# 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 represents the special character and 0 represents a normal character for the 26 English alphabets. The task is to find the length of the longest substring with at most K normal characters.

Examples:

Input : P = “normal”, Q = “00000000000000000000000000”, K=1
Output :
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 :
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.

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 the 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 it is 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`

## Javascript

 ``

Output
```3

```

Time Complexity: O(N*N)
Auxiliary Space: O(1)

#### Another Approach:

Initialize two variables, start and end, to zero, a variable max_len to zero, and an array freq of size 26 to zero.

Traverse the string from left to right:

a. Increment the frequency of the current character in the freq array.

b. If the frequency of the current character exceeds k, move the start pointer to the right and decrement the frequency of the character at the previous start position.

c. Update the max_len if the current substring length is greater than max_len.

Return max_len.

## C++

 `#include ``#include ` `using` `namespace` `std;` `int` `main() {``char` `str[] = ``"aabbccddd"``;``int` `k = 2;``int` `n = ``strlen``(str);``  ``int` `start = 0;``int` `end = 0;``int` `max_len = 0;``int` `freq[26] = {0};` `// Traverse the string from left to right and ``  ``// find the length of the longest substring with at most K normal characters``while` `(end < n) {``    ``freq[str[end] - ``'a'``]++;``    ``if` `(freq[str[end] - ``'a'``] > k) {``        ``freq[str[start] - ``'a'``]--;``        ``start++;``    ``}``    ``max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;``    ``end++;``}` `cout << ``"Length of longest substring with at most "` `<< k << ``" normal characters is: "` `<< max_len << endl;` `return` `0;``}`

## C

 `#include ``#include ` `int` `main() {``    ``char` `str[] = ``"aabbccddd"``;``    ``int` `k = 2;``    ``int` `n = ``strlen``(str);``    ` `    ``int` `start = 0;``    ``int` `end = 0;``    ``int` `max_len = 0;``    ``int` `freq[26] = {0};``    ` `    ``// Traverse the string from left to right and find the length of the longest substring with at most K normal characters``    ``while` `(end < n) {``        ``freq[str[end] - ``'a'``]++;``        ``if` `(freq[str[end] - ``'a'``] > k) {``            ``freq[str[start] - ``'a'``]--;``            ``start++;``        ``}``        ``max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;``        ``end++;``    ``}``    ` `    ``printf``(``"Length of longest substring with at most %d normal characters is: %d\n"``, k, max_len);``    ` `    ``return` `0;``}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `public` `class` `Main {``    ``public` `static` `void` `main(String[] args) {``        ``String str = ``"aabbccddd"``;``        ``int` `k = ``2``;``        ``int` `n = str.length();` `        ``int` `start = ``0``;``        ``int` `end = ``0``;``        ``int` `max_len = ``0``;``        ``int``[] freq = ``new` `int``[``26``];` `        ``// Traverse the string from left to right and``       ``// find the length of the longest substring with``      ``// at most K normal characters``        ``while` `(end < n) {``            ``freq[str.charAt(end) - ``'a'``]++;``            ``if` `(freq[str.charAt(end) - ``'a'``] > k) {``                ``freq[str.charAt(start) - ``'a'``]--;``                ``start++;``            ``}``            ``max_len = (end - start + ``1` `> max_len) ? end - start + ``1` `: max_len;``            ``end++;``        ``}` `        ``System.out.printf(``"Length of longest substring with at most %d normal characters is: %d\n"``, k, max_len);``    ``}``}` `// This code is contributed by rishabmalhdijo`

## Python3

 `# Python implementation of longest substring with at most k normal characters``def` `longest_substring(``str``, k):``    ``n ``=` `len``(``str``)``    ``start ``=` `0``    ``end ``=` `0``    ``max_len ``=` `0``    ``freq ``=` `[``0``] ``*` `26``    ` `    ``# Traverse the string from left to right and find the length of the longest substring with at most K normal characters``    ``while` `end < n:``        ``freq[``ord``(``str``[end]) ``-` `ord``(``'a'``)] ``+``=` `1``        ``if` `freq[``ord``(``str``[end]) ``-` `ord``(``'a'``)] > k:``            ``freq[``ord``(``str``[start]) ``-` `ord``(``'a'``)] ``-``=` `1``            ``start ``+``=` `1``        ``max_len ``=` `max``(max_len, end ``-` `start ``+` `1``)``        ``end ``+``=` `1``    ` `    ``print``(``"Length of longest substring with at most"``, k, ``"normal characters is:"``, max_len)` `# Driver code``str` `=` `"aabbccddd"``k ``=` `2``longest_substring(``str``, k)`

## C#

 `using` `System;` `class` `MainClass {``public` `static` `void` `Main (``string``[] args) {``char``[] str = ``"aabbccddd"``.ToCharArray();``int` `k = 2;``int` `n = str.Length;``int` `start = 0;``int` `end = 0;``int` `max_len = 0;``int``[] freq = ``new` `int``[26];``  ``// Traverse the string from left to right and``// find the length of the longest substring with at most K normal characters``while` `(end < n) {``  ``freq[str[end] - ``'a'``]++;``  ``if` `(freq[str[end] - ``'a'``] > k) {``    ``freq[str[start] - ``'a'``]--;``    ``start++;``  ``}``  ``max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;``  ``end++;``}` `Console.WriteLine(``"Length of longest substring with at most {0} normal characters is: {1}"``, k, max_len);` `}``}`

## Javascript

 `function` `findLongestSubstring(str, k) {``    ``const n = str.length;``    ``let start = 0;``    ``let end = 0;``    ``let max_len = 0;``    ``let freq = ``new` `Array(26).fill(0);` `    ``// Traverse the string from left to right and ``    ``// find the length of the longest substring with at most K ``    ``// normal characters``    ``while` `(end < n) {``        ``freq[str.charCodeAt(end) - ``'a'``.charCodeAt()]++;``        ``if` `(freq[str.charCodeAt(end) - ``'a'``.charCodeAt()] > k) {``            ``freq[str.charCodeAt(start) - ``'a'``.charCodeAt()]--;``            ``start++;``        ``}``        ``max_len = Math.max(end - start + 1, max_len);``        ``end++;``    ``}` `    ``return` `max_len;``}` `// Driver program``const str = ``"aabbccddd"``;``const k = 2;``const longestSubstringLength = findLongestSubstring(str, k);` `console.log(`Length of longest substring ``with` `at most \${k} normal characters is: \${longestSubstringLength}`);`

Output
```Length of longest substring with at most 2 normal characters is: 8

```

time complexity of O(n), where n is the string length

space complexity of O(1)

Previous
Next