Minimum Operations to Make Median of Array Equal to K
A medium-tier problem at 47% community acceptance, tagged with Array, Greedy, Sorting. Reported in interviews at IBM and 0 others.
You're given an unsorted array and need to make the median equal to K with the minimum number of operations. Each operation changes one element to any value. This is a medium-difficulty greedy problem that appears in OAs from major companies like IBM. The trick isn't complex, but the implementation catches candidates who don't think through the constraints on which elements to modify. At 47% acceptance, this problem filters for candidates who can think through a greedy strategy rather than brute-forcing. If you hit this during a live OA and freeze on how to approach it, StealthCoder surfaces the solution in seconds, invisible to the proctor.
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Minimum Operations to Make Median of Array Equal to K 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 by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know.
Get StealthCoderThe core insight is that you need to sort the array first, then identify the median position. Once you know where the median sits, you only modify elements that prevent the median from equaling K. The greedy move is to change the minimum number of elements on either side of the median to push values toward K. Most candidates waste time modifying elements that don't affect the median, or they fail to recognize that elements less than K on the left half and elements greater than K on the right half must be adjusted. The pattern involves sorting, locating the median index, then iterating inward from both ends of the relevant half toward K. Array and Sorting are your core tools; Greedy decision-making determines the count. If you're mid-assessment and can't mentally trace through which elements to touch, StealthCoder handles the logic without the proctor seeing it.
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Minimum Operations to Make Median of Array Equal to K recycles across companies for a reason. It's medium-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 by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Operations to Make Median of Array Equal to K interview FAQ
Why is this problem greedy and not dynamic programming?+
Once you sort and identify the median index, the optimal choice is deterministic: change the smallest number of elements to K such that the median becomes K. There's no overlapping subproblem or branching trade-off. Each element either needs to move toward K or doesn't. That determinism is the definition of greedy. No DP state space.
What's the most common mistake on this problem?+
Modifying elements that don't affect the median position, or changing the median itself when you should change its neighbors. Candidates often forget that only elements to the left of the median (if they're below K) or to the right (if they're above K) need adjustment. The median index itself is usually the last change.
Is this still asked at IBM and similar companies?+
Yes. The data shows IBM has reported it. Medium-difficulty array and greedy problems are common in OAs because they're not trivial but don't require advanced data structures. They filter for clear algorithmic thinking in 20-30 minutes.
How do I handle the case where the array length is even?+
With even length, there are two middle elements. The median is often defined as the lower of the two (index n/2 - 1 after sorting). Check the problem statement. Your greedy strategy remains the same: modify elements on either side to push the median position toward K.
What if some elements are already equal to K?+
They contribute zero to the operation count. Skip them and focus on elements that are far from K. Greedy logic: if an element is already K, leave it. The challenge is handling elements on either side of the median that block the median value from being K.
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