Skip to content
Related Articles

Related Articles

Improve Article

Minimum elements to be removed such that sum of adjacent elements is always even

  • Difficulty Level : Basic
  • Last Updated : 15 Apr, 2021

Given an array of N integers. The task is to eliminate the minimum number of elements such that in the resulting array the sum of any two adjacent values is even.

Examples: 

Input : arr[] = {1, 2, 3}
Output : 1
Remove 2 from the array.

Input : arr[] = {1, 3, 5, 4, 2}
Output : 2
Remove 4 and 2.

Approach: The sum of 2 numbers is even if either both of them is odd or both of them is even. This means for every pair of consecutive numbers that have the different parity, eliminate one of them.
So, to make the adjacent elements sum even, either all elements should be odd or even. So the following greedy algorithm works:

  • Go through all the elements in order.
  • Count the odd and even elements in the array.
  • Return the minimum count.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum number of eliminations
// such that sum of all adjacent elements is even
int min_elimination(int n, int arr[])
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2)
            countOdd++;
 
    // Return the minimum of even and
    // odd count
    return min(countOdd, n - countOdd);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 7, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << min_elimination(n, arr);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
     
// Function to find minimum number of
// eliminations such that sum of all
// adjacent elements is even
static int min_elimination(int n, int arr[])
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // Return the minimum of even
    // and odd count
    return Math.min(countOdd, n - countOdd);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 7, 9 };
    int n = arr.length;
 
    System.out.println(min_elimination(n, arr));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python 3 implementation of the
# above approach
 
# Function to find minimum number of
# eliminations such that sum of all
# adjacent elements is even
def min_elimination(n, arr):
    countOdd = 0
 
    # Stores the new value
    for i in range(n):
         
        # Count odd numbers
        if (arr[i] % 2):
            countOdd += 1
 
    # Return the minimum of even and
    # odd count
    return min(countOdd, n - countOdd)
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 7, 9]
    n = len(arr)
 
    print(min_elimination(n, arr))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function to find minimum number of
// eliminations such that sum of all
// adjacent elements is even
static int min_elimination(int n, int[] arr)
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // Return the minimum of even
    // and odd count
    return Math.Min(countOdd, n - countOdd);
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 7, 9 };
    int n = arr.Length;
 
    Console.WriteLine(min_elimination(n, arr));
}
}
 
// This code is contributed by Code_Mech

PHP




<?php
// PHP implementation of the above approach
 
// Function to find minimum number of
// eliminations such that sum of all
// adjacent elements is even
function min_elimination($n, $arr)
{
    $countOdd = 0;
 
    // Stores the new value
    for ($i = 0; $i < $n; $i++)
 
        // Count odd numbers
        if ($arr[$i] % 2 == 1)
            $countOdd++;
 
    // Return the minimum of even
    // and odd count
    return min($countOdd, $n - $countOdd);
}
 
// Driver code
$arr = array(1, 2, 3, 7, 9);
$n = sizeof($arr);
 
echo(min_elimination($n, $arr));
 
// This code is contributed by Code_Mech
?>

Javascript




<script>
 
 
// Function to find minimum number of eliminations
// such that sum of all adjacent elements is even
function min_elimination(n, arr)
{
    let countOdd = 0;
 
    // Stores the new value
    for (let i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2)
            countOdd++;
 
    // Return the minimum of even and
    // odd count
    return Math.min(countOdd, n - countOdd);
}
 
// Driver code
 
let arr= [1, 2, 3, 7, 9];
let n = arr.length;
 
document.write(min_elimination(n, arr));
 
</script>
Output: 



1

 

Time Complexity: O(N )

Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :