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NeetCode #183LC-283EasyTwo Pointers
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#183 · #283 · Move Zeroes(移动零)

📌 Problem Statement & Constraints

Given an integer array nums, move all 0s to the end while maintaining the relative order of the non-zero elements, in place. Constraints: 1 <= nums.length <= 10^4, -2^31 <= nums[i] <= 2^31 - 1.

💡 Core Algorithmic Approaches

  1. Use a write pointer k for the next non-zero slot. Scan with a read pointer and copy every non-zero value to nums[k].
  2. After the scan, fill the remaining slots from k to the end with zeros.
  3. This preserves the relative order of the non-zero elements, which a swap-based approach with a two-ended pointer would not.
  4. An alternative is to swap the non-zero value into position k immediately, which achieves the same result in one pass.

💻 Benchmark Python3 Implementation

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        k = 0                          # next slot for a non-zero value
        for i in range(len(nums)):
            if nums[i] != 0:
                nums[k] = nums[i]
                k += 1
        for i in range(k, len(nums)):
            nums[i] = 0                # pad the tail with zeros


# One-pass swap variant (also preserves order)
class Solution2:
    def moveZeroes(self, nums: List[int]) -> None:
        k = 0
        for i in range(len(nums)):
            if nums[i] != 0:
                nums[k], nums[i] = nums[i], nums[k]
                k += 1

⚡ Complexity Deep Dive

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

⚠️ Interview Pitfalls & Follow-ups

  • Using nums.remove(0) in a loop: O(n^2) and the indices shift underneath you.
  • Swapping zeros to the front: the relative order of the non-zero elements must be preserved, so zeros go to the end.
  • Forgetting the padding pass: the slots after k still hold stale values, so they must be zeroed (or, in the swap variant, they are already zeros).
  • Comparing with is 0: use == 0, since is compares identity.