# Minimize flips required such that string does not any pair of consecutive 0s

Given a binary string S, the task is to find the minimum count of flips required to modify a string such that it does not contain any pair of consecutive 0s.

Examples:

Input: S = “10001”
Output: 1
Explanation:
Flipping S modifies S to “10101”.
Therefore, the required output is 1.

Input: S = “100001”
Output: 2
Explanation:
Flipping S modifies S to “110001”.
Flipping S modifies S to “110101”.

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

• Iterate over the characters of the string. For every ith character, check if S[i] and S[i + 1] are equal to ‘0’ or not. If found to be true, then increment count and update S[i + 1] to ‘1’.
• Finally, print the count obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find minimum flips required` `// such that a string does not contain` `// any pair of consecutive 0s` `bool` `cntMinOperation(string S, ``int` `N)` `{`   `    ``// Stores minimum count of flips` `    ``int` `cntOp = 0;`   `    ``// Iterate over the characters` `    ``// of the string` `    ``for` `(``int` `i = 0; i < N - 1; i++) {`   `        ``// If two consecutive characters` `        ``// are equal to '0'` `        ``if` `(S[i] == ``'0'` `&& S[i + 1] == ``'0'``) {`   `            ``// Update S[i + 1]` `            ``S[i + 1] = ``'1'``;`   `            ``// Update cntOp` `            ``cntOp += 1;` `        ``}` `    ``}`   `    ``return` `cntOp;` `}`   `// Driver Code` `int` `main()` `{` `    ``string S = ``"10001"``;` `    ``int` `N = S.length();` `    ``cout << cntMinOperation(S, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG` `{`   `// Function to find minimum flips required` `// such that a String does not contain` `// any pair of consecutive 0s` `static` `int` `cntMinOperation(``char` `[]S, ``int` `N)` `{`   `    ``// Stores minimum count of flips` `    ``int` `cntOp = ``0``;`   `    ``// Iterate over the characters` `    ``// of the String` `    ``for` `(``int` `i = ``0``; i < N - ``1``; i++) ` `    ``{`   `        ``// If two consecutive characters` `        ``// are equal to '0'` `        ``if` `(S[i] == ``'0'` `&& S[i + ``1``] == ``'0'``) ` `        ``{`   `            ``// Update S[i + 1]` `            ``S[i + ``1``] = ``'1'``;`   `            ``// Update cntOp` `            ``cntOp += ``1``;` `        ``}` `    ``}` `    ``return` `cntOp;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``String S = ``"10001"``;` `    ``int` `N = S.length();` `    ``System.out.print(cntMinOperation(S.toCharArray(), N));` `}` `}`   `// This code is contributed by shikhasingrajput`

## Python3

 `# Python3 program for the above approach`   `# Function to find minimum flips required` `# such that a string does not contain` `# any pair of consecutive 0s` `def` `cntMinOperation(S, N):`   `    ``# Stores minimum count of flips` `    ``cntOp ``=` `0`   `    ``# Iterate over the characters` `    ``# of the string` `    ``for` `i ``in` `range``(N ``-` `1``):`   `        ``# If two consecutive characters` `        ``# are equal to '0'` `        ``if` `(S[i] ``=``=` `'0'` `and` `S[i ``+` `1``] ``=``=` `'0'``):`   `            ``# Update S[i + 1]` `            ``S[i ``+` `1``] ``=` `'1'`   `            ``# Update cntOp` `            ``cntOp ``+``=` `1` `    ``return` `cntOp`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``S ``=` `"10001"` `    ``N ``=` `len``(S)` `    ``print``(cntMinOperation([i ``for` `i ``in` `S], N))`   `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG` `{`   `  ``// Function to find minimum flips required` `  ``// such that a String does not contain` `  ``// any pair of consecutive 0s` `  ``static` `int` `cntMinOperation(``char` `[]S, ``int` `N)` `  ``{`   `    ``// Stores minimum count of flips` `    ``int` `cntOp = 0;`   `    ``// Iterate over the characters` `    ``// of the String` `    ``for` `(``int` `i = 0; i < N - 1; i++) ` `    ``{`   `      ``// If two consecutive characters` `      ``// are equal to '0'` `      ``if` `(S[i] == ``'0'` `&& S[i + 1] == ``'0'``) ` `      ``{`   `        ``// Update S[i + 1]` `        ``S[i + 1] = ``'1'``;`   `        ``// Update cntOp` `        ``cntOp += 1;` `      ``}` `    ``}` `    ``return` `cntOp;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``string` `S = ``"10001"``;` `    ``int` `N = S.Length;` `    ``Console.WriteLine(cntMinOperation(S.ToCharArray(), N));` `  ``}` `}`   `// This code is contributed by AnkThon`

## Javascript

 ``

Output:

`1`

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

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