Reduce insertion of 0 or 1 such that no adjoining pair has identical 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 will 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 factor 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:

Strategy: The issue may be solved primarily based 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 just the variety of pairs which are already adjoining and equal, i.e, positions i (1 ≤ i <N) such that Ai = Ai + 1, which may be computed with a easy for loop.

Comply with the under steps to resolve the issue:

  • Initialize a variable depend = 0.
  • Iterate a loop for every factor in A, and examine if it is the same as the following factor.
    • 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 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 fundamental(String[] args)

    {

        int[] A = { 0, 0, 1, 0, 0 };

        int N = A.size;

  

        

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

    }

}

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

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