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NeetCode #192LC-2460EasyTwo Pointers
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#192 · #2460 · Apply Operations to an Array(对数组执行操作)

📌 Problem Statement & Constraints

You are given a 0-indexed array nums of non-negative integers. For each index i from 0 to n-2: if nums[i] == nums[i+1], multiply nums[i] by 2 and set nums[i+1] to 0. Then move all zeros to the end while keeping the relative order of the non-zero elements. Return the resulting array. Constraints: 1 <= nums.length <= 2000, 0 <= nums[i] <= 1000.

💡 Core Algorithmic Approaches

  1. Perform the doubling pass first, comparing each element with its original successor before any modification.
  2. Then compact the array with a write pointer, exactly as in Move Zeroes.
  3. The order matters: doubling must complete before the compaction, since compaction changes the adjacency relationships.
  4. Note that nums[i+1] is zeroed after a doubling, which prevents a cascading double-double.

💻 Benchmark Python3 Implementation

class Solution:
    def applyOperations(self, nums: List[int]) -> List[int]:
        n = len(nums)
        for i in range(n - 1):
            if nums[i] == nums[i + 1]:
                nums[i] *= 2
                nums[i + 1] = 0        # prevent cascading
        k = 0
        for i in range(n):
            if nums[i] != 0:
                nums[k] = nums[i]
                k += 1
        for i in range(k, n):
            nums[i] = 0                # pad the tail
        return nums

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n): two passes.
💾 Space Complexity
O(1) extra: in place.

⚠️ Interview Pitfalls & Follow-ups

  • Zeroing the left element instead of the right: the doubling applies to nums[i] and the successor is zeroed.
  • Letting the doubled value cascade: after zeroing nums[i+1], the next comparison uses 0, which prevents a further doubling. Without the zeroing, [2, 2, 4] would behave differently.
  • Compacting before doubling: the adjacency would be destroyed.
  • Forgetting the padding pass: the tail would retain stale values.