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

• Difficulty Level : Easy
• Last Updated : 15 Sep, 2022

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 ` `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

 ` ``\$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

 ``

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)

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