# Maximize the given number by replacing a segment of digits with the alternate digits given

Given a large number of N digits. We are also given with 10 numbers which represents the alternate number for all the one-digit numbers from 0 to 9. We can replace any digit in N with the given alternate digit to it, but we are only allowed to replace any consecutive segment of numbers for once only, the task is to replace any consecutive segment of numbers such that the obtained number is the largest of all possible replacements.

Examples:

Input: n = 1337, a[] = {0, 1, 2, 5, 4, 6, 6, 3, 1, 9}
Output: 1557
1 can be replaced with 1 as a[1] = 1 (No effect)
3 can be replaced with 5
7 can be replaced with 3 (isn’t required if the number needs to be maximized)

Input: number = 11111, a[] = {0, 5, 2, 5, 4, 6, 6, 3, 1, 9}
Output: 55555

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since we need to get the largest number possible, hence iterate from the left and find the number whose alternate number is greater than the current one, and keep replacing the upcoming numbers till the alternate number is not smaller.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximized number ` `string get_maximum(string s, ``int` `a[]) ` `{ ` `    ``int` `n = s.size(); ` ` `  `    ``// Iterate till the end of the string ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Check if it is greater or not ` `        ``if` `(s[i] - ``'0'` `< a[s[i] - ``'0'``]) { ` `            ``int` `j = i; ` ` `  `            ``// Replace with the alternate till smaller ` `            ``while` `(j < n && (s[j] - ``'0'` `<= a[s[j] - ``'0'``])) { ` `                ``s[j] = ``'0'` `+ a[s[j] - ``'0'``]; ` `                ``j++; ` `            ``} ` ` `  `            ``return` `s; ` `        ``} ` `    ``} ` ` `  `    ``// Return original s in case ` `    ``// no change took place ` `    ``return` `s; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"1337"``; ` `    ``int` `a[] = { 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 }; ` `    ``cout << get_maximum(s, a); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to return the maximized number ` `static` `String get_maximum(``char``[] s, ``int` `a[]) ` `{ ` `    ``int` `n = s.length; ` ` `  `    ``// Iterate till the end of the string ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` ` `  `        ``// Check if it is greater or not ` `        ``if` `(s[i] - ``'0'` `< a[s[i] - ``'0'``])  ` `        ``{ ` `            ``int` `j = i; ` ` `  `            ``// Replace with the alternate till smaller ` `            ``while` `(j < n && (s[j] - ``'0'` `<= a[s[j] - ``'0'``]))  ` `            ``{ ` `                ``s[j] = (``char``) (``'0'` `+ a[s[j] - ``'0'``]); ` `                ``j++; ` `            ``} ` ` `  `            ``return` `String.valueOf(s); ` `        ``} ` `    ``} ` ` `  `    ``// Return original s in case ` `    ``// no change took place ` `    ``return` `String.valueOf(s); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s = ``"1337"``; ` `    ``int` `a[] = { ``0``, ``1``, ``2``, ``5``, ``4``, ``6``, ``6``, ``3``, ``1``, ``9` `}; ` `    ``System.out.println(get_maximum(s.toCharArray(), a)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the maximized number  ` `def` `get_maximum(s, a) :  ` `    ``s ``=` `list``(s) ` `    ``n ``=` `len``(s)  ` `     `  `    ``# Iterate till the end of the string  ` `    ``for` `i ``in` `range``(n) : ` `         `  `        ``# Check if it is greater or not ` `        ``if` `(``ord``(s[i]) ``-` `ord``(``'0'``) < a[``ord``(s[i]) ``-` `ord``(``'0'``)]) : ` `            ``j ``=` `i ` `             `  `            ``# Replace with the alternate till smaller  ` `            ``while` `(j < n ``and` `(``ord``(s[j]) ``-` `ord``(``'0'``) <``=` `                            ``a[``ord``(s[j]) ``-` `ord``(``'0'``)])) : ` `                ``s[j] ``=` `chr``(``ord``(``'0'``) ``+` `a[``ord``(s[j]) ``-` `ord``(``'0'``)]) ` `                ``j ``+``=` `1` `             `  `            ``return` `"".join(s); ` `     `  `    ``# Return original s in case  ` `    ``# no change took place  ` `    ``return` `s  ` ` `  ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``s ``=` `"1337"` `    ``a ``=` `[ ``0``, ``1``, ``2``, ``5``, ``4``, ``6``, ``6``, ``3``, ``1``, ``9` `]  ` `    ``print``(get_maximum(s, a)) ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the maximized number ` `static` `String get_maximum(``char``[] s, ``int``[] a) ` `{ ` `    ``int` `n = s.Length; ` ` `  `    ``// Iterate till the end of the string ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` ` `  `        ``// Check if it is greater or not ` `        ``if` `(s[i] - ``'0'` `< a[s[i] - ``'0'``])  ` `        ``{ ` `            ``int` `j = i; ` ` `  `            ``// Replace with the alternate till smaller ` `            ``while` `(j < n && (s[j] - ``'0'` `<= a[s[j] - ``'0'``]))  ` `            ``{ ` `                ``s[j] = (``char``) (``'0'` `+ a[s[j] - ``'0'``]); ` `                ``j++; ` `            ``} ` ` `  `            ``return` `String.Join(``""``,s); ` `        ``} ` `    ``} ` ` `  `    ``// Return original s in case ` `    ``// no change took place ` `    ``return` `String.Join(``""``,s); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String s = ``"1337"``; ` `    ``int``[] a = { 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 }; ` `    ``Console.WriteLine(get_maximum(s.ToCharArray(), a)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` `

Output:

```1557
```

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Striver(underscore)79 at Codechef and codeforces D

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