# Smallest alphabet greater than a given character

• Difficulty Level : Medium
• Last Updated : 27 May, 2021

Given a list of sorted characters consisting of both Uppercase and Lowercase Alphabets and a particular target value, say K, the task is to find the smallest element in the list that is larger than K.
Letters also wrap around. For example, if K = ‘z’ and letters = [‘A’, ‘r’, ‘z’], then the answer would be ‘A’.

Examples:

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```Input : Letters = ["D", "J", "K"]
K = "B"
Output: 'D'
Explanation:
The Next greater character of "B" is 'D'
since it is the smallest element from the
set of given letters, greater than "B".

Input:  Letters = ["h", "n", "s"]
K = "t"
Output: 'h'```

Prerequisites: Binary Search

Approach: Binary Search can be applied to find the index of the smallest character in the given Set of Letters such that the character at that index is greater than K. If the element at the current mid is smaller than or equal to K, binary search is applied on the Right half, else it is applied on the left half.

## C++

 `/* C++ Program to find the smallest character``from the given set of letter, which is greater``than the target element */``#include ``using` `namespace` `std;` `/* Returns the smallest character from the given``set of letters that is greater than K */` `// In this code we consider only uppercase characters or only lowercase characters``// Incase if we have mixed characters then we``//convert all either into lowercase or uppercase``char` `nextGreatestAlphabet(vector<``char``>& alphabets, ``char` `K)``{``    ``int` `n= alphabets.size();``    ``if``(K>=alphabets[n-1]) ``return` `alphabets[0];``  ` `  ` `    ``int` `l = 0, r = alphabets.size() - 1;` `    ``// Take the first element as l and``    ``// the rightmost element as r``    ``int` `ans = -1;``    ``while` `(l <= r) {` `        ``// if this while condition does not satisfy``        ``// simply return the first element.``        ``int` `mid = (l + r) / 2;``        ``if` `(alphabets[mid] > K)``        ``{``            ``r = mid - 1;``            ``ans = mid;``        ``}``        ``else``            ``l = mid + 1;``    ``}` `    ``// Return the smallest element``    ``return` `alphabets[ans];``}` `// Driver Code``int` `main()``{``    ``vector<``char``> letters{ ``'A'``, ``'K'``, ``'S'` `};``    ``char` `K = ``'L'``;` `    ``// Function call``    ``char` `result = nextGreatestAlphabet(letters, K);``    ``cout << result << endl;``    ``return` `0;``}`

## Java

 `/* Java Program to find the smallest character``from the given set of letter, which is greater``than the target element */` `class` `GFG``{` `    ``/* Returns the smallest character from the given``    ``set of letters that is greater than K */``    ``static` `char` `nextGreatestAlphabet(``char` `alphabets[],``                                     ``char` `K)``    ``{``      ``int` `n = alphabets.length;``      ``if``(K>=alphabets[n-``1``])``        ``return` `alphabets[``0``];` `        ``int` `l = ``0``, r = alphabets.length - ``1``;``     ` `        ``int` `ans = -``1``;``        ``// Take the first element as l and``        ``// the rightmost element as r``        ``while` `(l <= r)``        ``{``            ``// if this while condition does not``            ``// satisfy simply return the first``            ``// element.``            ``int` `mid = (l + r) / ``2``;``            ``if` `(alphabets[mid] > K)``            ``{``                ``r = mid - ``1``;``                ``ans = mid;``            ``}``            ``else``                ``l = mid + ``1``;``        ``}` `        ``// Return the smallest element``        ``return` `alphabets[ans];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``char` `letters[] = { ``'A'``, ``'r'``, ``'z'` `};``        ``char` `K = ``'z'``;``        ``char` `result = nextGreatestAlphabet(letters, K);` `        ``// Function call``        ``System.out.println(result);``    ``}``}` `// This code is contributed by Smitha.`

## Python 3

 `# Python 3 Program to find the smallest``# character from the given set of letter,``# which is greater than the target``# element */` `# Returns the smallest character from``# the given set of letters that is``# greater than K`  `def` `nextGreatestAlphabet(alphabets, K):` `    ``n ``=` `len``(alphabets)``    ``if``(K >``=` `alphabets[n``-``1``]):``       ``return` `alphabets[``0``]``    ``l ``=` `0``    ``r ``=` `len``(alphabets) ``-` `1``    ``ans ``=` `-``1``    ``# Take the first element as l and``    ``# the rightmost element as r``    ``while` `(l <``=` `r):` `        ``# if this while condition does``        ``# not satisfy simply return the``        ``# first element.``        ``mid ``=` `int``((l ``+` `r) ``/` `2``)` `        ``if` `(alphabets[mid] > K):``            ``r ``=` `mid ``-` `1``            ``ans ``=` `mid``        ``else``:``            ``l ``=` `mid ``+` `1` `    ``# Return the smallest element``    ``if` `(alphabets[ans] < K):``        ``return` `alphabets[``0``]``    ``else``:``        ``return` `alphabets[ans]`  `# Driver Code``letters ``=` `[``'A'``, ``'r'``, ``'z'``]``K ``=` `'z'` `# Function call``result ``=` `nextGreatestAlphabet(letters, K)``print``(result)` `# This code is contributed by Smitha`

## C#

 `/* C# Program to find the smallest character``from the given set of letter, which is greater``than the target element */``using` `System;` `class` `GFG {` `    ``/* Returns the smallest character from the given``    ``set of letters that is greater than K */``    ``static` `char` `nextGreatestAlphabet(``char``[] alphabets,``                                     ``char` `K)``    ``{` `        ``int` `n= alphabets.Length;``        ``if``(K >= alphabets[n-1])``          ``return` `alphabets[0];``        ``int` `l = 0, r = alphabets.Length - 1;``        ``int` `ans = -1;``        ``// Take the first element as l and``        ``// the rightmost element as r``        ``while` `(l <= r)``        ``{` `            ``// if this while condition does not``            ``// satisfy simply return the first``            ``// element.``            ``int` `mid = (l + r) / 2;` `            ``if` `(alphabets[mid] > K)``            ``{``                ``ans = mid;``                ``r = mid - 1;``            ``}``            ``else``                ``l = mid + 1;``        ``}` `        ``// Return the smallest element``        ``return` `alphabets[ans];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``char``[] letters = { ``'A'``, ``'r'``, ``'z'` `};``        ``char` `K = ``'z'``;``      ` `        ``// Function call``        ``char` `result = nextGreatestAlphabet(letters, K);` `        ``Console.Write(result);``    ``}``}` `// This code is contributed by Smitha`

## Javascript

 ``
Output
`S`

The Time Complexity of the above approach is, O(log N) where N is the number of characters in the given set of Letters.

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