# Minimum K such that every substring of length atleast K contains a character c

Given a string S containing lowercase latin letters. A character c is called K-amazing if every substring of S with length atleast K contains this character c. Find the minimum possible K such that there exists atleast one K-amazing character.

Examples:

Input : S = “abcde”
Output :
Explanation : Every substring of length atleast 3 contains character ‘c’, i.e.
{“abc”, “bcd”, “cde”, “abcd”, “bcde”, “abcde”}

Input :S = “aaaa”
Output :1

Prerequisites : Binary Search

Naive Solution : A simple approach is to iterate over all possible lengths of substrings i.e. from 1 to N (size of string) and for every current length substrings check whether some character appears in all of those substrings.

Efficient Solution :

The key idea is to perform binary search over the answer K, since if some character c appears in all substrings of length X, it will always appear in all substrings of length (X + 1). Hence, we can check for the current length and try to minimise it using divide and conquer algorithm. For checking if some character appears in all substrings of length X, iterate over all characters from ‘a’ to ‘z’ and inside another loop iteratively store the last occurrence of the last character.

Let the current position be j, so the last substring of length X will be from (j – X) to X. Check if the position of last occurrence of current K-amazing character is greater than (j – X) or not. If it is greater, then that substring is a valid string.

Below is the implementation of the above approach.

## C++

 `// CPP Program to find minimum K such that` `// every substring of length atleast K` `// contains some character c` `#include ` `using` `namespace` `std;`   `// This function checks if there exists some` `// character which appears in all K length` `// substrings` `int` `check(string s, ``int` `K)` `{` `    ``// Iterate over all possible characters` `    ``for` `(``int` `ch = 0; ch < 26; ch++) {` `        ``char` `c = ``'a'` `+ ch;`   `        ``// stores the last occurrence` `        ``int` `last = -1;`   `        ``// set answer as true;` `        ``bool` `found = ``true``;` `        ``for` `(``int` `i = 0; i < K; i++)` `            ``if` `(s[i] == c)` `                ``last = i;`   `        ``// No occurrence found of current` `        ``// character in first substring` `        ``// of length K` `        ``if` `(last == -1)` `            ``continue``;`   `        ``// Check for every last substring` `        ``// of length K where last occurr-` `        ``// ence exists in substring` `        ``for` `(``int` `i = K; i < s.size(); i++) {` `            ``if` `(s[i] == c)` `                ``last = i;`   `            ``// If last occ is not` `            ``// present in substring` `            ``if` `(last <= (i - K)) {` `                ``found = ``false``;` `                ``break``;` `            ``}` `        ``}` `        ``// current character is K amazing` `        ``if` `(found)` `            ``return` `1;` `    ``}` `    ``return` `0;` `}`   `// This function performs binary search over the` `// answer to minimise it` `int` `binarySearch(string s)` `{` `    ``int` `low = 1, high = (``int``)s.size();` `    ``int` `ans;` `    ``while` `(low <= high) {` `        ``int` `mid = (high + low) >> 1;`   `        ``// Check if answer is found try` `        ``// to minimise it` `        ``if` `(check(s, mid)) {` `            ``ans = mid;` `            ``high = mid - 1;` `        ``}` `        ``else` `            ``low = mid + 1;` `    ``}` `    ``return` `ans;` `}`   `// Driver Code to test above functions` `int32_t main()` `{` `    ``string s = ``"abcde"``;` `    ``cout << binarySearch(s) << endl;`   `    ``s = ``"aaaa"``;` `    ``cout << binarySearch(s) << endl;` `    ``return` `0;` `}`

