (int i = 0; i < N; i++) 
  {
    // If K is greater than
    // arr[i] 
    if (K > arr[i]) 
    {
      // If (K - arr[i]) is even 
      if ((K - arr[i]) % 2 == 0) 
      {
        // Update cntOpe 
        cntOpe += 2; 
      } 
      else
      {
        // Update cntOpe 
        cntOpe += 1; 
      } 
    } 
 
    // If K is less than 
    // arr[i] 
    else if (K < arr[i]) 
    {
      // If (arr[i] - K) is 
      // even 
      if ((K - arr[i]) % 2 == 0) 
      {
        // Update cntOpe 
        cntOpe += 1; 
      } 
      else
      {
        // Update cntOpe 
        cntOpe += 2; 
      } 
    } 
  } 
 
  return cntOpe; 
} 
 
// Driver code
public static void main(String[] args) 
{
  int arr[] = {8, 7, 2, 1, 3}; 
  int K = 5; 
  int N = arr.length; 
  System.out.println(
  MinOperation(arr, N, K));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement
# the above approach
  
# Function to find the minimum operations
# required to make array elements equal to K
def MinOperation(arr, N, K):
     
    # Stores minimum count of operations
    cntOpe = 0
  
    # Traverse the given array
    for i in range(N):
  
        # If K is greater than arr[i]
        if (K > arr[i]):
  
            # If (K - arr[i]) is even
            if ((K - arr[i]) % 2 == 0):
  
                # Update cntOpe
                cntOpe += 2
             
            else:
  
                # Update cntOpe
                cntOpe += 1
             
        # If K is less than arr[i]
        elif (K < arr[i]):
             
            # If (arr[i] - K) is even
            if ((K - arr[i]) % 2 == 0):
  
                # Update cntOpe
                cntOpe += 1
             
            else:
  
                # Update cntOpe
                cntOpe += 2
 
    return cntOpe
 
# Driver Code
arr = [ 8, 7, 2, 1, 3 ]
K = 5
N = len(arr)
 
print(MinOperation(arr, N, K))
 
# This code is contributed by sanjoy_62

C#

// C# program to implement 
// the above approach 
using System;
 
class GFG{
     
// Function to find the minimum 
// operations required to make 
// array elements equal to K 
public static int MinOperation(int []arr, 
                               int N, int K) 
{ 
   
  // Stores minimum count of 
  // operations 
  int cntOpe = 0; 
 
  // Traverse the given array 
  for(int i = 0; i < N; i++) 
  {
     
    // If K is greater than
    // arr[i] 
    if (K > arr[i]) 
    {
       
      // If (K - arr[i]) is even 
      if ((K - arr[i]) % 2 == 0) 
      {
         
        // Update cntOpe 
        cntOpe += 2; 
      } 
      else
      {
         
        // Update cntOpe 
        cntOpe += 1; 
      } 
    } 
 
    // If K is less than 
    // arr[i] 
    else if (K < arr[i]) 
    {
       
      // If (arr[i] - K) is 
      // even 
      if ((K - arr[i]) % 2 == 0) 
      {
         
        // Update cntOpe 
        cntOpe += 1; 
      } 
      else
      {
         
        // Update cntOpe 
        cntOpe += 2; 
      } 
    } 
  } 
  return cntOpe; 
} 
 
// Driver code
public static void Main(String[] args) 
{
  int []arr = {8, 7, 2, 1, 3}; 
  int K = 5; 
  int N = arr.Length;
   
  Console.WriteLine(
  MinOperation(arr, N, K));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
// Javascript program to implement
// the above approach
 
// Function to find the minimum
// operations required to make
// array elements equal to K
function MinOperation(arr, N, K)
{
  // Stores minimum count of
  // operations
  let cntOpe = 0;
  
  // Traverse the given array
  for (let i = 0; i < N; i++)
  {
    // If K is greater than
    // arr[i]
    if (K > arr[i])
    {
      // If (K - arr[i]) is even
      if ((K - arr[i]) % 2 == 0)
      {
        // Update cntOpe
        cntOpe += 2;
      }
      else
      {
        // Update cntOpe
        cntOpe += 1;
      }
    }
  
    // If K is less than
    // arr[i]
    else if (K < arr[i])
    {
      // If (arr[i] - K) is
      // even
      if ((K - arr[i]) % 2 == 0)
      {
        // Update cntOpe
        cntOpe += 1;
      }
      else
      {
        // Update cntOpe
        cntOpe += 2;
      }
    }
  }
  
  return cntOpe;
}
 
    // Driver Code
     
    let arr = [8, 7, 2, 1, 3];
  let K = 5;
  let N = arr.length;
  document.write(
  MinOperation(arr, N, K));
      
</script>

Output:

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

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