# Lexicographically smallest string which is not a subsequence of given string

Given a string **S**, the task is to find the string which is lexicographically smallest and not a subsequence of the given string **S**.

**Examples:**

Input:S = “abcdefghijklmnopqrstuvwxyz”Output:

aaExplanation:

String “aa” is the lexicographically smallest string which is not present in the given string as a subsequence.

Input:S = “aaaa”Output:

aaabExplanation:

String “aaa” is the lexicographically smallest string which is not present in the given string as a subsequence.

**Approach:** The idea is to solve the given problem is to find the frequency of the character **‘a’** in the given string (say **f**). If the value of **f** is less than the length of the string, then the answer would be the string formed by concatenating **‘a’** **f+1** number of times as **aaa….(f+1 times)** will be the required lexicographically smallest string which is not a subsequence. Otherwise, the answer would be the string formed by concatenating ‘**a’ (f-1 times)** and **‘b’** at the end of the string.

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find lexicographically smallest string that` `// is not a subsequence of S` `string smallestNonSubsequence(string S, ` `int` `N)` `{` ` ` `// variable to store frequency of 'a'` ` ` `int` `freq = 0;` ` ` `// calculate frequency of 'a'` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `if` `(S[i] == ` `'a'` `)` ` ` `freq++;` ` ` `// Initialize string consisting of freq number of 'a'` ` ` `string ans(freq, ` `'a'` `);` ` ` `// change the last digit to 'b'` ` ` `if` `(freq == N)` ` ` `ans[freq - 1] = ` `'b'` `;` ` ` `// add another 'a' to the string` ` ` `else` ` ` `ans += ` `'a'` `;` ` ` `// return the answer string` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `// Input` ` ` `string S = ` `"abcdefghijklmnopqrstuvwxyz"` `;` ` ` `int` `N = S.length();` ` ` `// Function call` ` ` `cout << smallestNonSubsequence(S, N) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `public` `class` `GFG` `{` `// Function to find lexicographically smallest string that` `// is not a subsequence of S` `static` `String smallestNonSubsequence(String S, ` `int` `N)` `{` ` ` ` ` `// variable to store frequency of 'a'` ` ` `int` `freq = ` `0` `;` ` ` ` ` `// calculate frequency of 'a'` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `if` `(S.charAt(i) == ` `'a'` `)` ` ` `freq++;` ` ` ` ` `// Initialize string consisting of freq number of 'a'` ` ` `String ans=` `""` `;` ` ` `for` `(` `int` `i = ` `0` `; i < f req; i++){` ` ` `ans += ` `'a'` `;` ` ` `}` ` ` ` ` `// change the last digit to 'b'` ` ` `if` `(freq == N)` ` ` `ans = ans.replace(ans.charAt(freq - ` `1` `), ` `'b'` `);` ` ` ` ` `// add another 'a' to the string` ` ` `else` ` ` `ans += ` `'a'` `;` ` ` ` ` `// return the answer string` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `String S = ` `"abcdefghijklmnopqrstuvwxyz"` `;` ` ` `int` `N = S.length();` ` ` `// Function call` ` ` `System.out.println(smallestNonSubsequence(S, N));` ` ` `}` `}` `// This code is contributed by SoumikMondal` |

## Python3

`# Python3 program for the above approach` `# Function to find lexicographically smallest` `# string that is not a subsequence of S` `def` `smallestNonSubsequence(S, N):` ` ` ` ` `# Variable to store frequency of 'a'` ` ` `freq ` `=` `0` ` ` ` ` `# Calculate frequency of 'a'` ` ` `for` `i ` `in` `range` `(N):` ` ` `if` `(S[i] ` `=` `=` `'a'` `):` ` ` `freq ` `+` `=` `1` ` ` ` ` `# Initialize string consisting` ` ` `# of freq number of 'a'` ` ` `ans ` `=` `""` ` ` `for` `i ` `in` `range` `(freq):` ` ` `ans ` `+` `=` `'a'` ` ` `# Change the last digit to 'b'` ` ` `if` `(freq ` `=` `=` `N):` ` ` `ans ` `=` `ans.replace(ans[freq ` `-` `1` `], ` `'b'` `)` ` ` ` ` `# Add another 'a' to the string` ` ` `else` `:` ` ` `ans ` `+` `=` `'a'` ` ` ` ` `# Return the answer string` ` ` `return` `ans` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Input` ` ` `S ` `=` `"abcdefghijklmnopqrstuvwxyz"` ` ` `N ` `=` `len` `(S)` ` ` `# Function call` ` ` `print` `(smallestNonSubsequence(S, N))` ` ` `# This code is contributed by SURENDRA_GANGWAR` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to find lexicographically smallest string that` `// is not a subsequence of S` `function` `smallestNonSubsequence(S,N)` `{` ` ` `// variable to store frequency of 'a'` ` ` `let freq = 0;` ` ` ` ` `// calculate frequency of 'a'` ` ` `for` `(let i = 0; i < N; i++)` ` ` `if` `(S[i] == ` `'a'` `)` ` ` `freq++;` ` ` ` ` `// Initialize string consisting of freq number of 'a'` ` ` `let ans = ` `new` `Array(freq).fill(` `'a'` `).join(` `''` `);` ` ` `// change the last digit to 'b'` ` ` `if` `(freq == N)` ` ` `ans[freq - 1] = ` `'b'` `;` ` ` ` ` `// add another 'a' to the string` ` ` `else` ` ` `ans += ` `'a'` `;` ` ` ` ` `// return the answer string` ` ` `return` `ans;` `}` `// Driver code` `// Input` `let S = ` `"abcdefghijklmnopqrstuvwxyz"` `;` `let N = S.length;` `// Function call` `document.write(smallestNonSubsequence(S, N),` `"</br>"` `);` `// This code is contributed by shinjanpatra` `</script>` |

**Output**

aa

**Time Complexity:** O(N)**Auxiliary Space:** O(1)