# Smallest alphabet greater than a given character

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:

```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'
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

## CPP

 `/* 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 */` `char` `nextGreatestAlphabet(vector<``char``>& alphabets,  ` `                                          ``char` `K) ` `{ ` ` `  `    ``int` `l = 0, r = alphabets.size() - 1; ` ` `  `    ``// Take the first element as l and ` `    ``// the rightmost element as r ` `    ``while` `(l < r && alphabets[r] > K) { ` ` `  `        ``// if this while condition does not satisfy ` `        ``// simply return the first element. ` `        ``int` `mid = (l + r) / 2; ` `        ``if` `(alphabets[mid] > K) ` `            ``r = mid; ` `        ``else` `            ``l = mid + 1; ` `    ``} ` ` `  `    ``// Return the smallest element ` `    ``return` `alphabets[l]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``vector<``char``> letters{ ``'A'``, ``'K'``, ``'S'` `}; ` `    ``char` `K = ``'L'``; ` `    ``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` `l = ``0``, r = alphabets.length - ``1``; ` `     `  `        ``// Take the first element as l and ` `        ``// the rightmost element as r ` `        ``while` `(l < r && alphabets[r] > K) ` `        ``{ ` `     `  `            ``// if this while condition does not ` `            ``// satisfy simply return the first ` `            ``// element. ` `            ``int` `mid = (l + r) / ``2``; ` `            ``if` `(alphabets[mid] > K) ` `                ``r = mid; ` `            ``else` `                ``l = mid + ``1``; ` `        ``} ` `     `  `        ``// Return the smallest element ` `        ``return` `alphabets[l]; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String [] args) ` `    ``{ ` `        ``char` `letters[] = { ``'A'``, ``'K'``, ``'S'` `}; ` `        ``char` `K = ``'L'``; ` `        ``char` `result =  ` `              ``nextGreatestAlphabet(letters, K); ` `               `  `        ``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): ` ` `  `    ``l ``=` `0` `    ``r ``=` `len``(alphabets) ``-` `1` ` `  `    ``# Take the first element as l and ` `    ``# the rightmost element as r ` `    ``while` `(l < r ``and` `alphabets[r] > K) : ` ` `  `        ``# if this while condition does ` `        ``# not satisfy simply return the ` `        ``# first element. ` `        ``mid ``=` `int``((l ``+` `r) ``/` `2``) ` `         `  `        ``if` `(alphabets[mid] > K): ` `            ``r ``=` `mid ` `        ``else``: ` `            ``l ``=` `mid ``+` `1` `     `  `    ``# Return the smallest element ` `    ``return` `alphabets[l] ` ` `  `# Driver Code ` `letters ``=` `[``'A'``, ``'K'``, ``'S'``] ` `K ``=` `'L'` `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` `l = 0, r = alphabets.Length - 1; ` `     `  `        ``// Take the first element as l and ` `        ``// the rightmost element as r ` `        ``while` `(l < r && alphabets[r] > K) ` `        ``{ ` `     `  `            ``// if this while condition does not ` `            ``// satisfy simply return the first ` `            ``// element. ` `            ``int` `mid = (l + r) / 2; ` `             `  `            ``if` `(alphabets[mid] > K) ` `                ``r = mid; ` `            ``else` `                ``l = mid + 1; ` `        ``} ` `     `  `        ``// Return the smallest element ` `        ``return` `alphabets[l]; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``char` `[]letters = { ``'A'``, ``'K'``, ``'S'` `}; ` `        ``char` `K = ``'L'``; ` `        ``char` `result = ` `             ``nextGreatestAlphabet(letters, K); ` `              `  `        ``Console.Write(result); ` `    ``} ` `} ` ` `  `// This code is contributed by Smitha `

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.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Smitha Dinesh Semwal

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.