Minimum operations required to modify the array such that parity of adjacent elements is different

Given an array A[], the task is to find the minimum number of operations required to convert the array into B[] such that for every index in B (except the last) parity(b[i]) != parity(b[i + 1]) where parity(x) = x % 3. 

Below is the operation to be performed: 

Any element from the set {1, 2} can be added to any element of the array.

Examples:

Input: A[] = {2, 1, 3, 0} 
Output: 1 
1 can be added to 0 in a single operation and the array becomes {2, 1, 3, 1}.
Input: A[] = {2, 2, 2, 2} 
Output: 2 

Approach: An optimal approach is to update the element which is in between two other elements so that it can be updated such that its parity can be different from the element previous to it and the element next to it in a single operation (as the number of operations needs to be minimized). 
So, start traversing the array from A[1] to A[n – 2] and for every element A[i] such that parity(A[i]) == parity(A[i – 1]) or parity(A[i]) == parity(A[i + 1]), update A[i] such that its parity is different from both the numbers surrounding it and keep a track of the count of operations performed. Print the count in the end.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the parity of a number
int parity(int a)
{
    return a % 3;
}

// Function to return the minimum
// number of operations required
int solve(int array[], int size)
{
    int operations = 0;
    for (int i = 0; i < size - 1; i++) {

        // Operation needs to be performed
        if (parity(array[i]) == parity(array[i + 1])) {

            operations++;
            if (i + 2 < size) {

                // Parity of previous element
                int pari1 = parity(array[i]);

                // Parity of next element
                int pari2 = parity(array[i + 2]);

                // Update parity of current element to be other than
                // the parities of the previous and the next number
                if (pari1 == pari2) {
                    if (pari1 == 0)
                        array[i + 1] = 1;
                    else if (pari1 == 1)
                        array[i + 1] = 0;
                    else
                        array[i + 1] = 1;
                }
                else {
                    if ((pari1 == 0 && pari2 == 1)
                        || (pari1 == 1 && pari2 == 0))
                        array[i + 1] = 2;
                    if ((pari1 == 1 && pari2 == 2)
                        || (pari1 == 2 && pari2 == 1))
                        array[i + 1] = 0;
                    if ((pari1 == 2 && pari2 == 0)
                        || (pari1 == 0 && pari2 == 2))
                        array[i + 1] = 1;
                }
            }
        }
    }

    return operations;
}

// Driver Code
int main()
{
    int array[] = { 2, 1, 3, 0 };
    int size = sizeof(array) / sizeof(array[0]);
    cout << solve(array, size) << endl;

  return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function to return the parity of a number
    static int parity(int a)
    {
        return a % 3;
    }

    // Function to return the minimum
    // number of operations required
    static int solve(int []array, int size)
    {
        int operations = 0;
        for (int i = 0; i < size - 1; i++)
        {

            // Operation needs to be performed
            if (parity(array[i]) == parity(array[i + 1]))
            {

                operations++;
                if (i + 2 < size)
                {

                    // Parity of previous element
                    int pari1 = parity(array[i]);

                    // Parity of next element
                    int pari2 = parity(array[i + 2]);

                    // Update parity of current
                    // element to be other than
                    // the parities of the previous
                    // and the next number
                    if (pari1 == pari2)
                    {
                        if (pari1 == 0)
                            array[i + 1] = 1;
                        else if (pari1 == 1)
                            array[i + 1] = 0;
                        else
                            array[i + 1] = 1;
                    }
                    else
                    {
                        if ((pari1 == 0 && pari2 == 1) ||
                            (pari1 == 1 && pari2 == 0))
                            array[i + 1] = 2;
                        if ((pari1 == 1 && pari2 == 2)
                            || (pari1 == 2 && pari2 == 1))
                            array[i + 1] = 0;
                        if ((pari1 == 2 && pari2 == 0)
                            || (pari1 == 0 && pari2 == 2))
                            array[i + 1] = 1;
                    }
                }
            }
        }

        return operations;
    }

    // Driver Code
    public static void main (String arg[])
    {
        int []array = { 2, 1, 3, 0 };
        int size = array.length;
        System.out.println(solve(array, size));
    }
}

// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach

# Function to return the parity
# of a number
def parity(a):
    return a % 3

# Function to return the minimum
# number of operations required
def solve(array, size):

    operations = 0
    for i in range(0, size - 1):

        # Operation needs to be performed
        if parity(array[i]) == parity(array[i + 1]):

            operations += 1
            if i + 2 < size:

                # Parity of previous element
                pari1 = parity(array[i])

                # Parity of next element
                pari2 = parity(array[i + 2])

                # Update parity of current element to
                # be other than the parities of the
                # previous and the next number
                if pari1 == pari2:
                    if pari1 == 0:
                        array[i + 1] = 1
                    elif pari1 == 1:
                        array[i + 1] = 0
                    else:
                        array[i + 1] = 1

                else:
                    if ((pari1 == 0 and pari2 == 1) or
                        (pari1 == 1 and pari2 == 0)):
                        array[i + 1] = 2
                    if ((pari1 == 1 and pari2 == 2) or
                        (pari1 == 2 and pari2 == 1)):
                        array[i + 1] = 0
                    if ((pari1 == 2 and pari2 == 0) and
                        (pari1 == 0 and pari2 == 2)):
                        array[i + 1] = 1

    return operations

# Driver Code
if __name__ == "__main__":

    array = [2, 1, 3, 0]
    size = len(array)
    print(solve(array, size))

