Longest Sub-string with at most K Distinct Characters

PROBLEM STATEMENT

You are given string S of length N, and an integer K. Your task is to find the length of the longest substring that contains at most K distinct characters.

Input Format:

The first line contains an Integer 'T' which denotes the number of test cases/queries to be run. 
Then the test cases follow. 

The first line of input for each test case/query contains an integer K.

The second line of input for each test case/query contains a string S.

Output Format:

For each test case, print the length of the longest substring that contains at most K distinct characters.

Output for every test case will be printed in a separate line.

Note:

You do not need to print anything, it has already been taken care of. Just implement the function.

Constraints:

1 <= T <= 10
1 <= K <= 26
1 <= N <= 10^4

Time Limit: 1sec

Approach 1

We will generate all the substrings using 2 nested for loops and we will have a ‘CHECK’ function which returns true if the number of distinct character in the substring is less than equal to K otherwise false.
We will have an ans variable initialize to 0. We will call the ‘CHECK’ function with every substring and if it returns true then
- ANS = MAX(ANS , CURRENT_SUBSTRING.SIZE())
To implement the check function we will have K and substring S as inputs and we will create an array of size 26 let’s call this ‘FREQ’ and initialize all position with 0.
- Increase the frequency of every character in the substring S.
  - FREQ[S[i]-’a’]++
- Have a variable ‘DISTINCT’ = 0
  - Then run a loop from i = 0 to i = 25 and increase the ‘DISTINCT’ if FREQ[i] is non-zero.
- Return true if DISTINCT <= K otherwise return false
Finally, return ANS which have our required answer.

Try Problem

Approach 2

We will run a loop from i = 0 to i = N - 1 where i will be starting position of the substring and j = i to j = N - 1 where j will be the end position of the substring. Before starting the second loop we will create a hash table or an array of size 26 let’s call this ‘FREQ’ array.
Inside inner loop increase the frequency of the j’th character. Now freq array will have the frequency of the substring starting from the index i and ending at the index j. Now we will check the number of distinct characters which will be simply the number of the non-zero valued position in the frequency array.
Now initialize a variable ANS = 0 and check if the number of distinct characters is less than equal to K then
- ANS = MAX(ANS, j - i + 1)
Now return the ans which has the required answer.

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Approach 3

We will be doing the binary search on the answer that is the length of the substring where LOW = 1 and HIGH = N will be the range of our answer and we will have a variable ANS = 0 to store the required answer.
Now, as usual, we will write a while loop till LOW <= HIGH and we will find the mid
- For the mid length, we will try to find if there is a substring of length equal to MID which has the distinct characters less than equal to K.
- To check the number of distinct characters in a substring of length equal to mid in O(N) we will use the sliding window with a hash table or an array of size 26 and initialize all the position with 0.
- We will run a loop from i = 0 to i = MID - 1 which has a length equal to mid and increase the count of the characters in the array let’s say FREQ i.e FREQ[S[i]-’a’]++.
- Now the number of distinct characters will be equal to the number of non-zero value in the freq array. We will have a variable let’s say possible = false we will make it to true of the number of distinct characters is less than equal to K for this first substring of length equal to MID.
- Now run a loop from i = MID to i = N - mid and increase the count of i’th character i.e
  - FREQ[S[i]-’a’]++
- And decrease the count of i-mid character i.e
  - FREQ[S[i-MID]-’a’]--
- It will slide our window by 1 length towards the right side so will need to decrease the count of the character which is removed from the window which i - MID character and we need to increase the count of the new character that is at index i.
- Now, will do the same procedure to count the distinct characters and update the variable possible.
- If the possible is true then we can try for bigger length substring so
  - LOW = MID + 1
  - ANS = MAX(ANS, MID)
- Otherwise, we need to decrease the substring size so
  - HIGH = MID - 1
Now after the binary search we will have our required answer in the variable ans so return it.

Try Problem

Approach 4

Create a hash table or a frequency array lets name it ‘FREQ’ and initialize all the position with 0.
We will run a loop from i = 0 to i = N - 1 which will be the right boundary of the sliding windows and we will have a variable start = 0 which will be the left boundary of the sliding window.
- Increase the frequency of i’th character by 1 i.e
  - FREQ[S[i]-’a’]++
- Now our window has expanded by 1 length but we need to check if the substring has less than equal to K distinct characters.
- So we will run a while loop and check if FREQ array has greater than K distinct character then
  - FREQ[S[START]-’a’]--
- Now once we come out of the while loop current window is bounded by start and i have less than equal to K distinct characters.
- We will maintain a variable ‘ANS’ which will be our required answer now since this substring is valid we can take a maximum of ans and current substring length.
  - ANS = MAX(ANS, i - START +1)
Now we have our required answer in ‘ANS’ so simply return it.

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