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 **0**s.

**Examples:**

Input:S = “10001”Output:1Explanation:

Flipping S[2] modifies S to “10101”.

Therefore, the required output is 1.

Input:S = “100001”Output:2Explanation:

Flipping S[1] modifies S to “110001”.

Flipping S[3] 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
**i**character, check if^{th}**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 <bits/stdc++.h>` `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

`<script>` ` ` `// JavaScript program for the above approach` ` ` ` ` `// Function to find minimum flips required` ` ` `// such that a String does not contain` ` ` `// any pair of consecutive 0s` ` ` `function` `cntMinOperation(S, N)` ` ` `{` ` ` `// Stores minimum count of flips` ` ` `let cntOp = 0;` ` ` `// Iterate over the characters` ` ` `// of the String` ` ` `for` `(let 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;` ` ` `}` ` ` ` ` `let S = ` `"10001"` `;` ` ` `let N = S.length;` ` ` `document.write(cntMinOperation(S.split(` `''` `), N));` `</script>` |

**Output:**

1

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

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