I knew there was a smarter way than looping through every range but my brain went to brute force first, which would've been O(n*k) for k updates.
Use a difference array to apply each range update in O(1) time, then compute the prefix sum to get the final array in O(n) time. This avoids the naive O(n*m) approach and demonstrates efficient algorithm design.
Pro tip: Mention that the difference array technique is a standard pattern for range update problems and can be extended to 2D or multiple updates. Also, clarify that the updates are inclusive of both endpoints, and handle edge cases like empty array or no updates.
Confirm the input format: array of zeros of size n, and a list of operations where each operation is (start, end, value). Ensure updates are inclusive and that the array is 0-indexed.
Explain that a difference array allows O(1) range updates by adding value at start and subtracting at end+1, then prefix sum yields the final array.
Initialize a difference array of size n+1 with zeros. For each operation (l, r, val), do diff[l] += val and diff[r+1] -= val (if r+1 < n).
Iterate through the difference array, maintaining a running sum, and assign each element to the result array. This gives the final values after all updates.
State time complexity O(n + m) and space O(n). Discuss handling of empty array, no operations, and large values (potential overflow).
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