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# Minimum flips required in a binary string such that all K-size substring contains 1

• Difficulty Level : Hard
• Last Updated : 26 May, 2021

Given a binary string str of size N and a positive integer K, the task is to find the minimum number of flips required to make all substring of size K contain at least one ‘1’.
Examples:

Input: str = “0001”, K = 2
Output:
Explanation:
Flipping the bit at index 1 modifies str to “0101”.
All substrings of size 2 are “01”, “10”, and “01”.
Each substring contains at least one 1.
Input: str = “101”, K = 2
Output:
Explanation:
All substrings of size 2 are “10” and “01”.
Since both of them already have at least one ‘1’, no flips required in the original string.

Approach:
Follow the steps below to solve the problem:

1. The idea is use Sliding Window Technique to check whether every substring of length K contains any ‘1’ or not.
2. Maintain a variable last_idx to store the last index of a window where the character was ‘1’. The value of this variable will be -1 if there is no ‘1’ present in the current window.
3. For any such window, we will increment the number of flips by flipping the character at last index of the current window to ‘1’ and update the index last_idx to that index.
4. Flipping the last character of the current window ensures that the following K-1 windows will have at least one ‘1’ as well. Thus, this approach minimizes the number of flips required.
5. Repeat this process for the rest of the string and print the final count of flips required.

Below is the implementation of the above approach :

## C++

 `// C++ program to find the``// minimum numbers of flips``// required in a binary string``// such that all substrings of``// size K has atleast one 1` `#include ``using` `namespace` `std;` `// Function to calculate and``// return the minimum number``// of flips to make string valid``int` `minimumMoves(string S, ``int` `K)``{``    ``int` `N = S.length();` `    ``// Stores the count``    ``// of required flips``    ``int` `ops = 0;` `    ``// Stores the last index``    ``// of '1' in the string``    ``int` `last_idx = -1;` `    ``// Check for the first``    ``// substring of length K``    ``for` `(``int` `i = 0; i < K; i++) {` `        ``// If i-th character``        ``// is '1'``        ``if` `(S[i] == ``'1'``)``            ``last_idx = i;``    ``}` `    ``// If the substring had``    ``// no '1'``    ``if` `(last_idx == -1) {` `        ``// Increase the``        ``// count of required``        ``// flips``        ``++ops;` `        ``// Flip the last``        ``// index of the``        ``// window``        ``S[K - 1] = ``'1'``;` `        ``// Update the last``        ``// index which``        ``// contains 1``        ``last_idx = K - 1;``    ``}` `    ``// Check for remaining substrings``    ``for` `(``int` `i = 1; i < N - K + 1; i++) {` `        ``// If last_idx does not``        ``// belong to current``        ``// window make it -1``        ``if` `(last_idx < i)``            ``last_idx = -1;` `        ``// If the last character of``        ``// the current substring``        ``// is '1', then update``        ``// last_idx to i+k-1;``        ``if` `(S[i + K - 1] == ``'1'``)``            ``last_idx = i + K - 1;` `        ``// If last_idx == -1, then``        ``// the current substring``        ``// has no 1``        ``if` `(last_idx == -1) {` `            ``// Increase the count``            ``// of flips``            ``++ops;` `            ``// Update the last``            ``// index of the``            ``// current window``            ``S[i + K - 1] = ``'1'``;` `            ``// Store the last``            ``// index of current``            ``// window as the``            ``// index of last '1'``            ``// in the string``            ``last_idx = i + K - 1;``        ``}``    ``}``    ``// Return the number``    ``// of operations``    ``return` `ops;``}` `// Driver Code``int` `main()``{``    ``string S = ``"001010000"``;``    ``int` `K = 3;``    ``cout << minimumMoves(S, K);``    ``return` `0;``}`

## Java

 `// Java program to find the``// minimum numbers of flips``// required in a binary string``// such that all substrings of``// size K has atleast one 1``class` `GFG{``    ` `// Function to calculate and``// return the minimum number``// of flips to make string valid``public` `static` `int` `minimumMoves(String s, ``int` `K)``{``    ``StringBuilder S = ``new` `StringBuilder(s);``    ``int` `N = S.length();` `    ``// Stores the count``    ``// of required flips``    ``int` `ops = ``0``;` `    ``// Stores the last index``    ``// of '1' in the string``    ``int` `last_idx = -``1``;` `    ``// Check for the first``    ``// substring of length K``    ``for``(``int` `i = ``0``; i < K; i++)``    ``{``        ` `    ``// If i-th character``    ``// is '1'``    ``if` `(S.charAt(i) == ``'1'``)``        ``last_idx = i;``    ``}` `    ``// If the substring had``    ``// no '1'``    ``if` `(last_idx == -``1``)``    ``{``        ` `        ``// Increase the count``        ``// of required flips``        ``++ops;` `        ``// Flip the last index``        ``// of the window``        ``S.setCharAt(K - ``1``, ``'1'``);` `        ``// Update the last index``        ``// which contains 1``        ``last_idx = K - ``1``;``    ``}` `    ``// Check for remaining substrings``    ``for``(``int` `i = ``1``; i < N - K + ``1``; i++)``    ``{``        ` `    ``// If last_idx does not``    ``// belong to current``    ``// window make it -1``    ``if` `(last_idx < i)``        ``last_idx = -``1``;``        ` `    ``// If the last character of``    ``// the current substring``    ``// is '1', then update``    ``// last_idx to i+k-1;``    ``if` `(S.charAt(i + K - ``1``) == ``'1'``)``        ``last_idx = i + K - ``1``;``        ` `    ``// If last_idx == -1, then``    ``// the current substring``    ``// has no 1``    ``if` `(last_idx == -``1``)``    ``{``            ` `        ``// Increase the count``        ``// of flips``        ``++ops;``            ` `        ``// Update the last index``        ``// of the current window``        ``S.setCharAt(i + K - ``1``, ``'1'``);``            ` `        ``// Store the last index``        ``// of current window as``        ``// the index of last '1'``        ``// in the string``        ``last_idx = i + K - ``1``;``    ``}``    ``}``    ` `    ``// Return the number``    ``// of operations``    ``return` `ops;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String S = ``"001010000"``;``    ``int` `K = ``3``;``    ` `    ``System.out.println(minimumMoves(S, K));``}``}` `// This code is contributed by jrishabh99`

