Minimum increments required to make array elements alternately even and odd

Given an array arr[] consisting of N positive integers, the task is to find the minimum number of increments required to make the array arr[] alternating sequence of even and odd numbers.

Examples:

Input: arr[] = {1, 4, 6, 8, 9, 5}
Output: 2
Explanation:
Incrementing arr[2] modifies arr[] to {1, 4, 7, 8, 9, 5}.
Incrementing arr[5] by 1 modifies arr[] to {1, 4, 7, 8, 9, 6}.

Input: arr[] = {3, 5, 7, 9, 4, 2}
Output: 3
Explanation:
Incrementing arr[0] by 1 modifies arr[] to {4, 5, 7, 9, 4, 2}.
Incrementing arr[2] by 1 modifies arr[] to {4, 5, 8, 9, 4, 2}.
Incrementing arr[5] by 1 modifies arr[] to {4, 5, 8, 9, 4, 3}.

Approach: To solve the given problem, the idea is to check the number of increments required for making the array elements odd and even alternately as well as even and odd alternately. Finally, print the minimum of the two counts of increments obtained. Follow the steps below to solve the problem:

• Initialize a variable, say cnt as 0, to store the count of increments required to convert the array into an alternating sequence of even and odd numbers or vice-versa.
• Traverse the array arr[] using the variable i and perform the following steps:
  • If i is even and arr[i] is odd, then increment cnt by 1.
  • If i is odd and arr[i] is even, then increment cnt by 1.
• After completing the above steps, the minimum of cnt and (N – cnt) is the required result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to find the minimum number of
// increments required to make the array
// even-odd alternately or vice-versa
int minIncr(int *arr,int n)
{

    // Store the minimum number of
    // increments required
    int forEven = 0;

    // Traverse the array arr[]
    for(int i = 0; i < n; i++)
    {

        // Increment forEven if even
        // element is present at an
        // odd index
        if(i % 2){
            if((arr[i] % 2) == 0)
                forEven += 1;
        }

        // Increment forEven if odd
        // element is present at an
        // even index
        else{
            if(arr[i] % 2)
                forEven += 1;
        }
    }

    // Return the minimum number of increments
    return min(forEven, n - forEven);
}

int main()
{

    // Driver Code
    int arr[] = {1, 4, 6, 8, 9, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout<<minIncr(arr,n);
    return 0;
}

// This code is contributed by rohitsingh07052.

Java

// Java program for the above approach
class GFG{

// Function to find the minimum number of
// increments required to make the array
// even-odd alternately or vice-versa
static int minIncr(int []arr, int n)
{

    // Store the minimum number of
    // increments required
    int forEven = 0;

    // Traverse the array arr[]
    for(int i = 0; i < n; i++)
    {

        // Increment forEven if even
        // element is present at an
        // odd index
        if (i % 2 == 1)
        {
            if ((arr[i] % 2) == 0)
                forEven += 1;
        }

        // Increment forEven if odd
        // element is present at an
        // even index
        else
        {
            if (arr[i] % 2 == 1)
                forEven += 1;
        }
    }

    // Return the minimum number of increments
    return Math.min(forEven, n - forEven);
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 1, 4, 6, 8, 9, 5 };
    int n = arr.length;

    System.out.println(minIncr(arr,n));
}
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to find the minimum number of
# increments required to make the array
# even-odd alternately or vice-versa
def minIncr(arr):

    # Store the minimum number of
    # increments required
    forEven = 0

    # Traverse the array arr[]
    for i in range(len(arr)):

        # Increment forEven if even
        # element is present at an
        # odd index
        if i % 2:
            if not arr[i] % 2:
                forEven += 1

        # Increment forEven if odd
        # element is present at an
        # even index
        else:
            if arr[i] % 2:
                forEven += 1

    # Return the minimum number of increments
    return min(forEven, len(arr)-forEven)


# Driver Code
arr = [1, 4, 6, 8, 9, 5]
print(minIncr(arr))

C#

// C# program for the above approach
using System;
public class GFG
{

  // Function to find the minimum number of
  // increments required to make the array
  // even-odd alternately or vice-versa
  static int minIncr(int []arr, int n)
  {

    // Store the minimum number of
    // increments required
    int forEven = 0;

    // Traverse the array []arr
    for(int i = 0; i < n; i++)
    {

      // Increment forEven if even
      // element is present at an
      // odd index
      if (i % 2 == 1)
      {
        if ((arr[i] % 2) == 0)
          forEven += 1;
      }

      // Increment forEven if odd
      // element is present at an
      // even index
      else
      {
        if (arr[i] % 2 == 1)
          forEven += 1;
      }
    }

    // Return the minimum number of increments
    return Math.Min(forEven, n - forEven);
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 1, 4, 6, 8, 9, 5 };
    int n = arr.Length;

    Console.WriteLine(minIncr(arr,n));
  }
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript program for the above approach

// Function to find the minimum number of
// increments required to make the array
// even-odd alternately or vice-versa
function minIncr(arr, n)
{

    // Store the minimum number of
    // increments required
    let forEven = 0;

    // Traverse the array arr[]
    for(let i = 0; i < n; i++)
    {

        // Increment forEven if even
        // element is present at an
        // odd index
        if(i % 2){
            if((arr[i] % 2) == 0)
                forEven += 1;
        }

        // Increment forEven if odd
        // element is present at an
        // even index
        else{
            if(arr[i] % 2)
                forEven += 1;
        }
    }

    // Return the minimum number of increments
    return Math.min(forEven, n - forEven);
}

// Driver Code
let arr = [1, 4, 6, 8, 9, 5];
let n = arr.length;
document.write(minIncr(arr,n));

</script>

Output:

2

Time Complexity: O(N)
Auxiliary Space: O(1)