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Minimum changes required such that the string satisfies the given condition

  • Last Updated : 19 May, 2021

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 <bits/stdc++.h>
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




<?php
// PHP implementation of the approach
 
// Function to return the minimum
// changes required
function minChanges($str, $n)
{
 
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    $count = $zeros = $ones = 0;
 
    // First character has to be '1'
    if ($str[0] != '1')
    {
        $count++;
        $ones++;
    }
 
    for ($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
$str = "0000";
$n = strlen($str);
echo minChanges($str, $n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the minimum
// changes required
function minChanges(str, n)
{
   
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    let count = 0, zeros = 0, ones = 0;
   
    // First character has to be '1'
    if (str[0] != '1')
    {
        count++;
        ones++;
    }
   
    for (let 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
 
     let str = "0000".split('');
    let n = str.length;
     document.write(minChanges(str, n));
             
</script>
Output: 
2

 


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