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 Codeint main(){ string str = "zyxedcba"; int k = 1; cout << largestString(str, k) << endl;} |
Java
// Java program to implement the// above approachclass 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 Codepublic 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 codestr = "zyxedcba"k = 1print(*largestString(str, k), sep = "")# This code is contributed by divyeshrabadiya07 |
C#
// C# program to implement the// above approachusing 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 Codepublic 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
