Minimum Moves to Equal Array Elements III
1. What is this problem about?
The "Minimum Moves to Equal Array Elements III" (often associated with variations involving weights or specific constraints) continues the theme of equalizing an array. In some versions of this problem, you are given an array and you want to make all elements equal, but the "cost" of moving an element might depend on its index or the distance moved.
Commonly, this specific title refers to problems where you are limited in which elements you can change or where you have to move elements to a specific "target" that isn't just a value but a position (like in the "Min Cost to Move Chips" problem).
2. Why is this asked in interviews?
Adobe and other firms use these variations to see if a candidate can adapt a known solution (like the Median rule) to new constraints.
Key evaluation points:
- Adaptability: Can the candidate handle "weighted" moves?
- Parity Logic: Realizing that moving an element 2 steps might be "free" or cheaper than 1 step (common in grid/chip problems).
- Cost Analysis: Identifying the most efficient target.
3. Algorithmic pattern used
Depending on the specific variant, the pattern is often Parity Counting or Weighted Median.
- If moving 2 steps costs 0 and 1 step costs 1: The goal is to move everyone to the same "parity" (all even or all odd positions). You just count how many numbers are even and how many are odd, and pick the smaller count.
- If moves have different costs: You find a weighted median.
4. Example explanation
Imagine the problem where moving 2 steps is free and 1 step costs 1.
Array of positions: [1, 2, 3]
- Odds: 1, 3 (Count = 2)
- Evens: 2 (Count = 1)
To make them all equal:
- Move everyone to position 1: ∣1−1∣=0,∣2−1∣=1,∣3−1∣=0 (since 2 steps is free). Total = 1.
- Move everyone to position 2: ∣1−2∣=1,∣2−2∣=0,∣3−2∣=1. Total = 2.
Minimum moves = 1.
5. Common mistakes candidates make
- Ignoring Parity: Trying to use a standard median when the cost function makes parity more important.
- Overcomplicating: Using BFS for a simple math/parity problem.
- Wrong Target: Picking a target position that isn't one of the existing positions (usually, the optimal position is always one of the inputs).
6. Interview preparation tip
Always clarify the cost of a "move." If the cost is constant regardless of distance, it's a counting problem. If the cost is proportional to distance, it's a median problem. If the cost is proportional to distance squared, it's a mean problem.
