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Minimum Subarray flips required to convert all elements of a Binary Array to K
• Last Updated : 07 May, 2021

Given a binary array arr[] consisting of N integers, the task is to calculate the minimum number of operations of the following type needed to convert all elements of the array equal to K:

• Select any index X from the given array.
• Flip all the elements of the subarray arr[X] … arr[N – 1], i.e., if arr[i] = 1, then set arr[i] as 0 and vice versa.

Examples:

Input: N = 8, arr[ ] = {1, 0, 1, 0, 0, 1, 1, 1}, K = 0
Output:
Explanation:
Operation 1: X = 0 (chosen index). After modifying values the updated array arr[] is [0, 1, 0, 1, 1, 0, 0, 0].
Operation 2: X = 1 (chosen index). After modifying values the updated array arr[] is [0, 0, 1, 0, 0, 1, 1, 1].
Operation 3: X = 2 (chosen index). After modifying values the updated array arr[] is [0, 0, 0, 1, 1, 0, 0, 0].
Operation 4: X = 3 (chosen index). After modifying values the updated array arr[] is [0, 0, 0, 0, 0, 1, 1, 1].
Operation 5: X = 5 (chosen index). After modifying values the updated array arr[] is [0, 0, 0, 0, 0, 0, 0, 0].
Input: N = 8, arr[ ] = {1, 0, 1, 0, 0, 1, 1, 1}, K = 1
Output:4

Approach:The following observations is to be made:

As any index X ( < N) can be chosen and each value from index X to the index N-1 can be modified, so it can be found that the approach is to count the number of changing points in the array.

Follow the steps below to solve the above problem:

1. Initialize a variable flag to inverse of K. i.e. 1 if K = 0 or vice versa, which denotes the current value.
2. Initialize a variable cnt to 0, that keeps the count of the number of changing points in the array arr[].
3. Traverse the array arr[] and for each index i apply the following steps:
• If the flag value and arr[i] value are different, go to the next iteration.
• If both the flag and arr[i] are equal, increase count and set flag to flag = (flag + 1) % 2.
4. Print the final value of count.

Below is the implementation of the above approach:

## C++14

 `// C++14 Program to implement``// the above appraoch``#include ``using` `namespace` `std;` `// Function to count the minimum``// number of subarray flips required``int` `minSteps(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `i, cnt = 0;``    ``int` `flag;``    ``if` `(k == 1)``        ``flag = 0;``    ``else``        ``flag = 1;` `    ``// Iterate the array``    ``for` `(i = 0; i < n; i++) {` `        ``// If arr[i] and flag are equal``        ``if` `(arr[i] == flag) {``            ``cnt++;``            ``flag = (flag + 1) % 2;``        ``}``    ``}` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 0, 1, 0, 0, 1, 1, 1 };``    ``int` `n = ``sizeof``(arr)``            ``/ ``sizeof``(arr);``    ``int` `k = 1;` `    ``cout << minSteps(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function to count the minimum``// number of subarray flips required``static` `int` `minSteps(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `i, cnt = ``0``;``    ``int` `flag;``    ` `    ``if` `(k == ``1``)``        ``flag = ``0``;``    ``else``        ``flag = ``1``;` `    ``// Iterate the array``    ``for``(i = ``0``; i < n; i++)``    ``{` `        ``// If arr[i] and flag are equal``        ``if` `(arr[i] == flag)``        ``{``            ``cnt++;``            ``flag = (flag + ``1``) % ``2``;``        ``}``    ``}` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = { ``1``, ``0``, ``1``, ``0``, ``0``, ``1``, ``1``, ``1` `};``    ``int` `n = arr.length;``    ``int` `k = ``1``;``    ` `    ``System.out.print(minSteps(arr, n, k));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to count the minimum``# number of subarray flips required``def` `minSteps(arr, n, k):` `    ``cnt ``=` `0``    ``if``(k ``=``=` `1``):``        ``flag ``=` `0``    ``else``:``        ``flag ``=` `1` `    ``# Iterate the array``    ``for` `i ``in` `range``(n):` `        ``# If arr[i] and flag are equal``        ``if``(arr[i] ``=``=` `flag):``            ``cnt ``+``=` `1``            ``flag ``=` `(flag ``+` `1``) ``%` `2` `    ``# Return the answer``    ``return` `cnt` `# Driver Code``arr ``=` `[ ``1``, ``0``, ``1``, ``0``, ``0``, ``1``, ``1``, ``1` `]``n ``=` `len``(arr)``k ``=` `1` `# Function call``print``(minSteps(arr, n, k))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to count the minimum``// number of subarray flips required``static` `int` `minSteps(``int``[] arr, ``int` `n, ``int` `k)``{``    ``int` `i, cnt = 0;``    ``int` `flag;``    ` `    ``if` `(k == 1)``        ``flag = 0;``    ``else``        ``flag = 1;` `    ``// Iterate the array``    ``for``(i = 0; i < n; i++)``    ``{``        ` `        ``// If arr[i] and flag are equal``        ``if` `(arr[i] == flag)``        ``{``            ``cnt++;``            ``flag = (flag + 1) % 2;``        ``}``    ``}` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main ()``{``    ``int``[] arr = { 1, 0, 1, 0, 0, 1, 1, 1 };``    ``int` `n = arr.Length;``    ``int` `k = 1;``    ` `    ``Console.Write(minSteps(arr, n, k));``}``}` `// This code is contributed by chitranayal`

## Javascript

 ``
Output:
`4`

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

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