Given string str containing lowercase characters, the task is to find the lexicographically largest sub-sequence of str.
Examples:
Input: str = “abc”
Output: c
All possible sub-sequences are “a”, “ab”, “ac”, “b”, “bc” and “c”
and “c” is the largest among them (lexicographically)Input: str = “neveropen”
Output: ss
Approach:
Let mx be the lexicographically largest character in the string. Since we want the lexicographically largest sub-sequence we should include all occurrences of mx. Now after all the occurrences have been used, the same process can be repeated for the remaining string (i.e. sub-string after the last occurrence of mx) and so on until the there are no more characters left.
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 lexicographically// largest sub-sequence of sstring getSubSeq(string s, int n){ string res = ""; int cr = 0; while (cr < n) { // Get the max character from the string char mx = s[cr]; for (int i = cr + 1; i < n; i++) mx = max(mx, s[i]); int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) if (s[i] == mx) { res += s[i]; lst = i; } // Repeat the steps for the remaining string cr = lst + 1; } return res;}// Driver codeint main(){ string s = "neveropen"; int n = s.length(); cout << getSubSeq(s, n);} |
Java
// Java implementation of the approachclass GFG{ // Function to return the lexicographically // largest sub-sequence of s static String getSubSeq(String s, int n) { String res = ""; int cr = 0; while (cr < n) { // Get the max character from the String char mx = s.charAt(cr); for (int i = cr + 1; i < n; i++) { mx = (char) Math.max(mx, s.charAt(i)); } int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) { if (s.charAt(i) == mx) { res += s.charAt(i); lst = i; } } // Repeat the steps for // the remaining String cr = lst + 1; } return res; } // Driver code public static void main(String[] args) { String s = "neveropen"; int n = s.length(); System.out.println(getSubSeq(s, n)); }}// This code is contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach# Function to return the lexicographically# largest sub-sequence of sdef getSubSeq(s, n): res = "" cr = 0 while (cr < n): # Get the max character from # the string mx = s[cr] for i in range(cr + 1, n): mx = max(mx, s[i]) lst = cr # Use all the occurrences of the # current maximum character for i in range(cr,n): if (s[i] == mx): res += s[i] lst = i # Repeat the steps for the # remaining string cr = lst + 1 return res# Driver codeif __name__ == '__main__': s = "neveropen" n = len(s) print(getSubSeq(s, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG { // Function to return the lexicographically // largest sub-sequence of s static String getSubSeq(String s, int n) { String res = ""; int cr = 0; while (cr < n) { // Get the max character from // the String char mx = s[cr]; for (int i = cr + 1; i < n; i++) { mx = (char) Math.Max(mx, s[i]); } int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) { if (s[i] == mx) { res += s[i]; lst = i; } } // Repeat the steps for // the remaining String cr = lst + 1; } return res; } // Driver code public static void Main(String[] args) { String s = "neveropen"; int n = s.Length; Console.WriteLine(getSubSeq(s, n)); } } // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the lexicographically// largest sub-sequence of sfunction getSubSeq(s, n){ var res = ""; var cr = 0; while (cr < n) { // Get the max character from the string var mx = s[cr].charCodeAt(0); for (var i = cr + 1; i < n; i++) mx = Math.max(mx, s[i].charCodeAt(0)); var lst = cr; // Use all the occurrences of the // current maximum character for (var i = cr; i < n; i++) if (s[i].charCodeAt(0) == mx) { res += s[i]; lst = i; } // Repeat the steps for the remaining string cr = lst + 1; } return res;}// Driver codevar s = "neveropen";var n = s.length;document.write( getSubSeq(s, n));// This code is contributed by famously.</script> |
PHP
<?php// PHP implementation of the approach// Function to return the lexicographically// largest sub-sequence of sfunction getSubSeq($s, $n){ $res = ""; $cr = 0; while ($cr < $n) { // Get the max character from the string $mx = $s[$cr]; for ($i = $cr + 1; $i < $n; $i++) $mx = max($mx, $s[$i]); $lst = $cr; // Use all the occurrences of the // current maximum character for ($i = $cr; $i < $n; $i++) if ($s[$i] == $mx) { $res .= $s[$i]; $lst = $i; } // Repeat the steps for the // remaining string $cr = $lst + 1; } return $res;}// Driver code$s = "neveropen";$n = strlen($s);echo getSubSeq($s, $n);// This code is contributed by // Akanksha Rai?> |
ss
Complexity Analysis:
- Time Complexity: O(N) where N is the length of the string.
- Auxiliary Space: O(N)
New Approach:- Another approach to solve this problem is using a stack. The idea is to traverse the given string from left to right and push the characters onto the stack. If the current character is greater than the top of the stack, we pop the elements from the stack until we encounter a character that is greater than the current character or the stack becomes empty. Then, we push the current character onto the stack. After traversing the entire string, the stack will contain the lexicographically largest sub-sequence.
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 lexicographically// largest sub-sequence of sstring getSubSeq(string s, int n){ stack<char> st; for (int i = 0; i < n; i++) { while (!st.empty() && s[i] > st.top()) st.pop(); st.push(s[i]); } string res = ""; while (!st.empty()) { res += st.top(); st.pop(); } reverse(res.begin(), res.end()); return res;}// Driver codeint main(){ string s = "neveropen"; int n = s.length(); cout << getSubSeq(s, n);} |
Java
import java.util.*;public class Main { public static String getSubSeq(String s, int n) { Stack<Character> st = new Stack<>(); for (int i = 0; i < n; i++) { while (!st.empty() && s.charAt(i) > st.peek()) st.pop(); st.push(s.charAt(i)); } StringBuilder res = new StringBuilder(); while (!st.empty()) { res.append(st.peek()); st.pop(); } return res.reverse().toString(); } public static void main(String[] args) { String s = "neveropen"; int n = s.length(); System.out.println(getSubSeq(s, n)); }} |
Python
# Python implementation of the approach# Function to return the lexicographically# largest sub-sequence of sdef getSubSeq(s): stack = [] for char in s: while stack and char > stack[-1]: stack.pop() stack.append(char) # Convert the stack to a string in reverse order res = ''.join(stack[::-1]) return res# Driver codeif __name__ == "__main__": s = "neveropen" result = getSubSeq(s) print(result) |
C#
using System;using System.Collections.Generic;using System.Linq;public class Program{ public static string GetSubSeq(string s, int n) { Stack<char> st = new Stack<char>(); for (int i = 0; i < n; i++) { while (st.Any() && s[i] > st.Peek()) { st.Pop(); } st.Push(s[i]); } string res = ""; while (st.Any()) { res += st.Peek(); st.Pop(); } return new string(res.Reverse().ToArray()); } public static void Main() { string s = "neveropen"; int n = s.Length; Console.WriteLine(GetSubSeq(s, n)); }} |
Javascript
function getSubSeq(s, n) { let st = []; for (let i = 0; i < n; i++) { while (st.length > 0 && s[i] > st[st.length - 1]) st.pop(); st.push(s[i]); } let res = ""; while (st.length > 0) { res += st[st.length - 1]; st.pop(); } return res.split("").reverse().join("");}// Driver codelet s = "neveropen";let n = s.length;console.log(getSubSeq(s, n)); |
ss
“Note that the time complexity of this approach is O(n) and space complexity is also O(n) due to the use of stack.”
Time complexity:- The time complexity of the given approach is O(n) as we are iterating over each character of the given string once.
Space complexity:-The space complexity of the given approach is also O(n) as we are using a stack to store the characters of the sub-sequence. In the worst-case scenario, where all the characters of the string are in decreasing order, the stack will contain all the characters of the string.
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