# Find length of longest substring with at most K normal characters

Given a string P consisting of small English letters and a 26-digit bit string Q, where 1 represent special character and 0 represent normal character for the 26 English alphabets. The task is to find the length of longest substring with at most K normal characters.

Examples:

Input : P = “normal”, Q = “00000000000000000000000000”, K=1
Output : 1
Explanation : In string Q all characters are normal.
Hence we can select any substring of length 1.

Input : P = “giraffe”, Q = “01111001111111111011111111”, K=2
Output : 3
Explanation : Normal characters in P from Q are {a, f, g, r}.
Therefore, possible substrings with at most 2 normal characters are {gir, ira, ffe}.
The maximum length of all substring is 3.

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

Approach:

To solve the problem mentioned above we will be using the concept of two pointers. Hence, maintain left and right pointers of the substring, and a count of normal characters. Increment right index till the count of normal characters is at most K. Then update the answer with a maximum length of substring encountered till now. Increment left index and decrement count till is it greater than K.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to Find ` `// length of longest substring ` `// with at most K normal characters ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find maximum ` `// length of normal substrings ` `int` `maxNormalSubstring(string& P, string& Q, ` `                       ``int` `K, ``int` `N) ` `{ ` ` `  `    ``if` `(K == 0) ` `        ``return` `0; ` ` `  `    ``// keeps count of normal characters ` `    ``int` `count = 0; ` ` `  `    ``// indexes of substring ` `    ``int` `left = 0, right = 0; ` ` `  `    ``// maintain length of longest substring ` `    ``// with at most K normal characters ` `    ``int` `ans = 0; ` ` `  `    ``while` `(right < N) { ` ` `  `        ``while` `(right < N && count <= K) { ` ` `  `            ``// get position of character ` `            ``int` `pos = P[right] - ``'a'``; ` ` `  `            ``// check if current character is normal ` `            ``if` `(Q[pos] == ``'0'``) { ` ` `  `                ``// check if normal characters ` `                ``// count exceeds K ` `                ``if` `(count + 1 > K) ` ` `  `                    ``break``; ` ` `  `                ``else` `                    ``count++; ` `            ``} ` ` `  `            ``right++; ` ` `  `            ``// update answer with substring length ` `            ``if` `(count <= K) ` `                ``ans = max(ans, right - left); ` `        ``} ` ` `  `        ``while` `(left < right) { ` ` `  `            ``// get position of character ` `            ``int` `pos = P[left] - ``'a'``; ` ` `  `            ``left++; ` ` `  `            ``// check if character is ` `            ``// normal then decrement count ` `            ``if` `(Q[pos] == ``'0'``) ` ` `  `                ``count--; ` ` `  `            ``if` `(count < K) ` `                ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// initialise the string ` `    ``string P = ``"giraffe"``, Q = ``"01111001111111111011111111"``; ` ` `  `    ``int` `K = 2; ` ` `  `    ``int` `N = P.length(); ` ` `  `    ``cout << maxNormalSubstring(P, Q, K, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to Find ` `// length of longest subString ` `// with at most K normal characters ` `class` `GFG{ ` `  `  `// Function to find maximum ` `// length of normal subStrings ` `static` `int` `maxNormalSubString(``char` `[]P, ``char` `[]Q, ` `                       ``int` `K, ``int` `N) ` `{ ` `  `  `    ``if` `(K == ``0``) ` `        ``return` `0``; ` `  `  `    ``// keeps count of normal characters ` `    ``int` `count = ``0``; ` `  `  `    ``// indexes of subString ` `    ``int` `left = ``0``, right = ``0``; ` `  `  `    ``// maintain length of longest subString ` `    ``// with at most K normal characters ` `    ``int` `ans = ``0``; ` `  `  `    ``while` `(right < N) { ` `  `  `        ``while` `(right < N && count <= K) { ` `  `  `            ``// get position of character ` `            ``int` `pos = P[right] - ``'a'``; ` `  `  `            ``// check if current character is normal ` `            ``if` `(Q[pos] == ``'0'``) { ` `  `  `                ``// check if normal characters ` `                ``// count exceeds K ` `                ``if` `(count + ``1` `> K) ` `  `  `                    ``break``; ` `  `  `                ``else` `                    ``count++; ` `            ``} ` `  `  `            ``right++; ` `  `  `            ``// update answer with subString length ` `            ``if` `(count <= K) ` `                ``ans = Math.max(ans, right - left); ` `        ``} ` `  `  `        ``while` `(left < right) { ` `  `  `            ``// get position of character ` `            ``int` `pos = P[left] - ``'a'``; ` `  `  `            ``left++; ` `  `  `            ``// check if character is ` `            ``// normal then decrement count ` `            ``if` `(Q[pos] == ``'0'``) ` `  `  `                ``count--; ` `  `  `            ``if` `(count < K) ` `                ``break``; ` `        ``} ` `    ``} ` `  `  `    ``return` `ans; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// initialise the String ` `    ``String P = ``"giraffe"``, Q = ``"01111001111111111011111111"``; ` `  `  `    ``int` `K = ``2``; ` `  `  `    ``int` `N = P.length(); ` `  `  `    ``System.out.print(maxNormalSubString(P.toCharArray(), Q.toCharArray(), K, N)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `   `  `# Function to find maximum  ` `# length of normal substrings  ` `def` `maxNormalSubstring(P, Q, K, N):  ` `     `  `    ``if` `(K ``=``=` `0``): ` `        ``return` `0`  `   `  `    ``# keeps count of normal characters  ` `    ``count ``=` `0`  `   `  `    ``# indexes of substring  ` `    ``left, right ``=` `0``, ``0` `     `  `    ``# maintain length of longest substring  ` `    ``# with at most K normal characters  ` `    ``ans ``=` `0` `   `  `    ``while` `(right < N): ` `   `  `        ``while` `(right < N ``and` `count <``=` `K): ` `   `  `            ``# get position of character  ` `            ``pos ``=` `ord``(P[right]) ``-` `ord``(``'a'``)  ` `   `  `            ``# check if current character is normal  ` `            ``if` `(Q[pos] ``=``=` `'0'``): ` `   `  `                ``# check if normal characters  ` `                ``# count exceeds K  ` `                ``if` `(count ``+` `1` `> K): ` `                    ``break` `                ``else``: ` `                    ``count ``+``=` `1`  `   `  `            ``right ``+``=` `1`  `   `  `            ``# update answer with substring length  ` `            ``if` `(count <``=` `K): ` `                ``ans ``=` `max``(ans, right ``-` `left) ` `   `  `        ``while` `(left < right):  ` `   `  `            ``# get position of character  ` `            ``pos ``=` `ord``(P[left]) ``-` `ord``(``'a'``) ` `   `  `            ``left ``+``=` `1` `   `  `            ``# check if character is  ` `            ``# normal then decrement count  ` `            ``if` `(Q[pos] ``=``=` `'0'``): ` `                ``count ``-``=` `1`  `   `  `            ``if` `(count < K): ` `                ``break`  `   `  `    ``return` `ans ` `   `  `# Driver code  ` `if``(__name__ ``=``=` `"__main__"``): ` `    ``# initialise the string  ` `    ``P ``=` `"giraffe"` `    ``Q ``=` `"01111001111111111011111111"` `   `  `    ``K ``=` `2` `   `  `    ``N ``=` `len``(P)  ` `   `  `    ``print``(maxNormalSubstring(P, Q, K, N))  ` ` `  `# This code is contributed by skylags `