## Java

 `// Java Program to find minimum K such that` `// every substring of length atleast K` `// contains some character c`   `class` `GFG` `{` `    `  `        `  `        ``// This function checks if there exists some` `        ``// character which appears in all K length` `        ``// substrings` `        ``static` `int` `check(String s, ``int` `K)` `        ``{` `            ``// Iterate over all possible characters` `            ``for` `(``int` `ch = ``0``; ch < ``26``; ch++) {` `                ``char` `c = (``char``)( ``'a'` `+ ch);` `        `  `                ``// stores the last occurrence` `                ``int` `last = -``1``;` `        `  `                ``// set answer as true;` `                ``boolean` `found = ``true``;` `                ``for` `(``int` `i = ``0``; i < K; i++)` `                    ``if` `(s.charAt(i) == c)` `                        ``last = i;` `        `  `                ``// No occurrence found of current` `                ``// character in first substring` `                ``// of length K` `                ``if` `(last == -``1``)` `                    ``continue``;` `        `  `                ``// Check for every last substring` `                ``// of length K where last occurr-` `                ``// ence exists in substring` `                ``for` `(``int` `i = K; i < s.length(); i++) {` `                    ``if` `(s.charAt(i) == c)` `                        ``last = i;` `        `  `                    ``// If last occ is not` `                    ``// present in substring` `                    ``if` `(last <= (i - K)) {` `                        ``found = ``false``;` `                        ``break``;` `                    ``}` `                ``}` `                ``// current character is K amazing` `                ``if` `(found)` `                    ``return` `1``;` `            ``}` `            ``return` `0``;` `        ``}` `        `  `        ``// This function performs binary search over the` `        ``// answer to minimise it` `        ``static` `int` `binarySearch(String s)` `        ``{` `            ``int` `low = ``1``, high = s.length();` `            ``int` `ans=``0``;` `            ``while` `(low <= high) {` `                ``int` `mid = (high + low) >> ``1``;` `        `  `                ``// Check if answer is found try` `                ``// to minimise it` `                ``if` `(check(s, mid)==``1``) {` `                    ``ans = mid;` `                    ``high = mid - ``1``;` `                ``}` `                ``else` `                    ``low = mid + ``1``;` `            ``}` `            ``return` `ans;` `        ``}` `        `  `        ``// Driver Code to test above functions` `        ``public` `static` `void` `main(String args[])` `        ``{` `            ``String s = ``"abcde"``;` `            ``System.out.println(binarySearch(s));` `        `  `            ``s = ``"aaaa"``;` `            ``System.out.println(binarySearch(s));` `    `  `        ``}`   `}`   `// This article is contributed ` `// by ihritik`

## Python3

 `# Python3 Program to find minimum K such ` `# that every substring of length atleast ` `# K contains some character c `   `# This function checks if there exists ` `# some character which appears in all ` `# K length substrings ` `def` `check(s, K): `   `    ``# Iterate over all possible characters ` `    ``for` `ch ``in` `range``(``0``, ``26``): ` `        ``c ``=` `chr``(``97` `+` `ch) ``# Ascii value of 'a' => 97`   `        ``# stores the last occurrence ` `        ``last ``=` `-``1`   `        ``# set answer as true ` `        ``found ``=` `True` `        ``for` `i ``in` `range``(``0``, K): ` `            ``if` `s[i] ``=``=` `c: ` `                ``last ``=` `i `   `        ``# No occurrence found of current character ` `        ``# in first substring of length K ` `        ``if` `last ``=``=` `-``1``: ` `            ``continue`   `        ``# Check for every last substring ` `        ``# of length K where last occurr- ` `        ``# ence exists in substring ` `        ``for` `i ``in` `range``(K, ``len``(s)):` `            ``if` `s[i] ``=``=` `c: ` `                ``last ``=` `i `   `            ``# If last occ is not ` `            ``# present in substring ` `            ``if` `last <``=` `(i ``-` `K): ` `                ``found ``=` `False` `                ``break` `            `  `        ``# current character is K amazing ` `        ``if` `found: ` `            ``return` `1` `    `  `    ``return` `0`   `# This function performs binary search ` `# over the answer to minimise it ` `def` `binarySearch(s): `   `    ``low, high, ans ``=` `1``, ``len``(s), ``None` `    `  `    ``while` `low <``=` `high: ` `        ``mid ``=` `(high ``+` `low) >> ``1`   `        ``# Check if answer is found ` `        ``# try to minimise it ` `        ``if` `check(s, mid): ` `            ``ans, high ``=` `mid, mid ``-` `1` `        `  `        ``else``:` `            ``low ``=` `mid ``+` `1` `    `  `    ``return` `ans `   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``: `   `    ``s ``=` `"abcde"` `    ``print``(binarySearch(s)) `   `    ``s ``=` `"aaaa"` `    ``print``(binarySearch(s)) `   `# This code is contributed by Rituraj Jain`

