Make all the elements of array even with given operations

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Given an array arr[] of positive integers, find the minimum number of operations required to make all the array elements even where:

If there is an odd number, then, increment the element and the next adjacent element by 1.
Each increment costs one operation.

Note: If there is any number in arr[] which is odd after all operations, then, print -1.

Examples:

Input: arr[] = {2, 3, 4, 5, 6}
Output: 4
Explanation:
Add 1 to 3 (at 1st index) and add 1 to its adjacent element 4(2nd index).
Now the array becomes {2, 4, 5, 5, 6}.
Add 1 to 5 (at 2nd index) and add 1 to its adjacent element 5(3rd index).
Now the array becomes {2, 4, 6, 6, 6}.
The resultant array has all even numbers.
The total number of operations for 4 increments is 4.

Input: arr[] = {5, 6}
Output: -1
Explanation:
Adding 1 to 5(0th index), then we have to increment 1 to its adjacent element 6(1st index).
Now the array becomes {6, 7}.
And we have 1 odd number left after all possible increments. Therefore, we can’t make all array elements even.

Approach:
This problem can be solved using Greedy Approach. The following are the steps:

Traverse the given array arr[].
If an odd element occurs, then increment that element by 1 to make it even and the next adjacent element by 1.
Repeat the above step for all the odd elements for the given array arr[].
If all the elements in arr[] are even, then print the number of operations.
Else print -1.

Below is the implementation of the above approach:

C++

// C++ program to make all array
// element even
#include "bits/stdc++.h"
using namespace std;
 
// Function to count the total
// number of operations needed to make
// all array element even
int countOperations(int arr[], int n)
{
    int count = 0;
 
    // Traverse the given array
    for (int i = 0; i < n - 1; i++) {
 
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] & 1) {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
 
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1)
            return -1;
    }
 
    // Returns the count of operations
    return count;
}
 
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(int);
    cout << countOperations(arr, n);
    return 0;
}

Java

// Java program to make all array
// element even
class GFG
{
 
// Function to count the total
// number of operations needed to make
// all array element even
static int countOperations(int arr[], int n)
{
    int count = 0;
 
    // Traverse the given array
    for (int i = 0; i < n - 1; i++)
    {
 
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] % 2 == 1) 
        {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
 
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 1)
            return -1;
    }
 
    // Returns the count of operations
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = arr.length;
    System.out.print(countOperations(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to make all array 
# element even 
 
# Function to count the total 
# number of operations needed to make 
# all array element even 
def countOperations(arr, n) :
 
    count = 0; 
 
    # Traverse the given array 
    for i in range(n - 1) :
 
        # If an odd element occurs 
        # then increment that element 
        # and next adjacent element 
        # by 1 
        if (arr[i] & 1) :
            arr[i] += 1; 
            arr[i + 1] += 1; 
            count += 2; 
 
    # Traverse the array if any odd 
    # element occurs then return -1 
    for i in range(n) :
        if (arr[i] & 1) :
            return -1; 
 
    # Returns the count of operations 
    return count; 
 
if __name__ == "__main__" : 
 
    arr = [ 2, 3, 4, 5, 6 ]; 
    n = len(arr); 
    print(countOperations(arr, n)); 
     
    # This code is contributed by AnkitRai01

C#

// C# program to make all array
// element even
using System;
 
class GFG
{
 
// Function to count the total
// number of operations needed to make
// all array element even
static int countOperations(int []arr, int n)
{
    int count = 0;
 
    // Traverse the given array
    for (int i = 0; i < n - 1; i++)
    {
 
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] % 2 == 1) 
        {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
 
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 1)
            return -1;
    }
 
    // Returns the count of operations
    return count;
}
 
// Driver code
public static void Main()
{
    int []arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;
    Console.Write(countOperations(arr, n));
}
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript program to make all array
// element even
 
// Function to count the total
// number of operations needed to make
// all array element even
function countOperations(arr, n)
{
    let count = 0;
 
    // Traverse the given array
    for (let i = 0; i < n - 1; i++) {
 
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] & 1) {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
 
    // Traverse the array if any odd
    // element occurs then return -1
    for (let i = 0; i < n; i++) {
        if (arr[i] & 1)
            return -1;
    }
 
    // Returns the count of operations
    return count;
}
 
 
let arr = [ 2, 3, 4, 5, 6 ];
let n = arr.length;
document.write(countOperations(arr, n));
 
// This code is contributed by _saurabh_jaiswal
</script>

Output:

Time Complexity: O(N) where N is the number of elements in the array.
Auxiliary Space: O(1)

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