# Minimum steps to convert one binary string to other only using negation

Given two binary strings A and B, the task is to convert A to B by choosing any sub-string of A and negating it (replace each 0 with 1 and each 1 with 0). Print the minimum number of operations required.

Examples:

Input: A = “101010”, B = “110011”
Output: 2
Choose the sub-string of length 2 from index 1 to 2 and negate it then A becomes “110010” and then take the last character and negate it.
The final string becomes “110011”

Input: A = “1010101”, B = “0011100”
Output: 3

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

Approach: Create a blank array and mark the indices which need to be negated. Then the answer will be number of blocks of consecutive 1’s in the array as a single block can be negated in a single operation.
For example, A = “101010” and B = “110011”
The newly created array will be {0, 1, 1, 0, 0, 1} so the answer will be 2,
A after first operation will be “110010”
After second operation “110011”

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum steps ` `// to convert string a to string b ` `void` `convert(``int` `n, string a, string b) ` `{ ` `    ``// array to mark the positions ` `    ``// needed to be negated ` `    ``int` `l[n]; ` `    ``int` `i; ` ` `  `    ``for` `(i = 0; i < n; i++) ` `        ``l[i] = 0; ` ` `  `    ``for` `(i = 0; i < n; i++) { ` ` `  `        ``// If two character are not same ` `        ``// then they need to be negated ` `        ``if` `(a[i] != b[i]) ` `            ``l[i] = 1; ` `    ``} ` ` `  `    ``// To count the blocks of 1 ` `    ``int` `cc = 0; ` ` `  `    ``// To count the number of 1's in ` `    ``// each block of 1's ` `    ``int` `vl = 0; ` `    ``for` `(i = 0; i < n; i++) { ` `        ``if` `(l[i] == 0) { ` `            ``if` `(vl != 0) ` `                ``cc += 1; ` ` `  `            ``vl = 0; ` `        ``} ` `        ``else` `            ``vl += 1; ` `    ``} ` ` `  `    ``// For the last block of 1's ` `    ``if` `(vl != 0) ` `        ``cc += 1; ` ` `  `    ``cout << cc << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string a = ``"101010"``; ` `    ``string b = ``"110011"``; ` ` `  `    ``int` `n = a.length(); ` `    ``convert(n, a, b); ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by ANKITRAI1 `

## Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `solution { ` ` `  `    ``// Function to find the minimum steps ` `    ``// to convert string a to string b ` `    ``static` `void` `convert(``int` `n, String a, String b) ` `    ``{ ` `        ``// array to mark the positions ` `        ``// needed to be negated ` `        ``int``[] l = ``new` `int``[n]; ` `        ``int` `i; ` ` `  `        ``for` `(i = ``0``; i < n; i++) ` `            ``l[i] = ``0``; ` ` `  `        ``for` `(i = ``0``; i < n; i++) { ` ` `  `            ``// If two character are not same ` `            ``// then they need to be negated ` `            ``if` `(a.charAt(i) != b.charAt(i)) ` `                ``l[i] = ``1``; ` `        ``} ` ` `  `        ``// To count the blocks of 1 ` `        ``int` `cc = ``0``; ` ` `  `        ``// To count the number of 1's in ` `        ``// each block of 1's ` `        ``int` `vl = ``0``; ` `        ``for` `(i = ``0``; i < n; i++) { ` `            ``if` `(l[i] == ``0``) { ` `                ``if` `(vl != ``0``) ` `                    ``cc += ``1``; ` ` `  `                ``vl = ``0``; ` `            ``} ` `            ``else` `                ``vl += ``1``; ` `        ``} ` ` `  `        ``// For the last block of 1's ` `        ``if` `(vl != ``0``) ` `            ``cc += ``1``; ` ` `  `        ``System.out.println(cc); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``String a = ``"101010"``; ` `        ``String b = ``"110011"``; ` ` `  `        ``int` `n = a.length(); ` `        ``convert(n, a, b); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function to find the minimum steps  ` `# to convert string a to string b ` `def` `convert(n, a, b): ` ` `  `    ``# List to mark the positions needed to  ` `    ``# be negated ` `    ``l ``=` `[``0``] ``*` `n ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# If two character are not same  ` `        ``# then they need to be negated ` `        ``if``(a[i] !``=` `b[i]): ` `            ``l[i] ``=` `1` ` `  `    ``# To count the blocks of 1 ` `    ``cc ``=` `0` ` `  `    ``# To count the number of 1's in each  ` `    ``# block of 1's ` `    ``vl ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``if` `(l[i] ``=``=` `0``): ` `            ``if``(vl !``=` `0``): ` `                ``cc ``+``=` `1` `            ``vl ``=` `0` `        ``else``: ` `            ``vl ``+``=` `1` ` `  `    ``# For the last block of 1's ` `    ``if``(vl !``=` `0``): ` `        ``cc ``+``=` `1` ` `  `    ``print``(cc) ` ` `  `# Driver code ` `a ``=` `"101010"` `b ``=` `"110011"` `n ``=` `len``(a) ` `convert(n, a, b) `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the minimum steps ` `    ``// to convert string a to string b ` `    ``static` `void` `convert(``int` `n, String a, String b) ` `    ``{ ` `        ``// array to mark the positions ` `        ``// needed to be negated ` `        ``int``[] l = ``new` `int``[n]; ` `        ``int` `i; ` ` `  `        ``for` `(i = 0; i < n; i++) ` `            ``l[i] = 0; ` ` `  `        ``for` `(i = 0; i < n; i++) { ` ` `  `            ``// If two character are not same ` `            ``// then they need to be negated ` `            ``if` `(a[i] != b[i]) ` `                ``l[i] = 1; ` `        ``} ` ` `  `        ``// To count the blocks of 1 ` `        ``int` `cc = 0; ` ` `  `        ``// To count the number of 1's in ` `        ``// each block of 1's ` `        ``int` `vl = 0; ` `        ``for` `(i = 0; i < n; i++) { ` `            ``if` `(l[i] == 0) { ` `                ``if` `(vl != 0) ` `                    ``cc += 1; ` ` `  `                ``vl = 0; ` `            ``} ` `            ``else` `                ``vl += 1; ` `        ``} ` ` `  `        ``// For the last block of 1's ` `        ``if` `(vl != 0) ` `            ``cc += 1; ` `        ``Console.WriteLine(cc); ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `Main() ` `    ``{ ` ` `  `        ``String a = ``"101010"``; ` `        ``String b = ``"110011"``; ` ` `  `        ``int` `n = a.Length; ` `        ``convert(n, a, b); ` `    ``} ` `} ` ` `  `// This code is contributed by jit_t. `

## PHP

 ` `

Output:

```2
```

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