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Reduce insertion of 0 or 1 such that no adjoining pair has similar worth

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Given a binary array A[] of size N, the duty is to seek out the minimal variety of operations required such that no adjoining pair has the identical worth the place in every operation we are able to insert both 0 or 1 at any place within the array.

Examples:

Enter: A[] = {0, 0, 1, 0, 0} 
Output: 2

Rationalization:  We are able to carry out the next operations to make consecutive component completely different in an array: 
Insert 1 at index 2 in A = {0, 0, 1, 0, 0} → {0, 1, 0, 1, 0, 0}
Insert 1 at index 6 in A = {0, 1, 0, 1, 0, 0} → {0, 1, 0, 1, 0, 1, 0} all consecutive components are completely different.

Enter: A[] = {0, 1, 1}
Output: 1 

Strategy: The issue might be solved based mostly on the next remark: 

A single transfer permits us to ‘break aside’ precisely one pair of equal adjoining components of an array, by inserting both 1 between 0, 0 or 0 between 1, 1. 

So, the reply is solely the variety of pairs which can be already adjoining and equal, i.e, positions i (1 ≤ i <N) such that Ai = Ai + 1, which might be computed with a easy for loop.

Comply with the under steps to unravel the issue:

  • Initialize a variable depend = 0.
  • Iterate a loop for every component in A, and verify if it is the same as the subsequent component.
    • If sure, increment the depend by 1.
  • Print the depend which provides the minimal variety of operations required to make consecutive components completely different in an array.

Under is the implementation of the above strategy.

Java

  

import java.io.*;

import java.util.*;

  

public class GFG {

  

    

    

    

    

    public static int minOperation(int arr[], int n)

    {

        int depend = 0;

  

        for (int i = 0; i < n - 1; i++) {

            if (arr[i] == arr[i + 1]) {

                depend++;

            }

        }

        return depend;

    }

  

    

    public static void major(String[] args)

    {

        int[] A = { 0, 0, 1, 0, 0 };

        int N = A.size;

  

        

        System.out.println(minOperation(A, N));

    }

}

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

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