Minimum Subarray reversals to sort given Binary Array

Given a binary array A[] of size N, the task is to find the minimum number of subarrays that need to be reversed to sort the binary array.

Examples:

Input: N = 4, A[]: {1, 0  , 0, 1}          
Output: 1
Explanation: Reverse the array from 0 to 2 to change the array to {0, 0, 1, 1}

Input: N = 4, A[]: {1, 0, 1  , 0}
Output: 2
Explanation: Reverse the array from 1 to 2 and then from 0 to 3

Approach: The idea to solve the problem is as follows:

To sort A[] iterate through the array and reverse every leftmost instance of a subarray of A[] with consecutive '1's and then consecutive '0's.

The count of all these subarrays can be found by counting the indices where A[i] = 1 and the next element i.e., A[i+1] = 0.

Follow the steps below to implement the idea:

• Initialize a count variable with 0.  
• Run a loop from 0 to N-2, and in each iteration do the following:
  • If the i^th element is 1 and (i+1)^th element is 0 increment count by one as there is a subarray of the type [. . .1100. . . ] that needs to be reversed to sort the array.
• Return the count variable. 

Below is the implementation of the above approach.

// C++ Code to Implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the minimum number of reversals
int minOperations(int n, int A[])
{
    // Declare variable to count the operations
    int count = 0;

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

        // Whenever there is a pattern of
        // consecutive 1's followed by 0's
        // It means you have to reverse that
        // subarray, so increase your count by 1
        if (A[i] == 1 && A[i + 1] == 0) {
            count++;
        }
    }

    // Return the count
    return count;
}

// Driver Code
int main()
{
    int A[] = { 1, 0, 1, 0 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function Call
    cout << minOperations(N, A);
    return 0;
}

// Java Code to Implement the approach
public class GFG {

  // Function to count the minimum number of reversals
  static int minOperations(int n, int A[])
  {

    // Declare variable to count the operations
    int count = 0;

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

      // Whenever there is a pattern of
      // consecutive 1's followed by 0's
      // It means you have to reverse that
      // subarray, so increase your count by 1
      if (A[i] == 1 && A[i + 1] == 0) {
        count++;
      }
    }

    // Return the count
    return count;
  }

  // Driver Code
  public static void main (String[] args)
  {
    int A[] = { 1, 0, 1, 0 };
    int N = A.length;

    // Function Call
    System.out.println(minOperations(N, A));
  }

}

// This code is contributed by AnkThon

# Python3 Code to Implement the approach

# Function to count the minimum number of reversals
def minOperations(n, A) :

    # Declare variable to count the operations
    count = 0;

    for i in range(n-1) :

        # Whenever there is a pattern of
        # consecutive 1's followed by 0's
        # It means you have to reverse that
        # subarray, so increase your count by 1
        if (A[i] == 1 and A[i + 1] == 0) :
            count += 1;

    # Return the count
    return count;

# Driver Code
if __name__ == "__main__" :

    A = [ 1, 0, 1, 0 ];
    N = len(A);

    # Function Call
    print(minOperations(N, A));
    
    # This code is contributed by AnkThon

// C# Code to Implement the approach
using System;

public class GFG {

  // Function to count the minimum number of reversals
  static int minOperations(int n, int []A)
  {

    // Declare variable to count the operations
    int count = 0;

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

      // Whenever there is a pattern of
      // consecutive 1's followed by 0's
      // It means you have to reverse that
      // subarray, so increase your count by 1
      if (A[i] == 1 && A[i + 1] == 0) {
        count++;
      }
    }

    // Return the count
    return count;
  }

  // Driver Code
  public static void Main (string[] args)
  {
    int []A = { 1, 0, 1, 0 };
    int N = A.Length;

    // Function Call
    Console.WriteLine(minOperations(N, A));
  }

}

// This code is contributed by AnkThon

    <script>
        // JavaScript Code to Implement the approach

        // Function to count the minimum number of reversals
        const minOperations = (n, A) => {
        
            // Declare variable to count the operations
            let count = 0;

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

                // Whenever there is a pattern of
                // consecutive 1's followed by 0's
                // It means you have to reverse that
                // subarray, so increase your count by 1
                if (A[i] == 1 && A[i + 1] == 0) {
                    count++;
                }
            }

            // Return the count
            return count;
        }

        // Driver Code
        let A = [1, 0, 1, 0];
        let N = A.length

        // Function Call
        document.write(minOperations(N, A));

    // This code is contributed by rakeshsahni

    </script>

2

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