# Longest sub-string having frequency of each character less than equal to k

Given a string str of length n. The problem is to find the length of the longest sub-string in str having frequency of each character less than equal to the given value k.

Examples :

```Input : str = "babcaag", k = 1
Output : 3
abc and bca are the two longest
sub-strings having frequency of each character
in them less than equal to '1'.

Input : str = "geeksforgeeks", k = 2
Output : 10
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach: Create an array freq[] of size 26 implemented as hash table to store the frequency of each character of str. Initialize all of its indexes with value ‘0’. Length of the string is n. Now implement the following algorithm.

```longSubstring(str, k)
Initialize start = 0
Initialize maxLen = 0
Declare ch

for i = 0 to n-1
ch = str[i]
freq[ch - 'a']++
if k < freq[ch - 'a'] then
if maxLen < (i - start) then
maxLen = i - start
while (k < freq[ch - 'a'])
freq[str[start] - 'a']--
start++

if maxLen < (n - start) then
maxLen = n - start

return maxLen
```

## C++

 `// C++ implementation to find ` `// the length of the longest  ` `// substring having frequency ` `// of each character less  ` `// than equal to k ` `#include ` `using` `namespace` `std; ` ` `  `#define SIZE 26 ` ` `  `// function to find the length ` `// of the longest substring  ` `// having frequency of each  ` `// character less than equal  ` `// to k ` `int` `longSubstring(string str, ``int` `k) ` `{ ` `    ``// hash table to store frequency ` `    ``// of each table ` `    ``int` `freq[SIZE]; ` ` `  `    ``// Initialize ` `    ``memset``(freq, 0, ``sizeof``(freq)); ` ` `  `    ``// 'start' index of the current ` `    ``// substring ` `    ``int` `start = 0; ` ` `  `    ``// to store the maximum length ` `    ``int` `maxLen = 0; ` `    ``char` `ch; ` ` `  `    ``int` `n = str.size(); ` ` `  `    ``// traverse the string 'str' ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``// get the current character ` `        ``// as 'ch' ` `        ``ch = str[i]; ` ` `  `        ``// increase frequency of  ` `        ``// 'ch' in 'freq[]' ` `        ``freq[ch - ``'a'``]++; ` ` `  `        ``// if frequency of 'ch' becomes ` `        ``// more than 'k' ` `        ``if` `(freq[ch - ``'a'``] > k)  ` `        ``{ ` `            ``// update 'maxLen' ` `            ``if` `(maxLen < (i - start)) ` `                ``maxLen = i - start; ` ` `  `            ``// decrease frequency of  ` `            ``// each character as they  ` `            ``// are encountered from  ` `            ``// the 'start' index until  ` `            ``// frequency of 'ch' is  ` `            ``// greater than 'k' ` `            ``while` `(freq[ch - ``'a'``] > k)  ` `            ``{ ` ` `  `                ``// decrement frequency  ` `                ``// by '1' ` `                ``freq[str[start] - ``'a'``]--; ` ` `  `                ``// increment 'start' ` `                ``start++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// update maxLen ` `    ``if` `(maxLen < (n - start)) ` `        ``maxLen = n - start; ` ` `  `    ``// required length ` `    ``return` `maxLen; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``string str = ``"babcaag"``; ` `    ``int` `k = 1; ` ` `  `    ``cout << ``"Length = "` `        ``<< longSubstring(str, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find ` `// the length of the longest  ` `// substring having frequency ` `// of each character less  ` `// than equal to k ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `public` `class` `GfG{ ` `     `  `    ``public` `final` `static` `int` `SIZE = ``26``; ` `     `  `    ``// function to find the length ` `    ``// of the longest substring  ` `    ``// having frequency of each  ` `    ``// character less than equal  ` `    ``// to k ` `    ``public` `static` `int` `longSubstring(String str1,  ` `                                           ``int` `k) ` `    ``{ ` `        ``// hash table to store frequency ` `        ``// of each table ` `        ``int``[] freq = ``new` `int` `[SIZE]; ` ` `  `        ``char``[] str = str1.toCharArray(); ` ` `  `        ``// 'start' index of the current ` `        ``// substring ` `        ``int` `start = ``0``; ` ` `  `        ``// to store the maximum length ` `        ``int` `maxLen = ``0``; ` `        ``char` `ch; ` ` `  `        ``int` `n = str1.length(); ` ` `  `        ``// traverse the string 'str' ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``// get the current character ` `            ``// as 'ch' ` `            ``ch = str[i]; ` ` `  `            ``// increase frequency of  ` `            ``// 'ch' in 'freq[]' ` `            ``freq[ch - ``'a'``]++; ` ` `  `            ``// if frequency of 'ch'  ` `            ``// becomes more than 'k' ` `            ``if` `(freq[ch - ``'a'``] > k)  ` `            ``{ ` `                ``// update 'maxLen' ` `                ``if` `(maxLen < (i - start)) ` `                    ``maxLen = i - start; ` ` `  `                ``// decrease frequency of  ` `                ``// each character as they  ` `                ``// are encountered from  ` `                ``// the 'start' index until  ` `                ``// frequency of 'ch' is  ` `                ``// greater than 'k' ` `                ``while` `(freq[ch - ``'a'``] > k)  ` `                ``{ ` ` `  `                    ``// decrement frequency ` `                    ``// by '1' ` `                    ``freq[str[start] - ``'a'``]--; ` ` `  `                    ``// increment 'start' ` `                    ``start++; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// update maxLen ` `        ``if` `(maxLen < (n - start)) ` `            ``maxLen = n - start; ` ` `  `        ``// required length ` `        ``return` `maxLen; ` `    ``} ` `     `  `    ``// Driver function  ` `    ``public` `static` `void` `main(String argc[]) ` `    ``{ ` `        ``String str = ``"babcaag"``; ` `        ``int` `k = ``1``; ` `         `  `        ``System.out.println(``"Length = "` `+ ` `        ``longSubstring(str, k)); ` `    ``} ` `} ` ` `  `/* This code is contributed by Sagar Shukla */`

## Python3

 `# Python3 implementation to find ` `# the length of the longest  ` `# substring having frequency ` `# of each character less than  ` `# equal to k ` ` `  `# import library ` `import` `numpy as np ` ` `  `SIZE ``=` `26` ` `  `# Function to find the length ` `# of the longest sub having ` `# frequency of each character ` `# less than equal to k ` `def` `longSub(``str``, k): ` `     `  `    ``# Hash table to store frequency ` `    ``# of each table ` `    ``freq ``=` `np.zeros(``26``, dtype ``=` `np.``int` `) ` ` `  `    ``# 'start' index of the  ` `    ``# current substring ` `    ``start ``=` `0` ` `  `    ``# To store the maximum length ` `    ``maxLen ``=` `0` `     `  `    ``n ``=` `len``(``str``) ` ` `  `    ``# Traverse the 'str' ` `    ``for` `i ``in` `range``(``0``, n): ` `         `  `        ``# Get the current character ` `        ``# as 'ch' ` `        ``ch ``=` `str``[i] ` ` `  `        ``# Increase frequency of  ` `        ``# 'ch' in 'freq[]' ` `        ``freq[``ord``(ch) ``-` `ord``(``'a'``) ] ``+``=` `1` ` `  `        ``# If frequency of 'ch'  ` `        ``# becomes more than 'k' ` `        ``if` `(freq[``ord``(ch) ``-` `ord``(``'a'``)] > k): ` `            ``# update 'maxLen' ` `            ``if` `(maxLen < (i ``-` `start)): ` `                ``maxLen ``=` `i ``-` `start ` ` `  `            ``# decrease frequency of  ` `            ``# each character as they  ` `            ``# are encountered from  ` `            ``# the 'start' index until  ` `            ``# frequency of 'ch' is  ` `            ``# greater than 'k' ` `            ``while` `(freq[``ord``(ch) ``-` `ord``(``'a'``)] > k): ` `                 `  `                ``# decrement frequency  ` `                ``# by '1' ` `                ``freq[``ord``(``str``[start]) ``-` `ord``(``'a'``)] ``-``=` `1` ` `  `                ``# increment 'start' ` `                ``start ``=` `start ``+` `1` ` `  `    ``# Update maxLen ` `    ``if` `(maxLen < (n ``-` `start)): ` `        ``maxLen ``=` `n ``-` `start ` ` `  `    ``# required length ` `    ``return` `maxLen; ` ` `  ` `  `# Driver Code ` `str` `=` `"babcaag"` `k ``=` `1` ` `  `print` `(``"Length ="``, longSub(``str``, k)) ` `  `  `# This code is contributed by 'saloni1297' `