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the parity of a number
static int parity(int a)
{
    return a % 3;
}

// Function to return the minimum
// number of operations required
static int solve(int []array, int size)
{
    int operations = 0;
    for (int i = 0; i < size - 1; i++)
    {

        // Operation needs to be performed
        if (parity(array[i]) == parity(array[i + 1]))
        {

            operations++;
            if (i + 2 < size)
            {

                // Parity of previous element
                int pari1 = parity(array[i]);

                // Parity of next element
                int pari2 = parity(array[i + 2]);

                // Update parity of current
                // element to be other than
                // the parities of the previous
                // and the next number
                if (pari1 == pari2)
                {
                    if (pari1 == 0)
                        array[i + 1] = 1;
                    else if (pari1 == 1)
                        array[i + 1] = 0;
                    else
                        array[i + 1] = 1;
                }
                else
                {
                    if ((pari1 == 0 && pari2 == 1) ||
                        (pari1 == 1 && pari2 == 0))
                        array[i + 1] = 2;
                    if ((pari1 == 1 && pari2 == 2)
                        || (pari1 == 2 && pari2 == 1))
                        array[i + 1] = 0;
                    if ((pari1 == 2 && pari2 == 0)
                        || (pari1 == 0 && pari2 == 2))
                        array[i + 1] = 1;
                }
            }
        }
    }

    return operations;
}

// Driver Code
public static void Main()
{
    int []array = { 2, 1, 3, 0 };
    int size = array.Length;
    Console.WriteLine(solve(array, size));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP implementation of the approach

// Function to return the parity of a number
function parity($a)
{
    return $a % 3;
}

// Function to return the minimum
// number of operations required
function solve($array, $size)
{
    $operations = 0;
    for ($i = 0; $i < $size - 1; $i++)
    {

        // Operation needs to be performed
        if (parity($array[$i]) == parity($array[$i + 1]))
        {

            $operations++;
            if ($i + 2 < $size)
            {

                // Parity of previous element
                $pari1 = parity($array[$i]);

                // Parity of next element
                $pari2 = parity($array[$i + 2]);

                // Update parity of current element to be
                // other than the parities of the previous
                // and the next number
                if ($pari1 == $pari2)
                {
                    if ($pari1 == 0)
                        $array[$i + 1] = 1;
                    else if ($pari1 == 1)
                        $array[$i + 1] = 0;
                    else
                        $array[$i + 1] = 1;
                }
                else
                {
                    if (($pari1 == 0 && $pari2 == 1) ||
                        ($pari1 == 1 && $pari2 == 0))
                        $array[$i + 1] = 2;
                    if (($pari1 == 1 && $pari2 == 2) ||
                        ($pari1 == 2 && $pari2 == 1))
                        $array[$i + 1] = 0;
                    if (($pari1 == 2 && $pari2 == 0) ||
                        ($pari1 == 0 && $pari2 == 2))
                        $array[$i + 1] = 1;
                }
            }
        }
    }

    return $operations;
}

// Driver Code
$array = array(2, 1, 3, 0 );
$size = count($array);
echo solve($array, $size);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the parity of a number
function parity(a)
{
    return a % 3;
}

// Function to return the minimum
// number of operations required
function solve(array, size)
{
    var operations = 0;
    for (var i = 0; i < size - 1; i++) {

        // Operation needs to be performed
        if (parity(array[i]) == parity(array[i + 1])) {

            operations++;
            if (i + 2 < size) {

                // Parity of previous element
                var pari1 = parity(array[i]);

                // Parity of next element
                var pari2 = parity(array[i + 2]);

                // Update parity of current element to be other than
                // the parities of the previous and the next number
                if (pari1 == pari2) {
                    if (pari1 == 0)
                        array[i + 1] = 1;
                    else if (pari1 == 1)
                        array[i + 1] = 0;
                    else
                        array[i + 1] = 1;
                }
                else {
                    if ((pari1 == 0 && pari2 == 1)
                        || (pari1 == 1 && pari2 == 0))
                        array[i + 1] = 2;
                    if ((pari1 == 1 && pari2 == 2)
                        || (pari1 == 2 && pari2 == 1))
                        array[i + 1] = 0;
                    if ((pari1 == 2 && pari2 == 0)
                        || (pari1 == 0 && pari2 == 2))
                        array[i + 1] = 1;
                }
            }
        }
    }

    return operations;
}

// Driver Code
var array = [2, 1, 3, 0];
var size = array.length;
document.write( solve(array, size));


</script>

Output:

1

Time Complexity: O(N), since there runs a loop from 0 to (n – 1).

Auxiliary Space: O(1), since no extra space has been taken.