Find K-Length Substrings With No Repeated Characters

Find K-Length Substrings With No Repeated Characters

What is this problem about?

In the Find K-Length Substrings With No Repeated Characters coding problem, you are given a string and an integer $k$. You need to count how many substrings of length $k$ consist of entirely unique characters. If $k$ is larger than the length of the string or the alphabet size (26), the answer is zero.

Why is this asked in interviews?

Amazon frequently uses this to test a candidate's proficiency with the Sliding Window interview pattern. It evaluations whether you can efficiently maintain a character frequency state as you slide through a string. It’s a "clean" problem that checks if you can avoid $O(n imes k)$ brute force by using a Hash Table or a frequency array to perform updates in $O(1)$ per character.

Algorithmic pattern used

This problem uses a Fixed-size Sliding Window with a Hash Table (or frequency array).

1. Initialize a frequency array of size 26 and a uniqueCount variable.
2. Build the first window of size $k$.
3. If uniqueCount == k, the window is valid.
4. Slide the window one character at a time:
   • Add the incoming character: update frequency and uniqueCount.
   • Remove the outgoing character: update frequency and uniqueCount.
   • Check if the window is valid.

Example explanation

s = "havefunonleetcode", k = 5

1. Window 1: "havef". All unique. Count = 1.
2. Window 2: "avefu". All unique. Count = 2.
3. ...
4. Window "unonl": 'n' is repeated. Not valid. ... and so on.

Common mistakes candidates make

• Re-calculating uniqueness: Using a Set to check every substring from scratch ($O(n imes k)$).
• Alphabet Constraint: Not realizing that if $k > 26$, the result must be 0 (Pigeonhole Principle).
• Hash Map overhead: Using a heavy Map object when a simple int[26] array would be much more performant.

Interview preparation tip

Always look for the "fixed-size" hint in sliding window problems. If the window size is constant, the logic is simpler than dynamic windows, but you must be precise with adding/removing elements simultaneously.