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Minimum number of operations required to obtain a given Binary String
• Last Updated : 26 Mar, 2021

Given a binary strings S of length N, the task is to obtain S from a string, say T, of length N consisting only of zeroes, by minimum number of operations. Each operation involves choosing any index i from string S and flipping all the bits at indices [i, N – 1] of the string T
Examples:

Input: S = “101”
Output:
Explanation:
“000” -> “111” -> “100” -> “101”.
Therefore, the minimum number of steps required is 3.
Input: S = “10111”
Output:
Explanation:
“00000” -> “11111” -> “10000” -> “10111”.
Therefore, the minimum number of steps required is 3.

Approach:
The idea is to find the first occurrence of 1 in the given string S and perform the given operation at that index. After this step, for every mismatch in the character of S and T at a particular index, repeat the opertion.
Follow the steps below:

1. Iterate over S and mark the first occurrence of 1.
2. Initialize two variables, say last and ans, where last stores the character (0 or 1) for which the last operation was performed and ans keeps count of the minimum number of steps required.
3. Iterate from the first occurrence of 1 till the end of the string.
4. If the current character is 0 and last = 1, then increment the count of ans, and set last = 0.
5. Otherwise, if last = 0 and the current character is 1, then set last = 1.
6. Return the final value of ans.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum``// number of operations required``// to obtain the string s``int` `minOperations(string s)``{``    ``int` `n = s.size();``    ``int` `pos = -1;` `    ``// Iterate the string s``    ``for` `(``int` `i = 0; i < s.size(); i++) {` `        ``// If first occurrence of 1``        ``// is found``        ``if` `(s[i] == ``'1'``) {` `            ``// Mark the index``            ``pos = i;``            ``break``;``        ``}``    ``}` `    ``// Base case: If no 1 occurred``    ``if` `(pos == -1) {` `        ``// No operations required``        ``return` `0;``    ``}` `    ``// Stores the character``    ``// for which last operation``    ``// was performed``    ``int` `last = 1;` `    ``// Stores minimum number``    ``// of operations``    ``int` `ans = 1;` `    ``// Iterate from pos to n``    ``for` `(``int` `i = pos + 1; i < n; i++) {` `        ``// Check if s[i] is 0``        ``if` `(s[i] == ``'0'``) {` `            ``// Check if last operation was``            ``// performed because of 1``            ``if` `(last == 1) {``                ``ans++;` `                ``// Set last to 0``                ``last = 0;``            ``}``        ``}``        ``else` `{` `            ``if` `(last == 0) {` `                ``// Check if last operation was``                ``// performed because of 0``                ``ans++;` `                ``// Set last to 1``                ``last = 1;``            ``}``        ``}``    ``}` `    ``// Return the answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``string s = ``"10111"``;` `    ``cout << minOperations(s);``    ``return` `0;``}`