## Python3

 `# Python3 program to find the minimum``# numbers of flips required in a binary``# string such that all substrings of``# size K has atleast one 1` `# Function to calculate and``# return the minimum number``# of flips to make string valid``def` `minimumMoves(S, K):``    ` `    ``N ``=` `len``(S)` `    ``# Stores the count``    ``# of required flips``    ``ops ``=` `0``    ` `    ``# Stores the last index``    ``# of '1' in the string``    ``last_idx ``=` `-``1` `    ``# Check for the first``    ``# substring of length K``    ``for` `i ``in` `range``(K):` `        ``# If i-th character``        ``# is '1'``        ``if` `(S[i] ``=``=` `'1'``):``            ``last_idx ``=` `i``    ` `    ``# If the substring had``    ``# no '1'``    ``if` `(last_idx ``=``=` `-``1``):` `        ``# Increase the count``        ``# of required flips``        ``ops ``+``=` `1` `        ``# Flip the last index``        ``# of the window``        ``S[K ``-` `1``] ``=` `'1'` `        ``# Update the last index``        ``# which contains 1``        ``last_idx ``=` `K ``-` `1` `    ``# Check for remaining substrings``    ``for` `i ``in` `range``(N ``-` `K ``+` `1``):``        ` `        ``# If last_idx does not``        ``# belong to current``        ``# window make it -1``        ``if` `(last_idx < i):``            ``last_idx ``=` `-``1` `        ``# If the last character of``        ``# the current substring``        ``# is '1', then update``        ``# last_idx to i + k-1;``        ``if` `(S[i ``+` `K ``-` `1``] ``=``=` `'1'``):``            ``last_idx ``=` `i ``+` `K ``-` `1` `        ``# If last_idx == -1, then``        ``# the current substring``        ``# has no 1``        ``if` `(last_idx ``=``=` `-``1``):` `            ``# Increase the count``            ``# of flips``            ``ops ``+``=` `1` `            ``# Update the last index``            ``# of the current window``            ``S ``=` `S[:i ``+` `K ``-` `1``] ``+` `'1'` `+` `S[i ``+` `K:]``        ` `            ``# Store the last index of``            ``# current window as the index``            ``# of last '1' in the string``            ``last_idx ``=` `i ``+` `K ``-` `1``            ` `    ``# Return the number``    ``# of operations``    ``return` `ops` `# Driver Code``S ``=` `"001010000"``K ``=` `3``;` `print``(minimumMoves(S, K))` `# This code is contributed by yatinagg`

## C#

 `// C# program to find the``// minimum numbers of flips``// required in a binary string``// such that all substrings of``// size K has atleast one 1``using` `System;``using` `System.Text;` `class` `GFG{``    ` `// Function to calculate and``// return the minimum number``// of flips to make string valid``public` `static` `int` `minimumMoves(String s, ``int` `K)``{``    ``StringBuilder S = ``new` `StringBuilder(s);``    ``int` `N = S.Length;` `    ``// Stores the count``    ``// of required flips``    ``int` `ops = 0;` `    ``// Stores the last index``    ``// of '1' in the string``    ``int` `last_idx = -1;` `    ``// Check for the first``    ``// substring of length K``    ``for``(``int` `i = 0; i < K; i++)``    ``{``        ` `        ``// If i-th character``        ``// is '1'``        ``if` `(S[i] == ``'1'``)``            ``last_idx = i;``    ``}` `    ``// If the substring had``    ``// no '1'``    ``if` `(last_idx == -1)``    ``{``        ` `        ``// Increase the count``        ``// of required flips``        ``++ops;` `        ``// Flip the last index``        ``// of the window``        ``S.Insert(K - 1, ``'1'``);` `        ``// Update the last index``        ``// which contains 1``        ``last_idx = K - 1;``    ``}` `    ``// Check for remaining substrings``    ``for``(``int` `i = 1; i < N - K + 1; i++)``    ``{``        ` `        ``// If last_idx does not``        ``// belong to current``        ``// window make it -1``        ``if` `(last_idx < i)``            ``last_idx = -1;``            ` `        ``// If the last character of``        ``// the current substring``        ``// is '1', then update``        ``// last_idx to i+k-1;``        ``if` `(S[i + K - 1] == ``'1'``)``            ``last_idx = i + K - 1;``            ` `        ``// If last_idx == -1, then``        ``// the current substring``        ``// has no 1``        ``if` `(last_idx == -1)``        ``{``                ` `            ``// Increase the count``            ``// of flips``            ``++ops;``                ` `            ``// Update the last index``            ``// of the current window``            ``S.Insert(i + K - 1, ``'1'``);``                ` `            ``// Store the last index``            ``// of current window as``            ``// the index of last '1'``            ``// in the string``            ``last_idx = i + K - 1;``        ``}``    ``}``    ` `    ``// Return the number``    ``// of operations``    ``return` `ops;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``String S = ``"001010000"``;``    ``int` `K = 3;``    ` `    ``Console.WriteLine(minimumMoves(S, K));``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output:

`1`

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

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