Cambios mínimos requeridos para que la string satisfaga la condición dada

Dada la string binaria str . En una sola operación, podemos cambiar cualquier ‘1’ a ‘0’ o cualquier ‘0’ a ‘1’ . La tarea es hacer un número mínimo de cambios en la string, de modo que si tomamos cualquier prefijo de la string, el número de 1 debe ser mayor o igual que el número de 0 .
Ejemplos: 
 

Entrada: str = “10001” 
Salida: 1 
Podemos cambiar str[2] de ‘0’ a ‘1’.
Entrada: str = “0000” 
Salida: 2 
 

Enfoque: El problema se puede resolver con avidez. El primer carácter de la string tiene que ser 1 . Luego, para el resto de la string, recorremos la string carácter por carácter y verificamos si la condición requerida se cumple o no, si no, aumentamos la cantidad de cambios requeridos.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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# 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 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
?>

<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>

2

Complejidad de tiempo: O(n), donde n es el tamaño de la string dada
Espacio auxiliar: O(1)