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# Lexicographically largest sub-sequence of the given string

Given string str containing lowercase characters, the task is to find the lexicographically largest sub-sequence of str.

Examples:

Input: str = “abc”
Output:
All possible sub-sequences are “a”, “ab”, “ac”, “b”, “bc” and “c”
and “c” is the largest among them (lexicographically)

Input: str = “geeksforgeeks”
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 ` `using` `namespace` `std;`   `// Function to return the lexicographically` `// largest sub-sequence of s` `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 = 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` `int` `main()` `{` `    ``string s = ``"geeksforgeeks"``;` `    ``int` `n = s.length();` `    ``cout << getSubSeq(s, n);` `}`

## Java

 `// Java implementation of the approach` `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.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 = ``"geeksforgeeks"``;` `        ``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 s` `def` `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 code` `if` `__name__ ``=``=` `'__main__'``:` `    ``s ``=` `"geeksforgeeks"` `    ``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 = ``"geeksforgeeks"``; ` `        ``int` `n = s.Length; ` `        ``Console.WriteLine(getSubSeq(s, n)); ` `    ``} ` `} `   `// This code is contributed by 29AjayKumar`

## PHP

 ``

## Javascript

 ``

Output

`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 ` `using` `namespace` `std;`   `// Function to return the lexicographically` `// largest sub-sequence of s` `string 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 code` `int` `main()` `{` `    ``string s = ``"geeksforgeeks"``;` `    ``int` `n = s.length();` `    ``cout << getSubSeq(s, n);` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `    ``public` `static` `String getSubSeq(String s, ``int` `n) {` `        ``Stack 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 = ``"geeksforgeeks"``;` `        ``int` `n = s.length();` `        ``System.out.println(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 code` `let s = ``"geeksforgeeks"``;` `let n = s.length;` `console.log(getSubSeq(s, n));`

## 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 = ``"geeksforgeeks"``;` `        ``int` `n = s.Length;` `        ``Console.WriteLine(GetSubSeq(s, n));` `    ``}` `}`

Output

`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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