# 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++

 `// C++ program to implement ` `// the above approach ` ` `  `#include ` `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; ` `} `

## Java

 `// 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)); ` `    ``} ` `} `

## Python3

 `# 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)) `

## C#

 `// 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)); ` `    ``} ` `} `

Output:

```658
```

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