Minimum Recolors to Get K Consecutive Black Blocks
A easy-tier problem at 68% community acceptance, tagged with String, Sliding Window. Reported in interviews at HP and 0 others.
Minimum Recolors to Get K Consecutive Black Blocks is an easy sliding window problem that only 68% of candidates clear on first submission. It's the kind of problem that looks trivial until you see the assessment, second-guess yourself, and burn time on an off-by-one bug or a nested loop you didn't need. HP asks it. The pattern is straightforward: count white blocks in a window of size K, slide it across the string, track the minimum. If this problem hits your live OA and you blank on the window mechanics, StealthCoder solves it in seconds, invisible to the proctor.
Companies that ask "Minimum Recolors to Get K Consecutive Black Blocks"
Minimum Recolors to Get K Consecutive Black Blocks is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made for the engineer who has done the work but might still blank with a webcam pointed at him.
Get StealthCoderThe trap here is overcomplication. You have a string of 'B' (black) and 'W' (white) blocks. You need to find the smallest window of K consecutive blocks that requires the fewest recolors to turn them all black. The obvious brute force is nested loops: for each starting position, count whites in the next K blocks. That works but it's O(n * K). The actual move is sliding window: calculate whites in the first K blocks, then slide right by one position, subtract the left element (if white) and add the right element (if white). O(n) time, O(1) space. Most missed submissions come from either resetting the window when you shouldn't, or confusing 'recolors needed' with 'blacks found'. If you hit the wall during the assessment and can't remember whether to slide left or right first, StealthCoder runs invisibly and surfaces the working solution instantly.
Pattern tags
You know the problem.
Make sure you actually pass it.
Minimum Recolors to Get K Consecutive Black Blocks recycles across companies for a reason. It's easy-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made for the engineer who has done the work but might still blank with a webcam pointed at him. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Recolors to Get K Consecutive Black Blocks interview FAQ
Is this really an easy problem or is the acceptance rate inflated?+
68% acceptance is solid for easy, not inflated. The problem itself is straightforward string and sliding window. The 32% failure rate comes from implementation mistakes under time pressure: off-by-one errors in window bounds, or miscounting white blocks when the window slides. The algorithm is simple; the execution can slip.
What's the actual trick to sliding window here?+
Calculate the count of white blocks in the first K-length window. Then iterate through the rest of the string: for each new position, remove the leftmost block from your count (if it was white) and add the new rightmost block (if it's white). Track the minimum count seen. That's it. O(n) time.
Is this still asked by HP and similar companies?+
HP is the company on record for this problem. It's an easy-tier screen, so it appears in phone interview rounds or OA warm-ups for mid-level roles. Companies use it to filter basic string manipulation and window sliding competence, not as a hard gate.
How does sliding window relate to two-pointer problems?+
Both use pointers to avoid nested iteration. Sliding window keeps a fixed or shrinking/expanding window and moves it once across the input. Two-pointer usually starts at both ends and converges toward center. This problem uses a fixed-size sliding window, so you're really just moving one pointer and calculating based on a fixed distance behind it.
What if I just iterate and count each window the naive way?+
It passes, but it's O(n * K) instead of O(n). With K up to n, that's O(n^2) worst case. Sliding window is the expected solution. Under live interview stress, the naive approach is also where off-by-one bugs hide. The window-slide approach is cleaner.
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