Find minimum operations needed to make an Array beautiful in C++

In this problem, we are given a binary array bin[] consisting n binary values which happen to be 0 and 1. Our task is to find minimum operations needed to make an Array beautiful.

Beautiful array is a special type of binary array which consists of a pattern of alternate 0’s and 1’s.

Problem Description − We need to find the number operations that are required to make the array beautiful. An operations consists of these steps −

Step 1 − Cut the array into two halves.

Step 2 − Reverse any one of the two halves.

Step 3 − Join then halves back.

We will be counting the number of operations that are required to make sure that the array becomes a beautiful array.

Let’s take an example to understand the problem,

Input

bin[] = {1, 0, 1, 0, 0, 1}

Output

1

Explanation

We will cut the array, make a subarray bin[4, 5], reverse it and join it back.

Solution Approach

The solution to the problem is based on finding the minimum number of switch operations which is equal to the number of consecutive zeros. The base cases are −

If the size of the array is 1, it is a beautiful array. If the size of the array is an odd number then it can never be a beautiful array.

For all even lengths, we will check the total number of consecutive zeros or ones which will be the number of operations to be performed.

Algorithm

Initialise − zeroCount , oneCount, consZero = 0

Step 1 − if ( n = 1), return 0

Step 2 − if (n%2 != 0), return -1.

Step 3 − Loop for i -> 0 to n-1.

Step 3.1 − if bin[i] == bin[i+1] == 0, consZero++.

Step 4 − if bin[n-1] == bin[0] == 0, consZero++.

Step 5 − return consZero.

Program to illustrate the working of our solution,

Example

 Live Demo

#include <iostream>
using namespace std;
int minOperations(int bin[], int n) {
   if(n == 1)
      return 0;
   if(n%2 != 0)
      return -1;
      int consZero = 0;
   for (int i = 0; i < n; ++i) {
      if (i + 1 < n) {
         if (bin[i] == 0 && bin[i + 1] == 0)
            consZero++;
      }
   }
   if (bin[0] == bin[n - 1] && bin[0] == 0)
      consZero++;
      return consZero;
}
int main() {
   int bin[] = { 1, 0, 1, 0, 0, 1};
   int n = sizeof(bin) / sizeof(bin[0]);
   cout<<"The minimum operations needed to make an Array beautiful is "<<minOperations(bin, n);
   return 0;
}

Output

The minimum operations needed to make an Array beautiful is 1

Sudhir sharma

Published on 12-Mar-2021 06:57:21

• Related Questions & Answers
• Find minimum number of merge operations to make an array palindrome in C++
• Program to find minimum operations to make array equal using Python
• Minimum operations to make XOR of array zero in C++
• Program to find minimum operations to make the array increasing using Python
• Minimum operations required to remove an array in C++
• Minimum number of operations on an array to make all elements 0 using C++.
• Minimum delete operations to make all elements of array same in C++.
• Minimum operations required to make all the array elements equal in C++
• Minimum operations to make GCD of array a multiple of k in C++
• Minimum number of Appends needed to make a string palindrome in C++
• Beautiful Array in C++
• Minimum number of letters needed to make a total of n in C++.
• Program to find minimum one bit operations to make integers zero in Python
• Minimum move to end operations to make all strings equal in C++
• Program to find number of operations needed to decrease n to 0 in C++