Is Find K-Length Substrings With No Repeated Characters easy or hard?

The problem is generally classified as Medium because it requires recognizing the sliding window optimization. The brute force idea is simple, but implementing an efficient O(n) solution with proper window updates and frequency tracking requires familiarity with string window patterns.

Find K-Length Substrings With No Repeated Characters Python/Java solution

Most implementations use the same sliding window logic across languages. Python typically uses a dictionary or Counter, while Java uses a HashMap or an int array for character counts. The algorithm iterates once through the string and maintains counts as the window moves.

How to solve Find K-Length Substrings With No Repeated Characters in O(n)?

Use a fixed-size sliding window of length k and maintain a hash map of character frequencies. Expand the window by adding the right pointer and shrink it when the size exceeds k by removing the left pointer character. When the window size equals k and all frequencies are 1, you found a valid substring. This ensures each character is processed only once.

What is the best approach for Find K-Length Substrings With No Repeated Characters?

The most efficient approach uses a sliding window with a hash table (frequency map). Move a window of length k across the string while tracking character counts. When the window size equals k and all characters are unique, increment the result. This runs in O(n) time and avoids recomputing uniqueness for every substring.

Is Find K-Length Substrings With No Repeated Characters asked at Google/Amazon/Meta?

Substring uniqueness and sliding window problems appear frequently in interviews at companies like Google, Amazon, and Meta. Variants such as longest substring without repeating characters or fixed-length substring constraints are common patterns used to evaluate string manipulation and window techniques.

What data structure is used in Find K-Length Substrings With No Repeated Characters?

A hash table (dictionary or map) stores the frequency of characters inside the current sliding window. This allows constant-time updates and checks when characters enter or leave the window. Some implementations also use a fixed-size array for ASCII characters.

What is the time complexity of Find K-Length Substrings With No Repeated Characters?

The optimal sliding window solution runs in O(n) time where n is the length of the string. Each character enters and leaves the window at most once. Space complexity is O(min(n, k)) for storing character frequencies in the hash map.

Description Hints 3 Solution Video 18:22

1100. Find K-Length Substrings With No Repeated Characters

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Given a string s and an integer k, return the number of substrings in s of length k with no repeated characters.

Example 1:

Input: s = "havefunonleetcode", k = 5
Output: 6
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

Example 2:

Input: s = "home", k = 5
Output: 0
Explanation: Notice k can be larger than the length of s. In this case, it is not possible to find any substring.

Constraints:

1 <= s.length <= 10⁴
s consists of lowercase English letters.
1 <= k <= 10⁴

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

How to check efficiently each K-length substring?

Hint 2

First store the first leftmost K-length substring in a hashTable or array of frequencies.

Hint 3

Then iterate through the rest of characters and erase the first element and add the next element from the right. If in the hashTable we have K different character we add 1 to the counter. After that return as answer the counter.

18:22

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Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

"havefunonleetcode"
5

Input

"home"
5

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