Número mínimo de operaciones requeridas para convertir una string en otra string

Dadas dos strings S y T de igual longitud. Ambas strings contienen solo los caracteres ‘0’ y ‘1’ . La tarea es encontrar el número mínimo de operaciones para convertir la string S en T. Hay 2 tipos de operaciones permitidas en la string S : 

• Intercambiar dos caracteres cualquiera de la string.
• Reemplace un ‘0’ con un ‘1’ o viceversa.

Ejemplos: 

Entrada: S = “011”, T = “101” 
Salida: 1 
Intercambia el primer y segundo carácter.

Entrada: S = “010”, T = “101” 
Salida: 2 
Intercambie el primer y segundo carácter y reemplace el tercer carácter con ‘1’. 
 

Enfoque: encuentre 2 valores para la string S , el número de índices que tienen 0 pero deberían ser 1 y el número de índices que tienen 1 pero deberían ser 0. El resultado sería el máximo de estos 2 valores ya que podemos usar intercambios en el mínimo de estos 2 valores y los caracteres no coincidentes restantes se pueden invertir, es decir , ‘0’ se puede cambiar a ‘1’ y ‘1’ se puede cambiar a ‘0’ .

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 operations
// of the given type required to convert
// string s to string t
int minOperations(string s, string t, int n)
{
    int ct0 = 0, ct1 = 0;
    for (int i = 0; i < n; i++) {
 
        // Characters are already equal
        if (s[i] == t[i])
            continue;
 
        // Increment count of 0s
        if (s[i] == '0')
            ct0++;
 
        // Increment count of 1s
        else
            ct1++;
    }
 
    return max(ct0, ct1);
}
 
// Driver code
int main()
{
    string s = "010", t = "101";
    int n = s.length();
    cout << minOperations(s, t, n);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the minimum
// operations of the given type required
// to convert string s to string t
static int minOperations(String s,
                        String t, int n)
{
    int ct0 = 0, ct1 = 0;
    for (int i = 0; i < n; i++)
    {
 
        // Characters are already equal
        if (s.charAt(i) == t.charAt(i))
            continue;
 
        // Increment count of 0s
        if (s.charAt(i) == '0')
            ct0++;
 
        // Increment count of 1s
        else
            ct1++;
    }
 
    return Math.max(ct0, ct1);
}
 
// Driver code
public static void main(String args[])
{
    String s = "010", t = "101";
    int n = s.length();
    System.out.println(minOperations(s, t, n));
}
}
 
// This code is contributed by
// Surendra_Gangwar

# Python 3 implementation of the approach
 
# Function to return the minimum operations
# of the given type required to convert
# string s to string t
def minOperations(s, t, n):
 
    ct0 = 0
    ct1 = 0
    for i in range(n):
 
        # Characters are already equal
        if (s[i] == t[i]):
            continue
 
        # Increment count of 0s
        if (s[i] == '0'):
            ct0 += 1
 
        # Increment count of 1s
        else:
            ct1 += 1
 
    return max(ct0, ct1)
 
# Driver code
if __name__ == "__main__":
     
    s = "010"
    t = "101"
    n = len(s)
    print(minOperations(s, t, n))
 
# This code is contributed by ita_c

// C# implementation of the approach
using System;
class GFG
{
 
// Function to return the minimum operations
// of the given type required to convert
// string s to string t
static int minOperations(string s, 
                         string t, int n)
{
    int ct0 = 0, ct1 = 0;
    for (int i = 0; i < n; i++)
    {
 
        // Characters are already equal
        if (s[i] == t[i])
            continue;
 
        // Increment count of 0s
        if (s[i] == '0')
            ct0++;
 
        // Increment count of 1s
        else
            ct1++;
    }
 
    return Math.Max(ct0, ct1);
}
 
// Driver code
public static void Main()
{
    string s = "010", t = "101";
    int n = s.Length;
    Console.Write(minOperations(s, t, n));
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the approach
 
// Function to return the minimum operations
// of the given type required to convert
// string s to string t
function minOperations($s, $t, $n)
{
    $ct0 = 0 ; $ct1 = 0;
    for ($i = 0; $i < $n; $i++)
    {
 
        // Characters are already equal
        if ($s[$i] == $t[$i])
            continue;
 
        // Increment count of 0s
        if ($s[$i] == '0')
            $ct0++;
 
        // Increment count of 1s
        else
            $ct1++;
    }
 
    return max($ct0, $ct1);
}
 
// Driver code
$s = "010"; $t = "101";
$n = strlen($s);
echo minOperations($s, $t, $n);
 
// This code is contributed by Ryuga
?>

<script>
 
// JavaScript implementation of the approach
 
// Function to return the minimum operations
// of the given type required to convert
// string s to string t
function minOperations(s, t, n)
{
    var ct0 = 0,
    ct1 = 0;
    for(var i = 0; i < n; i++)
    {
        // Characters are already equal
        if (s[i] === t[i])
            continue;
         
        // Increment count of 0s
        if (s[i] === "0")
            ct0++;
             
        // Increment count of 1s
        else
            ct1++;
    }
    return Math.max(ct0, ct1);
}
 
// Driver code
var s = "010",
t = "101";
var n = s.length;
 
document.write(minOperations(s, t, n));
 
// This code is contributed by rdtank
 
</script>

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Complejidad de tiempo: O(N)