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


Given a binary array A[] of size N, the duty is to search 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 will carry out the next operations to make consecutive ingredient 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 parts are completely different.

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

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

A single transfer permits us to ‘break aside’ precisely one pair of equal adjoining parts 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 might be already adjoining and equal, i.e, positions i (1 ≤ i <N) such that Ai = Ai + 1, which could be computed with a easy for loop.

Comply with the beneath steps to resolve the issue:

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

Under is the implementation of the above method.

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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