Largest number possible after removal of K digits

Given a positive number N, the target is to find the largest number that can be formed after removing any K digits from N.

Examples:

Input: N = 6358, K = 1
Output: 658

Input: N = 2589, K = 2
Output: 89

Approach:



  • Iterate a loop K times.
  • During every iteration, remove every digit from the current value of N once and store the maximum of all the numbers obtained.
  • In order to achieve this we store the maximum of (N / (i * 10)) * i + (N % i) where i ranges from [1, 10l – 1] where l denotes the number of digits of the current value of N.
  • Consider this maximum as the current value of N and proceed to the next iteration and repeat the above step.
  • Thus, after every iteration, we have the least digit from the current value of N removed. On repeating the process K times, we obtain the largest number possible.

For example:

Let us analyze this approach for N = 6358, K = 1
The different possibilities after removal of every digit once are as follows:
(6358 / 10) * 1 + 6358 % 1 = 635 + 0 = 635
(6358 / 100) * 10 + 6358 % 10 = 630 + 8 = 638
(6358 / 1000) * 100 + 6358 % 100 = 600 + 58 = 658
(6358 / 10000) * 1000 + 6358 % 1000 = 0 + 358 = 358

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// largest number possible
int maxnumber(int n, int k)
{
    // Generate the largest number
    // after removal of the least
    // K digits one by one
    for (int j = 0; j < k; j++) {
  
        int ans = 0;
        int i = 1;
  
        // Remove the least digit
        // after every iteration
        while (n / i > 0) {
  
            // Store the numbers formed after
            // removing every digit once
            int temp = (n / (i * 10))
                           * i
                       + (n % i);
            i *= 10;
  
            // Compare and store the maximum
            ans = max(ans, temp);
        }
  
        // Store the largest
        // number remaining
        n = ans;
    }
  
    // Return the remaining number
    // after K removals
    return n;
}
  
// Driver code
int main()
{
    int n = 6358;
    int k = 1;
  
    cout << maxnumber(n, k) << endl;
    return 0;
}

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Java

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// Java program to implement
// the above approach
  
import java.util.*;
import java.math.*;
  
class GFG {
  
    // Function to return the
    // largest number possible
    static int maxnumber(int n, int k)
    {
        // Generate the largest number
        // after removal of the least
        // K digits one by one
        for (int j = 0; j < k; j++) {
  
            int ans = 0;
            int i = 1;
  
            // Remove the least digit
            // after every iteration
            while (n / i > 0) {
  
                // Store the numbers formed after
                // removing every digit once
                int temp = (n / (i * 10))
                               * i
                           + (n % i);
                i *= 10;
  
                // Compare and store the maximum
                ans = Math.max(ans, temp);
            }
            n = ans;
        }
  
        // Return the remaining number
        // after K removals
        return n;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 6358;
        int k = 1;
  
        System.out.println(maxnumber(n, k));
    }
}

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Python3

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# Python program to implement
# the above approach
  
def maxnumber(n, k):
# Function to return the
# largest number possible
   
    for i in range(0, k):
        # Generate the largest number 
        # after removal of the least K digits
        # one by one
        ans = 0
        i = 1
        
        while n // i > 0:
        # Remove the least digit 
        # after every iteration
            temp = (n//(i * 10))*i + (n % i)
            i *= 10
        # Store the numbers formed after 
        # removing every digit once
          
        # Compare and store the maximum
            if temp > ans:
                ans = temp
        n = ans        
    
    # Return the remaining number
    # after K removals
    return ans;
    
   
n = 6358
k = 1
print(maxnumber(n, k))

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C#

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// C# program to implement
// the above approach
  
using System;
  
class GFG {
  
    // Function to return the
    // largest number possible
    static int maxnumber(int n, int k)
    {
        // Generate the largest number
        // after removal of the least
        // K digits one by one
        for (int j = 0; j < k; j++) {
  
            int ans = 0;
            int i = 1;
  
            // Remove the least digit
            // after every iteration
            while (n / i > 0) {
  
                // Store the numbers formed after
                // removing every digit once
                int temp = (n / (i * 10))
                               * i
                           + (n % i);
                i *= 10;
  
                // Compare and store the maximum
                if (temp > ans)
                    ans = temp;
            }
  
            // Store the largest
            // number remaining
            n = ans;
        }
  
        // Return the remaining number
        // after K removals
        return n;
    }
  
    // Driver code
    static public void Main()
    {
        int n = 6358;
        int k = 1;
        Console.WriteLine(maxnumber(n, k));
    }
}

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Output:

658

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