# Minimum removals to make array sum even in C++

C++Server Side ProgrammingProgramming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

66 Lectures 5.5 hours

## Problem statement

Given an array arr[] of N integers. We need to write a program to find minimum number of elements needed to be removed from the array, so that sum of remaining element is even.

## Example

If input array is {10, 20, 30, 5} then we need to remove one element i.e. 5 to make array sum even

## Algorithm

1. Sum of any number of even numbers is always even
2. Sum of odd numbers of odd numbers is always odd
3. Sum of odd numbers of even times is always even
4. Count the number of odd elements in the array. If the count of odd elements in the array is even, then we do not need to remove any element from the array but if the count of odd elements in the array is odd then by removing any one of the odd elements from the array

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinRemovals(int *arr, int n) {
int cnt = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] % 2 == 1) {
++cnt;
}
}
return (cnt % 2 == 0) ? 0 : 1;
}
int main() {
int arr[] = {10, 20, 30, 5};
int n = sizeof(arr) / sizeof(arr);
cout << "Minimum required removals = " << getMinRemovals(arr, n) << endl;
return 0;
}

When you compile and execute above program. It generates following output

## Output

Minimum required removals = 1