# Minimum flips required to generate continuous substrings of 0’s and 1’s

Given a binary string S of length N, the task is to find the minimum number of bit flips required to convert the given string such that it contains only continuous substrings of 0’s and 1’s such that the final string is in the form of 000..000, 111..111, 111…000 or 000…111.

Examples:

Input: S = 000100101, N = 9
Output: 2
Explanation:
000100101 -> 000000001

Input: S = 01100, N = 5
Output: 1
Explanation:
01100 -> 11100

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The minimum number of flips can be calculated efficiently in two linear traversals.
In the first traversal, we will calculate what can be the minimum number of flips required in the worst case, as it can be equal to the number of total 0’s initially.
In the second traversal, at every step, the total number of flip required will be the sum of total 1’s before that point and total 0’s after that point. we will take a minimum of all values calculated at every step.

Hence, to solve the problem, follow the steps below:

• Initialize variables count0 = 0, count1 = 0 and res = 0. where, count0 strores count of 0 and count1 stores count of 1 and res stores the bit flips required.
• Traverse the input string, calculate 0’s and store it in res variable.
• Traverse the input string and subtract the count of 0 if character 0 is found and store the count of character 1 in variable count1 and updates the res as min(res, count0+count1).

Below is the implementation of the above approach.

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `int` `minChanges(string str, ``int` `N) ` `{ ` `    ``int` `res; ` `    ``int` `count0 = 0, count1 = 0; ` ` `  `    ``// Traverse input string ` `    ``// and store the count of 0 ` `    ``for` `(``char` `x : str) { ` `        ``count0 += (x == ``'0'``); ` `    ``} ` `    ``res = count0; ` ` `  `    ``// Traverse the input string again ` `    ``// to find minimum number of flips ` `    ``for` `(``char` `x : str) { ` `        ``count0 -= (x == ``'0'``); ` `        ``count1 += (x == ``'1'``); ` `        ``res = min(res, count1 + count0); ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 9; ` `    ``string str = ``"000101001"``; ` ` `  `    ``cout << minChanges(str, N); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `static` `int` `minChanges(String str, ``int` `N) ` `{ ` `    ``int` `res; ` `    ``int` `count0 = ``0``, count1 = ``0``; ` ` `  `    ``// Traverse input string ` `    ``// and store the count of 0 ` `    ``for``(``char` `x : str.toCharArray())  ` `    ``{ ` `       ``if` `(x == ``'0'``) ` `           ``count0++; ` `    ``} ` `    ``res = count0; ` ` `  `    ``// Traverse the input string again ` `    ``// to find minimum number of flips ` `    ``for``(``char` `x : str.toCharArray())  ` `    ``{ ` `       ``if` `(x == ``'0'``) ` `           ``count0--; ` `       ``if` `(x == ``'1'``) ` `           ``count1++; ` `            `  `       ``res = Math.min(res, count1 + count0); ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``9``; ` `    ``String str = ``"000101001"``; ` ` `  `    ``System.out.println(minChanges(str, N)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 implementation of the above approach ` `def` `minChanges(``str``, N): ` `     `  `    ``count0 ``=` `0` `    ``count1 ``=` `0` `     `  `    ``# Traverse input string ` `    ``# and store the count of 0 ` `    ``for` `x ``in` `str``: ` `        ``count0 ``+``=` `(x ``=``=` `'0'``) ` ` `  `    ``res ``=` `count0 ` `             `  `    ``# Traverse the input string again ` `    ``# to find minimum number of flips ` `    ``for` `x ``in` `str``: ` `        ``count0 ``-``=` `(x ``=``=` `'0'``) ` `        ``count1 ``+``=` `(x ``=``=` `'1'``) ` `        ``res ``=` `min``(res, count1 ``+` `count0) ` `         `  `    ``return` `res ` ` `  ` `  `# Driver code ` `N ``=` `9` `str` `=` `"000101001"` ` `  `print``(minChanges(``str``, N)) ` ` `  `# This code is contributed by shubhamsingh10 `

Output:

```2
```

Time complexity: O(k), where, k is length of binary string.
Space complexity: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : SHUBHAMSINGH10, offbeat