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NeetCode #88LC-1243EasyArrays & Hashing
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#88 · #1243 · Array Transformation(数组变换)

📌 Problem Statement & Constraints

Given an integer array arr, repeatedly apply this transformation simultaneously to every interior element: increase it by 1 if it is smaller than both neighbours, decrease it by 1 if it is larger than both, otherwise leave it unchanged. Return the array once no further changes occur. Constraints: 3 <= arr.length <= 100, 1 <= arr[i] <= 100.

💡 Core Algorithmic Approaches

  1. This is a direct simulation: build the next state from the current one, and stop when the state stops changing.
  2. The update must be simultaneous, so write into a copy rather than mutating arr in place.
  3. Only interior indices 1..n-2 are affected, since the endpoints have no two neighbours.
  4. Termination is guaranteed because each element is bounded and the changes are monotone in a potential sense -- in practice the array settles within a few rounds.

💻 Benchmark Python3 Implementation

class Solution:
    def transformArray(self, arr: List[int]) -> List[int]:
        while True:
            nxt = arr[:]                   # simultaneous update needs a copy
            for i in range(1, len(arr) - 1):
                if arr[i] < arr[i - 1] and arr[i] < arr[i + 1]:
                    nxt[i] += 1
                elif arr[i] > arr[i - 1] and arr[i] > arr[i + 1]:
                    nxt[i] -= 1
            if nxt == arr:                 # fixed point reached
                return arr
            arr = nxt

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(rounds * n). The number of rounds is bounded by the value range (at most 100), so this is O(100 * n) = O(n) in practice.
💾 Space Complexity
O(n) for the working copy each round.

⚠️ Interview Pitfalls & Follow-ups

  • Updating arr in place: neighbours would see partially updated values, which is not what the simultaneous rule means.
  • Applying the rule to the endpoints: indices 0 and n-1 have no two neighbours and must never change.
  • Returning nxt instead of arr: both are equal at the fixed point, but returning the original reference is clearer.
  • Comparing nxt != arr with is: is compares identity, not contents.