Lexicographically largest possible String after removal of K characters

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Given a string S consisting of only lowercase letters, the task is to find the lexicographically largest string that can be obtained by removing K characters from the given string.

Examples:

Input: s = “zyxedcba”, K=1
Output: zyxedcb
Explanation: The character with the smallest ASCII value from the given string is ‘a’.
Removal of ‘a’ generates the lexicographically largest possible string.

Input: s = “abcde”, K=2
Output: cde

Approach:
The idea is to use Stack Data Structure it to solve the problem. Follow the steps below to solve the problem:

Traverse the string.
For every character, check if it is greater than the character at the top of the stack. If found to be true, pop the top element of the stack if K > 0.
Insert the character into the stack.
After completing the traversal of the string, if K > 0, then remove the top K elements of the stack.
Finally, store the characters in the stack as the answer. Print the answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
string largestString(string num, int k)
{
    // final result string
    string ans = "";
 
    for (auto i : num) {
 
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.length() && ans.back() < i
               && k > 0) {
 
            // Remove from the end of the string
            ans.pop_back();
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans.push_back(i);
    }
 
    // Perform remaining K deletions
    // from the end of the string
    while (ans.length() and k--) {
        ans.pop_back();
    }
 
    // Return the string
    return ans;
}
 
// Driver Code
int main()
{
    string str = "zyxedcba";
    int k = 1;
 
    cout << largestString(str, k) << endl;
}

Java

// Java program to implement the
// above approach
class GFG{
 
static String largestString(String num, int k)
{
     
    // Final result String
    String ans = "";
 
    for(char i : num.toCharArray())
    {
         
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.length() > 0 && 
               ans.charAt(ans.length() - 1) < i &&
                                          k > 0) 
        {
             
            // Remove from the end of the String
            ans = ans.substring(0, ans.length() - 1);
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans += i;
    }
 
    // Perform remaining K deletions
    // from the end of the String
    while (ans.length() > 0 && k-- > 0)
    {
        ans = ans.substring(0, ans.length() - 1);
    }
     
    // Return the String
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "zyxedcba";
    int k = 1;
 
    System.out.print(largestString(str, k) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement the 
# above approach 
def largestString(num, k):
     
    # Final result string 
    ans = []
     
    for i in range(len(num)):
         
        # If the current char exceeds the 
        # character at the top of the stack 
        while(len(ans) and ans[-1] < num[i] and
                                 k > 0):
             
            # Remove from the end of the string 
            ans.pop()
             
            # Decrease k for the removal 
            k -= 1
         
        # Insert current character 
        ans.append(num[i])
     
    # Perform remaining K deletions 
    # from the end of the string 
    while(len(ans) and k):
        k -= 1
        ans.pop()
     
    # Return the string 
    return ans
     
# Driver code
str = "zyxedcba"
k = 1
 
print(*largestString(str, k), sep = "")
 
# This code is contributed by divyeshrabadiya07

C#

// C# program to implement the
// above approach
using System;
 
class GFG{
 
static String largestString(String num, int k)
{
     
    // Final result String
    String ans = "";
 
    foreach(char i in num.ToCharArray())
    {
         
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.Length > 0 && 
           ans[ans.Length - 1] < i && k > 0) 
        {
             
            // Remove from the end of the String
            ans = ans.Substring(0, ans.Length - 1);
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans += i;
    }
 
    // Perform remaining K deletions
    // from the end of the String
    while (ans.Length > 0 && k-- > 0)
    {
        ans = ans.Substring(0, ans.Length - 1);
    }
     
    // Return the String
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "zyxedcba";
    int k = 1;
 
    Console.Write(largestString(str, k) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
    // JavaScript program to implement the
    // above approach
    function largestString(num, k) {
        // Final result String
        var ans = "";
 
        var str = num.split("");
 
        for (const i of str) {
          // If the current char exceeds the
          // character at the top of the stack
          while (ans.length > 0 && ans[ans.length - 1] < i && k > 0) {
            // Remove from the end of the String
            ans = ans.substring(0, ans.length - 1);
 
            // Decrease k for the removal
            k--;
          }
 
          // Insert current character
          ans += i;
        }
 
        // Perform remaining K deletions
        // from the end of the String
        while (ans.length > 0 && k-- > 0) {
          ans = ans.substring(0, ans.length - 1);
        }
 
        // Return the String
        return ans;
    }
 
     // Driver Code
     var str = "zyxedcba";
     var k = 1;
 
     document.write(largestString(str, k) + "<br>");
</script>

Output:

zyxedcb

Time Complexity: O(N)
Auxiliary Space: O(N)

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