Minimum Cost to Move Chips to The Same Position is an interesting puzzle-like problem. You have several chips at different positions on a 1D coordinate axis. You can move a chip from position pos to pos + 2 or pos - 2 with a cost of 0. Alternatively, you can move a chip from pos to pos + 1 or pos - 1 with a cost of 1. The objective is to move all chips to a single, identical position (any integer position) with the minimum total cost. This problem emphasizes observation over complex calculation.
This is a popular "Easy" level question at companies like Microsoft and Amazon. The Minimum Cost to Move Chips to The Same Position interview question is designed to see if a candidate can spot a simple mathematical property rather than jumping straight into a complex algorithm. It tests your ability to simplify a problem by identifying "free" moves, which is a key skill in both coding and system architecture.
The algorithmic pattern used here is Greedy and Math. The key observation is that moving a chip by any even distance costs 0. This means all chips at even positions can be moved to position 0 for free, and all chips at odd positions can be moved to position 1 for free. Now, you only have two groups: one at position 0 and one at position 1. To move everyone to the same spot, you either move all chips from 0 to 1 (cost = number of chips at 0) or all from 1 to 0 (cost = number of chips at 1). Naturally, you choose the smaller group to move. This "Array, Math, Greedy interview pattern" shows that counting is often faster than simulating movements.
Suppose you have chips at positions [1, 2, 3, 4, 100].
All even-positioned chips can move to position 0 for free. All odd-positioned chips can move to position 1 for free.
In the Minimum Cost to Move Chips to The Same Position coding problem, a common mistake is trying to use Breadth-First Search (BFS) or simulating the moves step-by-step. Since the number of positions can be very large, simulation will be too slow. Another mistake is overthinking the "target position" and trying to find a median or average, which is unnecessary here because the parity (even/odd) is all that matters.
Always look for "parity" in problems involving movements by fixed distances. If you can move 2 steps for free, the problem almost always boils down to a comparison between even and odd indices. This "Math interview pattern" is a great way to save time during an interview. Before coding, ask yourself: "What part of this problem is free?" The answer usually points directly to the most efficient solution.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Minimum Operations to Make Array Equal II | Medium | Solve | |
| Minimum Replacements to Sort the Array | Hard | Solve | |
| Largest Perimeter Triangle | Easy | Solve | |
| Rabbits in Forest | Medium | Solve | |
| Append K Integers With Minimal Sum | Medium | Solve |