## Java

 `// Java program to implement the``// above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the minimum``// number of operations required``// to obtain the string s``static` `int` `minOperations(String s)``{``    ``int` `n = s.length();``    ``int` `pos = -``1``;` `    ``// Iterate the string s``    ``for``(``int` `i = ``0``; i < s.length(); i++)``    ``{``        ` `        ``// If first occurrence of 1``        ``// is found``        ``if` `(s.charAt(i) == ``'1'``)``        ``{``            ` `            ``// Mark the index``            ``pos = i;``            ``break``;``        ``}``    ``}` `    ``// Base case: If no 1 occurred``    ``if` `(pos == -``1``)``    ``{` `        ``// No operations required``        ``return` `0``;``    ``}` `    ``// Stores the character``    ``// for which last operation``    ``// was performed``    ``int` `last = ``1``;` `    ``// Stores minimum number``    ``// of operations``    ``int` `ans = ``1``;` `    ``// Iterate from pos to n``    ``for``(``int` `i = pos + ``1``; i < n; i++)``    ``{``        ` `        ``// Check if s[i] is 0``        ``if` `(s.charAt(i) == ``'0'``)``        ``{` `            ``// Check if last operation was``            ``// performed because of 1``            ``if` `(last == ``1``)``            ``{``                ``ans++;` `                ``// Set last to 0``                ``last = ``0``;``            ``}``        ``}``        ``else``        ``{``            ``if` `(last == ``0``)``            ``{` `                ``// Check if last operation was``                ``// performed because of 0``                ``ans++;` `                ``// Set last to 1``                ``last = ``1``;``            ``}``        ``}``    ``}``    ` `    ``// Return the answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String s = ``"10111"``;``    ` `    ``System.out.println(minOperations(s));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the minimum``# number of operations required``# to obtain the string s``def` `minOperations(s):` `    ``n ``=` `len``(s)``    ``pos ``=` `-``1` `    ``# Iterate the string s``    ``for` `i ``in` `range``(``len``(s)):` `        ``# If first occurrence of 1``        ``# is found``        ``if` `(s[i] ``=``=` `'1'``):` `            ``# Mark the index``            ``pos ``=` `i``            ``break` `    ``# Base case: If no 1 occurred``    ``if` `(pos ``=``=` `-``1``):` `        ``# No operations required``        ``return` `0` `    ``# Stores the character``    ``# for which last operation``    ``# was performed``    ``last ``=` `1` `    ``# Stores minimum number``    ``# of operations``    ``ans ``=` `1` `    ``# Iterate from pos to n``    ``for` `i ``in` `range``(pos ``+` `1``, n):` `        ``# Check if s[i] is 0``        ``if` `(s[i] ``=``=` `'0'``):` `            ``# Check if last operation was``            ``# performed because of 1``            ``if` `(last ``=``=` `1``):``                ``ans ``+``=` `1` `                ``# Set last to 0``                ``last ``=` `0``        ` `        ``else``:` `            ``if` `(last ``=``=` `0``):` `                ``# Check if last operation was``                ``# performed because of 0``                ``ans ``+``=` `1` `                ``# Set last to 1``                ``last ``=` `1` `    ``# Return the answer``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``s ``=` `"10111"` `    ``print``(minOperations(s))` `# This code is contributed by chitranayal`

## C#

 `// C# program to implement the``// above approach``using` `System;``class` `GFG{`` ` `// Function to find the minimum``// number of operations required``// to obtain the string s``static` `int` `minOperations(String s)``{``    ``int` `n = s.Length;``    ``int` `pos = -1;`` ` `    ``// Iterate the string s``    ``for``(``int` `i = 0; i < s.Length; i++)``    ``{``         ` `        ``// If first occurrence of 1``        ``// is found``        ``if` `(s[i] == ``'1'``)``        ``{``             ` `            ``// Mark the index``            ``pos = i;``            ``break``;``        ``}``    ``}`` ` `    ``// Base case: If no 1 occurred``    ``if` `(pos == -1)``    ``{`` ` `        ``// No operations required``        ``return` `0;``    ``}`` ` `    ``// Stores the character``    ``// for which last operation``    ``// was performed``    ``int` `last = 1;`` ` `    ``// Stores minimum number``    ``// of operations``    ``int` `ans = 1;`` ` `    ``// Iterate from pos to n``    ``for``(``int` `i = pos + 1; i < n; i++)``    ``{``         ` `        ``// Check if s[i] is 0``        ``if` `(s[i] == ``'0'``)``        ``{`` ` `            ``// Check if last operation was``            ``// performed because of 1``            ``if` `(last == 1)``            ``{``                ``ans++;`` ` `                ``// Set last to 0``                ``last = 0;``            ``}``        ``}``        ``else``        ``{``            ``if` `(last == 0)``            ``{`` ` `                ``// Check if last operation was``                ``// performed because of 0``                ``ans++;`` ` `                ``// Set last to 1``                ``last = 1;``            ``}``        ``}``    ``}``     ` `    ``// Return the answer``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``String s = ``"10111"``;``     ` `    ``Console.Write(minOperations(s));``}``}`` ` `// This code is contributed by rock_cool`

## Javascript

 ``
Output:
`3`

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

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