Minimum Cost to Move Chips to The Same Position
A easy-tier problem at 72% community acceptance, tagged with Array, Math, Greedy. Reported in interviews at Morgan Stanley and 0 others.
You've got chips scattered across positions on a line, and you need to move them all to one spot with minimal cost. The catch: moving a chip costs 1 per position, but only if it crosses an odd-numbered position. This is the kind of problem that looks like it needs dynamic programming or binary search until you see the pattern. Morgan Stanley has asked it. With a 72% acceptance rate, most candidates solve it, but many waste time building the wrong solution. If this hits your live OA and you freeze on the insight, StealthCoder surfaces the working approach instantly while you stay invisible to the proctor.
Companies that ask "Minimum Cost to Move Chips to The Same Position"
Minimum Cost to Move Chips to The Same Position 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 engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.
Get StealthCoderThe trap is thinking you need to simulate movement or calculate total distance. The real trick is realizing that the cost only depends on how many chips start on even positions versus odd positions. To move a chip from even to odd (or vice versa) costs exactly 1. So the answer is just the minimum of chips starting on even positions or chips starting on odd positions. That's it. The greedy insight here is that you pick the parity with fewer chips, because moving the minority group always costs less than moving the majority. No sorting, no heap, no DP. This problem tests whether you can spot the mathematical structure hiding under an algorithmic wrapper. Most wrong solutions overcomplicate by calculating distances between positions. StealthCoder catches this early in the live assessment if you're heading down the greedy-wrong path.
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Minimum Cost to Move Chips to The Same Position 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. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Cost to Move Chips to The Same Position interview FAQ
Is this really an easy, or is the acceptance rate misleading?+
The 72% acceptance rate is honest. The problem is genuinely easy once you see that parity is the only thing that matters. The hard part is unlearning the instinct to calculate distances or simulate moves. Most people who fail miss the mathematical insight entirely, not the implementation.
Why does moving across an odd position cost 1, and how does that become the parity trick?+
A chip at position 0 (even) to position 3 (odd) crosses position 1 (odd), cost 1. Position 0 to 4 (even) costs 0. So you only pay when crossing from even parity to odd or vice versa. Count chips on each parity, move the smaller group. Total cost is the minimum count.
Should I try to optimize which position all chips move to?+
No. The problem is asking for minimum cost, and the cost is determined entirely by parity. Any target position of the same parity as your chosen majority costs zero additional moves for those chips. The position itself doesn't matter.
How does this relate to the Greedy and Math topics?+
Greedy: you pick the cheaper direction (move fewer chips). Math: the cost formula reduces to min(even_count, odd_count). The Array topic just means input is an array of positions. There's no complex data structure play here.
Morgan Stanley asked this once. Does that mean it'll show up again in their OA?+
Unknown. One company report doesn't predict future frequency. But if it does appear, you're looking at a 5-minute solve if you know the parity insight, or 20+ minutes of dead ends if you don't. That's where the edge matters most.
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