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

Given a 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: 1
We can change str[2] from ‘0’ to ‘1’.

Input: str = “0000”
Output: 2

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

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 ?>

Output:

2

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