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Minimum removals to make array sum odd

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  • Difficulty Level : Medium
  • Last Updated : 23 May, 2022

Given an array Arr[] of N integers. The task is to find the minimum number of elements needed to be removed from the array so that the sum of the remaining elements is odd. Considering there is at least one odd number.
 

Examples: 

Input:  arr[] = {1, 2, 3, 4}
Output: 1
Remove 1 to make array sum odd.

Input: arr[] =  {4, 2, 3, 4}
Output: 0
Sum is already odd.

Approach: The idea to solve this problem is to first recall the below properties of ODDs and EVENs: 

  • odd + odd = even
  • odd + even = odd
  • even + even = even
  • odd * even = even
  • even * even = even
  • odd * odd = odd

As the sum of any number of even numbers is even. So the count of even doesn’t matter. 
And the sum of the odd numbers of an odd number is always odd. So to make the array sum odd, the count of odd numbers must be odd. 
So, count the number of odd numbers and check if the count is odd then no removals required. Else the count is even then 1 odd number needs to be removed.
 

Below is the implementation of the above approach: 

C++




// C++ implementation of the above approach
#include <iostream>
using namespace std;
 
// Function to find minimum removals
int findCount(int arr[], int n)
{
    // Count odd numbers
    int countOdd = 0;
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // If the counter is odd return 0
    // otherwise return 1
    if (countOdd % 2 == 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 5, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << findCount(arr, n);
 
    return 0;
}

Java




// Java implementation of the above approach
class GfG
{
 
// Function to find minimum removals
static int findCount(int arr[], int n)
{
    // Count odd numbers
    int countOdd = 0;
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // If the counter is odd return 0
    // otherwise return 1
    if (countOdd % 2 == 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 5, 1 };
    int n = arr.length;
 
    System.out.println(findCount(arr, n));
}
}
 
// This code is contributed by
// Prerna Saini

Python3




# Python3 implementation of the
# above approach
 
# Function to find minimum removals
def findCount(arr, n) :
     
    # Count odd numbers
    countOdd = 0;
    for i in range(n) :
        if (arr[i] % 2 == 1) :
            countOdd += 1;
 
    # If the counter is odd return 0
    # otherwise return 1
    if (countOdd % 2 == 0) :
        return 1;
    else :
        return 0;
 
# Driver Code
if __name__ == "__main__" :
    arr = [ 1, 2, 3, 5, 1 ];
    n = len(arr) ;
 
    print(findCount(arr, n));
 
# This code is contributed by Ryuga

C#




// C# implementation of the approach
using System;
 
class GfG
{
 
// Function to find minimum removals
static int findCount(int []arr, int n)
{
    // Count odd numbers
    int countOdd = 0;
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // If the counter is odd return 0
    // otherwise return 1
    if (countOdd % 2 == 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 5, 1 };
    int n = arr.Length;
 
    Console.WriteLine(findCount(arr, n));
}
}
 
// This code has been contributed by 29AjayKumar

PHP




<?php
// PHP implementation of the above approach
 
// Function to find minimum removals
function findCount($arr, $n)
{
    // Count odd numbers
    $countOdd = 0;
    for ($i = 0; $i < $n; $i++)
    if ($arr[$i] % 2 == 1)
        $countOdd++;
 
    // If the counter is odd return 0
    // otherwise return 1
    if ($countOdd % 2 == 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
$arr = array(1, 2, 3, 5, 1);
$n = sizeof($arr);
 
echo(findCount($arr, $n));
 
// This code is contributed by
// Code_Mech.
?>

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to find minimum removals
function findCount(arr, n)
{
    // Count odd numbers
    var countOdd = 0;
    for (var i = 0; i < n; i++)
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // If the counter is odd return 0
    // otherwise return 1
    if (countOdd % 2 == 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
var arr = [ 1, 2, 3, 5, 1 ];
var n = arr.length;
document.write( findCount(arr, n));
 
</script>

Output: 

1

 

Time Complexity: O(n), where n is the size of the array.

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


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