# Smallest string which not a subsequence of the given string

• Difficulty Level : Medium
• Last Updated : 09 Jun, 2021

Given a string str, consisting of lowercase alphabets, the task is to find the shortest string which is not a subsequence of the given string. If multiple strings exist, then print any one of them.

Examples:

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Input: str = “abaabcc”
Output:
Explanation:
One of the possible shortest string which is not a subsequence of the given string is “d”. Therefore, the required output is “d”.

Input: str = “abcdefghijklmnopqrstuvwxyzaabbccdd”
Output: ze

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find shortest string which``// not a subsequence of the given string``string ShortestSubsequenceNotPresent(string str)``{` `    ``// Stores the shortest string which is``    ``// not a subsequence of the given string``    ``string shortestString;` `    ``// Stores length of string``    ``int` `N = str.length();` `    ``// Stores distinct character of subsegments``    ``unordered_set<``char``> subsegments;` `    ``// Traverse the given string``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Insert current character``        ``// into subsegments``        ``subsegments.insert(str[i]);` `        ``// If all lowercase alphabets``        ``// present in the subsegment``        ``if` `(subsegments.size() == 26) {` `            ``// Insert the last character of``            ``// subsegment into shortestString``            ``shortestString.push_back(str[i]);` `            ``// Remove all elements from``            ``// the current subsegment``            ``subsegments.clear();``        ``}``    ``}` `    ``// Traverse all lowercase alphabets``    ``for` `(``char` `ch = ``'a'``; ch <= ``'z'``; ch++) {` `        ``// If current character is not``        ``// present in the subsegment``        ``if` `(subsegments.count(ch) == 0) {``            ``shortestString.push_back(ch);` `            ``// Return shortestString``            ``return` `shortestString;``        ``}``    ``}``    ``return` `shortestString;``}` `// Driver Code``int` `main()``{` `    ``// Given String``    ``string str``        ``= ``"abcdefghijklmnopqrstuvwxyzaabbccdd"``;` `    ``cout << ShortestSubsequenceNotPresent(str);``    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``public` `static` `String``    ``ShortestSubsequenceNotPresent(String str)``    ``{``        ``// Stores the shortest string which is``        ``// not a subsequence of the given string``        ``String shortestString = ``""``;` `        ``// Stores length of string``        ``int` `N = str.length();` `        ``// Stores distinct character of subsegments``        ``HashSet subsegments = ``new` `HashSet<>();` `        ``// Traverse the given string``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// Insert current character``            ``// into subsegments``            ``subsegments.add(str.charAt(i));` `            ``// If all lowercase alphabets``            ``// present in the subsegment``            ``if` `(subsegments.size() == ``26``) {` `                ``// Insert the last character of``                ``// subsegment into shortestString``                ``shortestString``                    ``= shortestString + str.charAt(i);` `                ``// Remove all elements from``                ``// the current subsegment``                ``subsegments.clear();``            ``}``        ``}` `        ``// Traverse all lowercase alphabets``        ``for` `(``char` `ch = ``'a'``; ch <= ``'z'``; ch++) {` `            ``// If current character is not``            ``// present in the subsegment``            ``if` `(!subsegments.contains(ch)) {``                ``shortestString = shortestString + ch;` `                ``// Return shortestString``                ``return` `shortestString;``            ``}``        ``}``        ``return` `shortestString;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"abcdefghijklmnopqrstuvwxyzaabbccdd"``;` `        ``System.out.print(``            ``ShortestSubsequenceNotPresent(str));``    ``}``}``// This code is contributed by Manu Pathria`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find shortest string which``# not a subsequence of the given string``def` `ShortestSubsequenceNotPresent(``Str``):` `    ``# Stores the shortest string which is``    ``# not a subsequence of the given string``    ``shortestString ``=` `""` `    ``# Stores length of string``    ``N ``=` `len``(``Str``)` `    ``# Stores distinct character of subsegments``    ``subsegments ``=` `set``()` `    ``# Traverse the given string``    ``for` `i ``in` `range``(N):` `        ``# Insert current character``        ``# into subsegments``        ``subsegments.add(``Str``[i])` `        ``# If all lowercase alphabets``        ``# present in the subsegment``        ``if` `(``len``(subsegments) ``=``=` `26``) :` `            ``# Insert the last character of``            ``# subsegment into shortestString``            ``shortestString ``+``=` `Str``[i]` `            ``# Remove all elements from``            ``# the current subsegment``            ``subsegments.clear()``        ` `    ``# Traverse all lowercase alphabets``    ``for` `ch ``in` `range``(``int``(``26``)):``        ` `        ``# If current character is not``        ``# present in the subsegment``        ``if` `(``chr``(ch ``+` `97``) ``not` `in` `subsegments) :``            ` `            ``shortestString ``+``=` `chr``(ch ``+` `97``)``            ` `            ``# Return shortestString``            ``return` `shortestString``    ` `    ``return` `shortestString` `# Driver code``# Given String``Str` `=` `"abcdefghijklmnopqrstuvwxyzaabbccdd"` `print``(ShortestSubsequenceNotPresent(``Str``))` `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# program to implement``// the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``    ` `public` `static` `String ShortestSubsequenceNotPresent(``    ``String str)``{``    ` `    ``// Stores the shortest string which is``    ``// not a subsequence of the given string``    ``String shortestString = ``""``;` `    ``// Stores length of string``    ``int` `N = str.Length;` `    ``// Stores distinct character of subsegments``    ``HashSet<``char``> subsegments = ``new` `HashSet<``char``>();` `    ``// Traverse the given string``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// Insert current character``        ``// into subsegments``        ``subsegments.Add(str[i]);` `        ``// If all lowercase alphabets``        ``// present in the subsegment``        ``if` `(subsegments.Count == 26)``        ``{``            ` `            ``// Insert the last character of``            ``// subsegment into shortestString``            ``shortestString = shortestString + str[i];``            ` `            ``// Remove all elements from``            ``// the current subsegment``            ``subsegments.Clear();``        ``}``    ``}` `    ``// Traverse all lowercase alphabets``    ``for``(``char` `ch = ``'a'``; ch <= ``'z'``; ch++)``    ``{``        ` `        ``// If current character is not``        ``// present in the subsegment``        ``if` `(!subsegments.Contains(ch))``        ``{``            ``shortestString = shortestString + ch;``            ` `            ``// Return shortestString``            ``return` `shortestString;``        ``}``    ``}``    ``return` `shortestString;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``String str = ``"abcdefghijklmnopqrstuvwxyzaabbccdd"``;``    ` `    ``Console.Write(``        ``ShortestSubsequenceNotPresent(str));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output
`ze`

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

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