## C#

 `// C# implementation to Find ` `// length of longest subString ` `// with at most K normal characters ` `using` `System; ` ` `  `public` `class` `GFG{ ` ` `  `// Function to find maximum ` `// length of normal subStrings ` `static` `int` `maxNormalSubString(``char` `[]P, ``char` `[]Q, ` `                    ``int` `K, ``int` `N) ` `{ ` ` `  `    ``if` `(K == 0) ` `        ``return` `0; ` ` `  `    ``// keeps count of normal characters ` `    ``int` `count = 0; ` ` `  `    ``// indexes of subString ` `    ``int` `left = 0, right = 0; ` ` `  `    ``// maintain length of longest subString ` `    ``// with at most K normal characters ` `    ``int` `ans = 0; ` ` `  `    ``while` `(right < N) { ` ` `  `        ``while` `(right < N && count <= K) { ` ` `  `            ``// get position of character ` `            ``int` `pos = P[right] - ``'a'``; ` ` `  `            ``// check if current character is normal ` `            ``if` `(Q[pos] == ``'0'``) { ` ` `  `                ``// check if normal characters ` `                ``// count exceeds K ` `                ``if` `(count + 1 > K) ` ` `  `                    ``break``; ` ` `  `                ``else` `                    ``count++; ` `            ``} ` ` `  `            ``right++; ` ` `  `            ``// update answer with subString length ` `            ``if` `(count <= K) ` `                ``ans = Math.Max(ans, right - left); ` `        ``} ` ` `  `        ``while` `(left < right) { ` ` `  `            ``// get position of character ` `            ``int` `pos = P[left] - ``'a'``; ` ` `  `            ``left++; ` ` `  `            ``// check if character is ` `            ``// normal then decrement count ` `            ``if` `(Q[pos] == ``'0'``) ` ` `  `                ``count--; ` ` `  `            ``if` `(count < K) ` `                ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// initialise the String ` `    ``String P = ``"giraffe"``, Q = ``"01111001111111111011111111"``; ` ` `  `    ``int` `K = 2; ` ` `  `    ``int` `N = P.Length; ` ` `  `    ``Console.Write(maxNormalSubString(P.ToCharArray(), ` `                     ``Q.ToCharArray(), K, N)); ` `} ` `} ` ` `  `// This code contributed by Princi Singh `

Output:

```3
```

Time Complexity: The above method takes O(N) time.

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Improved By : skylags, princi singh