## C#

 `// C# implementation to find ` `// the length of the longest  ` `// substring having frequency ` `// of each character less  ` `// than equal to k ` `using` `System; ` ` `  `class` `GfG{ ` `     `  `    ``public` `static` `int` `SIZE = 26; ` `     `  `    ``// function to find the length ` `    ``// of the longest substring  ` `    ``// having frequency of each  ` `    ``// character less than equal  ` `    ``// to k ` `    ``public` `static` `int` `longSubstring(String str1,  ` `                                          ``int` `k) ` `    ``{ ` `        ``// hash table to store ` `        ``// frequency of each table ` `        ``int` `[]freq = ``new` `int` `[SIZE]; ` ` `  `        ``char` `[]str = str1.ToCharArray(); ` ` `  `        ``// 'start' index of the ` `        ``// current substring ` `        ``int` `start = 0; ` ` `  `        ``// to store the maximum length ` `        ``int` `maxLen = 0; ` `        ``char` `ch; ` ` `  `        ``int` `n = str1.Length; ` ` `  `        ``// traverse the string 'str' ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``// get the current character ` `            ``// as 'ch' ` `            ``ch = str[i]; ` ` `  `            ``// increase frequency of  ` `            ``// 'ch' in 'freq[]' ` `            ``freq[ch - ``'a'``]++; ` ` `  `            ``// if frequency of 'ch'  ` `            ``// becomes more than 'k' ` `            ``if` `(freq[ch - ``'a'``] > k)  ` `            ``{ ` `                ``// update 'maxLen' ` `                ``if` `(maxLen < (i - start)) ` `                    ``maxLen = i - start; ` ` `  `                ``// decrease frequency of  ` `                ``// each character as they  ` `                ``// are encountered from  ` `                ``// the 'start' index until  ` `                ``// frequency of 'ch' is  ` `                ``// greater than 'k' ` `                ``while` `(freq[ch - ``'a'``] > k)  ` `                ``{ ` `                    ``// decrement frequency ` `                    ``// by '1' ` `                    ``freq[str[start] - ``'a'``]--; ` ` `  `                    ``// increment 'start' ` `                    ``start++; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// update maxLen ` `        ``if` `(maxLen < (n - start)) ` `            ``maxLen = n - start; ` ` `  `        ``// required length ` `        ``return` `maxLen; ` `    ``} ` `     `  `    ``// Driver function  ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``String str = ``"babcaag"``; ` `        ``int` `k = 1; ` `         `  `        ``Console.Write(``"Length = "` `+ ` `                       ``longSubstring(str, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal `

## PHP

 ` ``\$k``)  ` `        ``{ ` `            ``// update 'maxLen' ` `            ``if` `(``\$maxLen` `< (``\$i` `- ``\$start``)) ` `                ``\$maxLen` `= ``\$i` `- ``\$start``; ` ` `  `            ``// decrease frequency of  ` `            ``// each character as they  ` `            ``// are encountered from  ` `            ``// the 'start' index until  ` `            ``// frequency of 'ch' is  ` `            ``// greater than 'k' ` `            ``while` `(``\$freq``[ord(``\$ch``) -  ` `                         ``ord(``'a'``)] > ``\$k``)  ` `            ``{ ` ` `  `                ``// decrement frequency  ` `                ``// by '1' ` `                ``\$freq``[ord(``\$str``[``\$start``]) -  ` `                      ``ord(``'a'``)]--; ` ` `  `                ``// increment 'start' ` `                ``\$start``++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// update maxLen ` `    ``if` `(``\$maxLen` `< (``\$n` `- ``\$start``)) ` `        ``\$maxLen` `= ``\$n` `- ``\$start``; ` ` `  `    ``// required length ` `    ``return` `\$maxLen``; ` `} ` ` `  `// Driver Code ` `\$str` `= ``"babcaag"``; ` `\$k` `= 1; ` ` `  `echo` `(``"Length = "` `. ` `       ``longSubstring(``\$str``, ``\$k``)); ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) ` `?> `

Output:

```Length = 3
```

Time Complexity: O(n).
Auxiliary Space: O(1).
Because of the while loop the complexity might seem quadratic but if we look closely the inner while loop will traverse the string single time only.”

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