# Longest substring with atmost K characters from the given set of characters

Given a string S, an integer K and set of characters Q[], the task is to find the longest substring in string S which contains atmost K characters from the given character set Q[].

Examples:

Input: S = “normal”, Q = {“a”, “o”, “n”, “b”, “r”, “l”}, K = 1
Output: 1
Explanation:
All the characters in the given string S are present in array.
Therefore, we can select any substring of length 1.

Input: S = “giraffe”, Q = {“a”, “f”, “g”, “r”}, K = 2
Output : 3
Explanation:
Possible substrings with atmost 2 characters
From the given set are {“gir”, “ira”, “ffe”}
The maximum length of all the substrings is 3.

Approach: The idea is to use the concept of two pointers to consider the substrings of maximum length, such that it contains at most K character from the given set. Below is the illustration of the approach:

• Maintain two pointers left and right as 0, to consider the string in between these pointers.
• Increment the right pointer untill the characters from the given set is at most K.
• Update the longest length substring to be difference between the right pointer and left pointer.
```cur_max = max(cur_max, right - left)
```
• Increment the left pointer and if the character which moved out from the two pointers is the character of the given set, then decrement the count of the characters from the set by 1.
• Similarly, repeat the steps above until the right pointer is not equal to the length of the string.

Below is the implmentation of the above approach:

## C++

 `// C++ implementation to find the ` `// longest substring in the string  ` `// which contains atmost K characters ` `// from the given set of characters ` ` `  `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to find the longest ` `// substring in the string ` `// which contains atmost K characters ` `// from the given set of characters ` `int` `maxNormalSubstring(string& P,  ` `         ``set<``char``> Q, ``int` `K, ``int` `N) ` `{ ` ` `  `    ``// Base Condition ` `    ``if` `(K == 0) ` `        ``return` `0; ` ` `  `    ``// Count for Characters  ` `    ``// from set in substring ` `    ``int` `count = 0; ` `     `  `    ``// Two pointers ` `    ``int` `left = 0, right = 0; ` `    ``int` `ans = 0; ` ` `  `    ``// Loop to iterate until  ` `    ``// right pointer is not  ` `    ``// equal to N ` `    ``while` `(right < N) { ` `         `  `        ``// Loop to increase the substring ` `        ``// length until the characters from ` `        ``// set are at most K ` `        ``while` `(right < N && count <= K) { ` ` `  `            ``// Check if current pointer ` `            ``// points a character from set ` `            ``if` `(Q.find(P[right]) != Q.end()){ ` `                 `  `                ``// If the count of the  ` `                ``// char is exceeding the limit ` `                ``if` `(count + 1 > K) ` `                    ``break``; ` `                ``else` `                    ``count++; ` `            ``} ` ` `  `            ``right++; ` ` `  `            ``// update answer with  ` `            ``// substring length ` `            ``if` `(count <= K) ` `                ``ans = max(ans, right - left); ` `        ``} ` `         `  `        ``// Increment the left pointer until ` `        ``// the count is less than or equal to K ` `        ``while` `(left < right) { ` `            ``left++; ` `             `  `            ``// If the charcter which comes out ` `            ``// then decrment the count by 1 ` `            ``if` `(Q.find(P[left]) != Q.end()) ` `                ``count--; ` ` `  `            ``if` `(count < K) ` `                ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string P = ``"giraffe"``; ` `    ``set<``char``> Q; ` `     `  `    ``// Construction of set ` `    ``Q.insert(``'a'``); ` `    ``Q.insert(``'f'``); ` `    ``Q.insert(``'g'``); ` `    ``Q.insert(``'r'``); ` `    ``int` `K = 2; ` `    ``int` `N = P.length(); ` ` `  `    ``// output result ` `    ``cout << maxNormalSubstring(P, Q, K, N); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to find the  ` `# longest substring in the string  ` `# which contains atmost K characters  ` `# from the given set of characters  ` ` `  `# Function to find the longest  ` `# substring in the string  ` `# which contains atmost K characters  ` `# from the given set of characters  ` `def` `maxNormalSubstring(P, Q, K, N): ` ` `  `    ``# Base Condition  ` `    ``if` `(K ``=``=` `0``): ` `        ``return` `0` ` `  `    ``# Count for Characters  ` `    ``# from set in substring  ` `    ``count ``=` `0` `     `  `    ``# Two pointers  ` `    ``left ``=` `0` `    ``right ``=` `0` `    ``ans ``=` `0` ` `  `    ``# Loop to iterate until  ` `    ``# right pointer is not  ` `    ``# equal to N  ` `    ``while` `(right < N): ` `         `  `        ``# Loop to increase the substring  ` `        ``# length until the characters from  ` `        ``# set are at most K  ` `        ``while` `(right < N ``and` `count <``=` `K):  ` ` `  `            ``# Check if current pointer  ` `            ``# points a character from set  ` `            ``if` `(P[right] ``in` `Q): ` `                 `  `                ``# If the count of the  ` `                ``# char is exceeding the limit  ` `                ``if` `(count ``+` `1` `> K): ` `                    ``break` `                ``else``: ` `                    ``count ``+``=` `1` ` `  `            ``right ``+``=` `1` ` `  `            ``# update answer with  ` `            ``# substring length  ` `            ``if` `(count <``=` `K): ` `                ``ans ``=` `max``(ans, right ``-` `left)  ` `         `  `        ``# Increment the left pointer until  ` `        ``# the count is less than or equal to K  ` `        ``while` `(left < right): ` `            ``left ``+``=` `1` `             `  `            ``# If the charcter which comes out  ` `            ``# then decrment the count by 1  ` `            ``if` `(P[left] ``in` `Q): ` `                ``count ``-``=` `1` ` `  `            ``if` `(count < K): ` `                ``break` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `P ``=` `"giraffe"` `Q ``=` `{``chr``} ` ` `  `# Construction of set  ` `Q.add(``'a'``) ` `Q.add(``'f'``) ` `Q.add(``'g'``) ` `Q.add(``'r'``) ` `K ``=` `2` `N ``=` `len``(P) ` ` `  `# Output result  ` `print``(maxNormalSubstring(P, Q, K, N)) ` ` `  `# This code is contributed by Sanjit_Prasad `

Output:

```3
```

Performance Analysis:

• Time Complexity: O(N)
• Auxillary Space: O(1)

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