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

  • Last Updated : 11 May, 2021

Given many N digits. We are also given 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  

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 <bits/stdc++.h>
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




<?php
// PHP implementation of the approach
// Function to return the maximized number
 
function get_maximum($s, $a)
{
    $n = strlen($s);
 
    // Iterate till the end of the string
    for ($i = 0; $i < $n; $i++)
    {
 
        // Check if it is greater or not
        if ($s[$i] - '0' < $a[$s[$i] - '0'])
        {
            $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
    $s = "1337";
    $a = array( 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 );
    echo get_maximum($s, $a);
     
    // This Code is contributed is Tushill.
?>

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the maximized number
    function get_maximum(s, a)
    {
        let n = s.length;
 
        // Iterate till the end of the string
        for (let i = 0; i < n; i++)
        {
 
            // Check if it is greater or not
            if (s[i].charCodeAt() - '0'.charCodeAt()
            < a[s[i].charCodeAt() - '0'.charCodeAt()])
            {
                let j = i;
 
                // Replace with the alternate till smaller
                while (j < n && (s[j].charCodeAt() - '0'.charCodeAt()
                <= a[s[j].charCodeAt() - '0'.charCodeAt()]))
                {
                    s[j] = String.fromCharCode('0'.charCodeAt()
                    + a[s[j].charCodeAt() - '0'.charCodeAt()]);
                    j++;
                }
 
                return s.join("");
            }
        }
 
        // Return original s in case
        // no change took place
        return s.join("");
    }
     
    let s = "1337";
    let a = [ 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 ];
    document.write(get_maximum(s.split(''), a));
 
</script>
Output: 
1557

 


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