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NeetCode #237LC-1652EasySliding Window
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#237 · #1652 · Defuse the Bomb(拆炸弹)

📌 Problem Statement & Constraints

You have a bomb with a circular code array nums and a key k. If k > 0, replace each element with the sum of the next k elements; if k < 0, with the sum of the previous |k| elements; if k == 0, replace every element with 0. Return the resulting array. Constraints: 1 <= nums.length <= 100, 1 <= nums[i] <= 100, -(n-1) <= k <= n-1.

💡 Core Algorithmic Approaches

  1. The array is circular, so indices are taken modulo n.
  2. For k > 0, sum the k elements after index i; for k < 0, sum the |k| elements before it.
  3. A direct implementation is O(n * |k|), which is acceptable given n <= 100.
  4. For large inputs a sliding window would make it O(n), but the straightforward version is clearer here.

💻 Benchmark Python3 Implementation

class Solution:
    def decrypt(self, code: List[int], k: int) -> List[int]:
        n = len(code)
        if k == 0:
            return [0] * n
        res = []
        for i in range(n):
            if k > 0:
                s = sum(code[(i + j) % n] for j in range(1, k + 1))
            else:
                s = sum(code[(i - j) % n] for j in range(1, -k + 1))
            res.append(s)
        return res

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n * |k|): with n <= 100 this is at most 10^4 operations. A sliding window would reduce it to O(n).
💾 Space Complexity
O(n) for the result.

⚠️ Interview Pitfalls & Follow-ups

  • Forgetting the modulo when indexing: the array is circular, so (i + j) % n is required.
  • Using range(k) instead of range(1, k+1): the window excludes the element itself.
  • Handling k == 0 inside the loop: returning an all-zero array upfront is simpler.
  • Assuming k is positive: it can be negative, in which case the sum looks backwards.