Minimum Subarray reversals to sort given Binary Array
Last Updated :
28 Jul, 2022
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 ith 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++
#include <bits/stdc++.h>
using namespace std;
int minOperations( int n, int A[])
{
int count = 0;
for ( int i = 0; i < n - 1; i++) {
if (A[i] == 1 && A[i + 1] == 0) {
count++;
}
}
return count;
}
int main()
{
int A[] = { 1, 0, 1, 0 };
int N = sizeof (A) / sizeof (A[0]);
cout << minOperations(N, A);
return 0;
}
|
Java
public class GFG {
static int minOperations( int n, int A[])
{
int count = 0 ;
for ( int i = 0 ; i < n - 1 ; i++) {
if (A[i] == 1 && A[i + 1 ] == 0 ) {
count++;
}
}
return count;
}
public static void main (String[] args)
{
int A[] = { 1 , 0 , 1 , 0 };
int N = A.length;
System.out.println(minOperations(N, A));
}
}
|
Python3
def minOperations(n, A) :
count = 0 ;
for i in range (n - 1 ) :
if (A[i] = = 1 and A[i + 1 ] = = 0 ) :
count + = 1 ;
return count;
if __name__ = = "__main__" :
A = [ 1 , 0 , 1 , 0 ];
N = len (A);
print (minOperations(N, A));
|
C#
using System;
public class GFG {
static int minOperations( int n, int []A)
{
int count = 0;
for ( int i = 0; i < n - 1; i++) {
if (A[i] == 1 && A[i + 1] == 0) {
count++;
}
}
return count;
}
public static void Main ( string [] args)
{
int []A = { 1, 0, 1, 0 };
int N = A.Length;
Console.WriteLine(minOperations(N, A));
}
}
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Javascript
<script>
const minOperations = (n, A) => {
let count = 0;
for (let i = 0; i < n - 1; i++) {
if (A[i] == 1 && A[i + 1] == 0) {
count++;
}
}
return count;
}
let A = [1, 0, 1, 0];
let N = A.length
document.write(minOperations(N, A));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
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