## C#

 `// C# Program to find minimum K such that` `// every substring of length atleast K` `// contains some character c`   `using` `System;` `class` `GFG` `{` `    `  `        `  `        ``// This function checks if there exists some` `        ``// character which appears in all K length` `        ``// substrings` `        ``static` `int` `check(String s, ``int` `K)` `        ``{` `            ``// Iterate over all possible characters` `            ``for` `(``int` `ch = 0; ch < 26; ch++) {` `                ``char` `c = (``char``)( ``'a'` `+ ch);` `        `  `                ``// stores the last occurrence` `                ``int` `last = -1;` `        `  `                ``// set answer as true;` `                ``bool` `found = ``true``;` `                ``for` `(``int` `i = 0; i < K; i++)` `                    ``if` `(s[i] == c)` `                        ``last = i;` `        `  `                ``// No occurrence found of current` `                ``// character in first substring` `                ``// of length K` `                ``if` `(last == -1)` `                    ``continue``;` `        `  `                ``// Check for every last substring` `                ``// of length K where last occurr-` `                ``// ence exists in substring` `                ``for` `(``int` `i = K; i < s.Length; i++) {` `                    ``if` `(s[i] == c)` `                        ``last = i;` `        `  `                    ``// If last occ is not` `                    ``// present in substring` `                    ``if` `(last <= (i - K)) {` `                        ``found = ``false``;` `                        ``break``;` `                    ``}` `                ``}` `                ``// current character is K amazing` `                ``if` `(found)` `                    ``return` `1;` `            ``}` `            ``return` `0;` `        ``}` `        `  `        ``// This function performs binary search over the` `        ``// answer to minimise it` `        ``static` `int` `binarySearch(String s)` `        ``{` `            ``int` `low = 1, high = s.Length;` `            ``int` `ans=0;` `            ``while` `(low <= high) {` `                ``int` `mid = (high + low) >> 1;` `        `  `                ``// Check if answer is found try` `                ``// to minimise it` `                ``if` `(check(s, mid)==1) {` `                    ``ans = mid;` `                    ``high = mid - 1;` `                ``}` `                ``else` `                    ``low = mid + 1;` `            ``}` `            ``return` `ans;` `        ``}` `        `  `        ``// Driver Code to test above functions` `        ``public` `static` `void` `Main()` `        ``{` `            ``String s = ``"abcde"``;` `            ``Console.WriteLine(binarySearch(s));` `        `  `            ``s = ``"aaaa"``;` `            ``Console.WriteLine(binarySearch(s));` `    `  `        ``}`   `}`   `// This article is contributed ` `// by ihritik`

## PHP

 `> 1; `   `        ``// Check if answer is found try ` `        ``// to minimise it ` `        ``if` `(check(``\$s``, ``\$mid``))` `        ``{ ` `            ``\$ans` `= ``\$mid``; ` `            ``\$high` `= ``\$mid` `- 1; ` `        ``} ` `        ``else` `            ``\$low` `= ``\$mid` `+ 1; ` `    ``} ` `    ``return` `\$ans``; ` `} `   `// Driver Code` `\$s` `= ``"abcde"``; ` `echo` `binarySearch(``\$s``) . ``"\n"``;`   `\$s` `= ``"aaaa"``; ` `echo` `binarySearch(``\$s``) . ``"\n"``; `   `// This code is contributed by Ryuga` `?>`

## Javascript

 ``

Output

```3
1
```

Complexity Analysis:

• Time Complexity: O(N * logN * 26), where N is the size of the given string.
• Auxiliary Space: O(1) because constant space is used.

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