Minimum Adjacent Swaps for K Consecutive Ones
A hard-tier problem at 42% community acceptance, tagged with Array, Greedy, Sliding Window. Reported in interviews at Flipkart and 0 others.
You're given an array of 0s and 1s. The problem asks: what's the minimum number of adjacent swaps to group K ones together consecutively. Flipkart has asked this one. It's a hard problem with a 42% acceptance rate, which means most people who see it either don't finish or walk away without the right insight. The trick isn't obvious on first read. You can stare at greedy approaches that don't work or brute force that times out. If this lands in your live assessment and you blank, StealthCoder runs invisibly and surfaces a working solution in seconds.
Companies that ask "Minimum Adjacent Swaps for K Consecutive Ones"
Minimum Adjacent Swaps for K Consecutive Ones 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. Built by an Amazon engineer who used it to pass JPMorgan's OA and system design loop.
Get StealthCoderThe core pattern: you need a sliding window to track K ones, but the cost function is unintuitive. For any window containing exactly K ones, the cost to consolidate them isn't just counting the zeros between them. It's the sum of distances each 1 must move to reach a target position. The greedy insight is that you want to minimize this sum. Use a prefix sum array to track cumulative ones and their positions, then slide a window of size K ones (not array indices, but the Kth occurrence). Calculate the cost at each position and find the minimum. Array, Greedy, Sliding Window, and Prefix Sum work together here. Most fail because they try to count swaps directly instead of modeling it as a positioning problem. This is where StealthCoder's edge matters: the pattern requires all four topics to click at once.
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Minimum Adjacent Swaps for K Consecutive Ones recycles across companies for a reason. It's hard-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. Built by an Amazon engineer who used it to pass JPMorgan's OA and system design loop. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Adjacent Swaps for K Consecutive Ones interview FAQ
Is this actually asked at interviews or is it a LeetCode-only problem?+
Flipkart has reported asking it. Hard-tier sliding window variants are standard at tier-1 companies, though this specific problem's acceptance rate is 42%, so it's not a household problem. It's exactly the kind of hard problem that separates prepared from unprepared candidates.
What's the first approach that fails?+
Counting zeros between ones and thinking swaps equal zero count. That doesn't work because the cost model is positional. You need to think about where each 1 ends up, not just what's between them. Prefix sums let you calculate this efficiently without iterating through every position.
How do I know when to use prefix sum here?+
When you need to quickly answer 'what's the sum of positions or indices in a range'. Here, you're summing the distances ones must travel. Prefix sum precomputes this so sliding window queries are O(1) instead of O(K) per position.
Can I solve this with just greedy and sliding window?+
Greedy chooses which window of K ones to consolidate. Sliding Window iterates candidate positions. But without Prefix Sum to quickly calculate the cost at each position, you'll time out on large arrays. All four topics are load-bearing.
If I don't nail this in my OA, am I guaranteed to fail the interview?+
Not necessarily. Interviewers value how you approach the problem, not just the final answer. But if you run out of time or hit a wall, that's where the hedge matters. Knowing StealthCoder has your back if you blank lets you stay calm and attempt the problem without panic.
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