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Maximize the value of A by replacing some of its digits with digits of B

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Given two string A and B which represents two integers, the task is to print the maximized value of A after replacing 0 or more digits of A with any digit of B

Note: A digit in B can only be used once.

Examples: 

Input: A = “1234”, B = “4321” 
Output: 4334 
1 can be replaced with 4 and 2 can be replaced with 3.

Input: A = “1002”, B = “100” 
Output: 1102 
The first 0 can be replaced with a 1. 

Approach: Since the value of A has to maximized, any digit will be replaced by only digits of greater value. The digits on the left have more significance in contributing to the value, so they should be replaced with as large values as possible. Sort B and iterate from left to right in A and try replacing the current digit with the maximum of available options if possible.

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 maximized value of A
string maxValue(string a, string b)
{
    // Sort digits in ascending order
    sort(b.begin(), b.end());
    int n = a.length();
    int m = b.length();
 
    // j points to largest digit in B
    int j = m - 1;
    for (int i = 0; i < n; i++) {
 
        // If all the digits of b have been used
        if (j < 0)
            break;
 
        if (b[j] > a[i]) {
            a[i] = b[j];
 
            // Current digit has been used
            j--;
        }
    }
 
    // Return the maximized value
    return a;
}
 
// Driver code
int main()
{
    string a = "1234";
    string b = "4321";
 
    cout << maxValue(a, b);
 
    return 0;
}

                    

Java

// Java implementation of the approach
  
import java.util.*;
 
class GFG{
 
// Function to return the maximized value of A
static String maxValue(char []a, char []b)
{
    // Sort digits in ascending order
    Arrays.sort(b);
    int n = a.length;
    int m = b.length;
 
    // j points to largest digit in B
    int j = m - 1;
    for (int i = 0; i < n; i++) {
 
        // If all the digits of b have been used
        if (j < 0)
            break;
 
        if (b[j] > a[i]) {
            a[i] = b[j];
 
            // Current digit has been used
            j--;
        }
    }
 
    // Return the maximized value
    return String.valueOf(a);
}
 
// Driver code
public static void main(String[] args)
{
    String a = "1234";
    String b = "4321";
 
    System.out.print(maxValue(a.toCharArray(), b.toCharArray()));
}
}
 
// This code is contributed by PrinciRaj1992

                    

Python3

# Python3 implementation of the approach
 
# Function to return the maximized
# value of A
def maxValue(a, b):
     
    # Sort digits in ascending order
    b = sorted(b)
    bi = [i for i in b]
    ai = [i for i in a]
     
    n = len(a)
    m = len(b)
 
    # j points to largest digit in B
    j = m - 1
    for i in range(n):
 
        # If all the digits of b
        # have been used
        if (j < 0):
            break
 
        if (bi[j] > ai[i]):
            ai[i] = bi[j]
 
            # Current digit has been used
            j -= 1
         
    # Return the maximized value
    x = "" . join(ai)
    return x
 
# Driver code
a = "1234"
b = "4321"
 
print(maxValue(a, b))
 
# This code is contributed
# by mohit kumar

                    

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the maximized value of A
static String maxValue(char []a, char []b)
{
    // Sort digits in ascending order
    Array.Sort(b);
    int n = a.Length;
    int m = b.Length;
 
    // j points to largest digit in B
    int j = m - 1;
    for (int i = 0; i < n; i++)
    {
 
        // If all the digits of b have been used
        if (j < 0)
            break;
 
        if (b[j] > a[i])
        {
            a[i] = b[j];
 
            // Current digit has been used
            j--;
        }
    }
 
    // Return the maximized value
    return String.Join("",a);
}
 
// Driver code
public static void Main(String[] args)
{
    String a = "1234";
    String b = "4321";
 
    Console.Write(maxValue(a.ToCharArray(), b.ToCharArray()));
}
}
 
// This code is contributed by PrinciRaj1992

                    

PHP

<?php
// PHP implementation of the approach
 
// Function to return the maximized value of A
function maxValue($a, $b)
{
    // Sort digits in ascending order
    sort($b);
    $n = sizeof($a);
    $m = sizeof($b);
 
    // j points to largest digit in B
    $j = $m - 1;
    for ($i = 0; $i < $n; $i++)
    {
 
        // If all the digits of b have been used
        if ($j < 0)
            break;
 
        if ($b[$j] > $a[$i])
        {
            $a[$i] = $b[$j];
 
            // Current digit has been used
            $j--;
        }
    }
 
    // Convert array into string
    $a = implode("",$a);
     
    // Return the maximized value
    return $a ;
}
 
    // Driver code
    # convert string into array
    $a = str_split("1234");
    $b = str_split("4321");
 
    echo maxValue($a, $b);
     
// This code is contributed by Ryuga
?>

                    

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the maximized value of A
function maxValue(a,b)
{
    // Sort digits in ascending order
    b.sort(function(x,y){return x-y;});
    let n = a.length;
    let m = b.length;
   
    // j points to largest digit in B
    let j = m - 1;
    for (let i = 0; i < n; i++) {
   
        // If all the digits of b have been used
        if (j < 0)
            break;
   
        if (b[j] > a[i]) {
            a[i] = b[j];
   
            // Current digit has been used
            j--;
        }
    }
   
    // Return the maximized value
    return (a).join("");
}
 
// Driver code
let a = "1234";
let b = "4321";
 
document.write(maxValue(a.split(""), b.split("")));
 
 
// This code is contributed by patel2127
</script>

                    

Output
4334

Complexity Analysis:

  • Time Complexity: O(n+m*log(m)) where n is the size of string a and m is the size of string b.
  • Auxiliary Space: O(1)


Last Updated : 15 Sep, 2022
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