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# Minimum changes required such that the string satisfies the given condition

Given binary string str. In a single operation, we can change any ‘1’ to ‘0’ or any ‘0’ to ‘1’. The task is to make minimum number of changes in the string such that if we take any prefix of the string, the number of 1’s should be greater than or equal number of 0’s.
Examples:

Input: str = “10001”
Output:
We can change str[2] from ‘0’ to ‘1’.
Input: str = “0000”
Output:

Approach: The problem can be solved greedily. The first character of the string has to be 1. Then for the rest of the string, we traverse through the string character by character and check if the required condition is fulfilled or not, if not then we increase the count of changes required.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum``// changes required``int` `minChanges(string str, ``int` `n)``{` `    ``// To store the count of minimum changes,``    ``// number of ones and the number of zeroes``    ``int` `count = 0, zeros = 0, ones = 0;` `    ``// First character has to be '1'``    ``if` `(str[0] != ``'1'``) {``        ``count++;``        ``ones++;``    ``}` `    ``for` `(``int` `i = 1; i < n; i++) {``        ``if` `(str[i] == ``'0'``)``            ``zeros++;``        ``else``            ``ones++;` `        ``// If condition fails``        ``// changes need to be made``        ``if` `(zeros > ones) {``            ``zeros--;``            ``ones++;``            ``count++;``        ``}``    ``}` `    ``// Return the required count``    ``return` `count;``}` `// Driver code``int` `main()``{``    ``string str = ``"0000"``;``    ``int` `n = str.length();``    ``cout << minChanges(str, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the minimum``// changes required``static` `int` `minChanges(``char``[] str, ``int` `n)``{` `    ``// To store the count of minimum changes,``    ``// number of ones and the number of zeroes``    ``int` `count = ``0``, zeros = ``0``, ones = ``0``;` `    ``// First character has to be '1'``    ``if` `(str[``0``] != ``'1'``)``    ``{``        ``count++;``        ``ones++;``    ``}` `    ``for` `(``int` `i = ``1``; i < n; i++)``    ``{``        ``if` `(str[i] == ``'0'``)``            ``zeros++;``        ``else``            ``ones++;` `        ``// If condition fails``        ``// changes need to be made``        ``if` `(zeros > ones)``        ``{``            ``zeros--;``            ``ones++;``            ``count++;``        ``}``    ``}` `    ``// Return the required count``    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``char` `[]str = ``"0000"``.toCharArray();``    ``int` `n = str.length;``    ``System.out.print(minChanges(str, n));``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the minimum``# changes required``def` `minChanges(``str``, n):``    ` `    ``# To store the count of minimum changes,``    ``# number of ones and the number of zeroes``    ``count, zeros, ones ``=` `0``, ``0``, ``0` `    ``# First character has to be '1'``    ``if` `(``ord``(``str``[``0``])!``=` `ord``(``'1'``)):``        ``count ``+``=` `1``        ``ones ``+``=` `1` `    ``for` `i ``in` `range``(``1``, n):``        ``if` `(``ord``(``str``[i]) ``=``=` `ord``(``'0'``)):``            ``zeros ``+``=` `1``        ``else``:``            ``ones ``+``=` `1` `        ``# If condition fails``        ``# changes need to be made``        ``if` `(zeros > ones):``            ``zeros ``-``=` `1``            ``ones ``+``=` `1``            ``count ``+``=` `1` `    ``# Return the required count``    ``return` `count` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``str` `=` `"0000"``    ``n ``=` `len``(``str``)``    ``print``(minChanges(``str``, n))` `# This code contributed by PrinciRaj1992`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the minimum``// changes required``static` `int` `minChanges(``char``[] str, ``int` `n)``{` `    ``// To store the count of minimum changes,``    ``// number of ones and the number of zeroes``    ``int` `count = 0, zeros = 0, ones = 0;` `    ``// First character has to be '1'``    ``if` `(str[0] != ``'1'``)``    ``{``        ``count++;``        ``ones++;``    ``}` `    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``if` `(str[i] == ``'0'``)``            ``zeros++;``        ``else``            ``ones++;` `        ``// If condition fails``        ``// changes need to be made``        ``if` `(zeros > ones)``        ``{``            ``zeros--;``            ``ones++;``            ``count++;``        ``}``    ``}` `    ``// Return the required count``    ``return` `count;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``char` `[]str = ``"0000"``.ToCharArray();``    ``int` `n = str.Length;``    ``Console.Write(minChanges(str, n));``}``}` `// This code contributed by Rajput-Ji`

## PHP

 ` ``\$ones``)``        ``{``            ``\$zeros``--;``            ``\$ones``++;``            ``\$count``++;``        ``}``    ``}` `    ``// Return the required count``    ``return` `\$count``;``}` `// Driver code``\$str` `= ``"0000"``;``\$n` `= ``strlen``(``\$str``);``echo` `minChanges(``\$str``, ``\$n``);` `// This code is contributed by mits``?>`

## Javascript

 ``

Output:

`2`

Time Complexity: O(n), where n is the size of the given string
Auxiliary Space: O(1)

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