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Make all the elements of array even with given operations
  • Last Updated : 23 Mar, 2020

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

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

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. Following are the steps:

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

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

4

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

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