Problem Description

You start with an array of zeros with a given length. You're given a series of updates where each update is a triplet [startIdx, endIdx, inc] that tells you to:

• Add inc to every element from index startIdx to endIdx (inclusive)

Your task is to apply all these updates to the array and return the final result.

Example walkthrough:

• If length = 5 and updates = [[1, 3, 2], [2, 4, 3]]
• Start with: [0, 0, 0, 0, 0]
• After first update (add 2 to indices 1-3): [0, 2, 2, 2, 0]
• After second update (add 3 to indices 2-4): [0, 2, 5, 5, 3]
• Return: [0, 2, 5, 5, 3]

The challenge is to find an efficient way to handle potentially many updates without actually iterating through each range for every update, which would be inefficient. The solution uses a difference array technique:

1. Instead of updating ranges directly, mark the boundaries:

   • When adding c to range [l, r], mark d[l] += c (start of increase)
   • Mark d[r+1] -= c (end of increase) if r+1 is within bounds

2. After marking all boundaries, compute the prefix sum of this difference array to get the final result. The prefix sum naturally propagates the increments across the ranges.

This approach reduces the time complexity from O(n × m) for naive range updates to O(n + m) where n is the array length and m is the number of updates.

Intuition

Think about what happens when you add a value to a range. If you were to track the changes between consecutive elements rather than the elements themselves, adding a constant to a range creates a very specific pattern:

• The difference increases by c at the start of the range
• The difference decreases by c right after the range ends
• All differences within the range remain unchanged

For example, if we add 3 to indices [1, 3] in array [0, 0, 0, 0, 0]:

• Original: [0, 0, 0, 0, 0]
• After update: [0, 3, 3, 3, 0]
• Differences between consecutive elements: [0, +3, 0, 0, -3]

Notice how we only needed to mark two positions: where the increment starts (+3 at index 1) and where it ends (-3 at index 4).

The key insight is that instead of laboriously updating every element in each range, we can just mark these "boundary events" in a difference array. When we later compute the prefix sum of this difference array, the increments naturally "flow" across their ranges:

• The +c at the start propagates through the prefix sum
• The -c after the end cancels it out
• Multiple overlapping ranges automatically combine through addition

This transforms the problem from "update many elements many times" to "mark boundaries, then reconstruct once" - a much more efficient approach. The prefix sum acts like a running total that accumulates all the boundary changes we've marked, giving us the final array values.

Solution Approach

The implementation uses a difference array technique to efficiently handle range updates:

Step 1: Initialize the difference array

d = [0] * length

Create a difference array d of the same size as our target array, initialized with zeros.

Step 2: Mark boundaries for each update

for l, r, c in updates:
    d[l] += c
    if r + 1 < length:
        d[r + 1] -= c

For each update [l, r, c]:

• Add c to d[l] - this marks where the increment starts
• Subtract c from d[r + 1] (if within bounds) - this marks where the increment ends

The boundary check if r + 1 < length prevents index out of bounds when the range extends to the last element.

Step 3: Compute prefix sum to get final array

return list(accumulate(d))

Use Python's accumulate function to compute the running sum of the difference array. This reconstructs the final array by propagating all the marked changes.

Example walkthrough:

• length = 5, updates = [[1,3,2], [2,4,3]]
• Initial difference array: [0, 0, 0, 0, 0]
• After first update [1,3,2]: d = [0, 2, 0, 0, -2]
• After second update [2,4,3]: d = [0, 2, 3, 0, -2] (note: d[5] would be -3 but it's out of bounds)
• Prefix sum: [0, 2, 5, 5, 3]

Time Complexity: O(n + m) where n is the array length and m is the number of updates Space Complexity: O(n) for the difference array

Example Walkthrough

Let's trace through a small example to see how the difference array technique works.

Given: length = 4, updates = [[0,2,3], [1,3,2]]

Step 1: Initialize difference array

d = [0, 0, 0, 0]

Step 2: Process first update [0,2,3]

• Add 3 to range [0,2]
• Mark d[0] += 3 (start of range)
• Mark d[3] -= 3 (end of range + 1)

d = [3, 0, 0, -3]

Step 3: Process second update [1,3,2]

• Add 2 to range [1,3]
• Mark d[1] += 2 (start of range)
• d[4] would be -= 2, but index 4 is out of bounds, so skip

d = [3, 2, 0, -3]

Step 4: Compute prefix sum

• Position 0: 3 → result[0] = 3
• Position 1: 3 + 2 = 5 → result[1] = 5
• Position 2: 5 + 0 = 5 → result[2] = 5
• Position 3: 5 + (-3) = 2 → result[3] = 2

Final result: [3, 5, 5, 2]

Verification with naive approach:

• Start: [0, 0, 0, 0]
• After [0,2,3]: [3, 3, 3, 0]
• After [1,3,2]: [3, 5, 5, 2] ✓

The difference array elegantly handles overlapping ranges - notice how position 1 and 2 automatically combine both updates through the prefix sum computation.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m), where n is the length of the array and m is the number of updates. The algorithm iterates through all m updates to populate the difference array (taking O(m) time), then performs a prefix sum operation using accumulate() which takes O(n) time to traverse the entire array.

The space complexity is O(n). The algorithm creates a difference array d of size n to store the incremental changes, and the final result array returned by accumulate() also requires O(n) space. No additional data structures that scale with input size are used.

Common Pitfalls

1. Index Out of Bounds When Marking End Boundary

The Pitfall: The most common mistake is forgetting to check if end_idx + 1 is within the array bounds before updating difference_array[end_idx + 1]. This leads to an IndexError when the update range extends to the last element of the array.

Incorrect Code:

for start_idx, end_idx, increment in updates:
    difference_array[start_idx] += increment
    difference_array[end_idx + 1] -= increment  # ❌ IndexError when end_idx == length - 1

Correct Solution:

for start_idx, end_idx, increment in updates:
    difference_array[start_idx] += increment
    if end_idx + 1 < length:  # ✅ Check bounds first
        difference_array[end_idx + 1] -= increment

2. Misunderstanding the Difference Array Concept

The Pitfall: Some developers try to directly update the range in each iteration, missing the entire point of the difference array optimization.

Incorrect Approach:

result = [0] * length
for start_idx, end_idx, increment in updates:
    for i in range(start_idx, end_idx + 1):  # ❌ O(n*m) complexity
        result[i] += increment
return result

Why It's Wrong: This defeats the purpose of using a difference array and results in O(n × m) time complexity instead of O(n + m).

3. Off-by-One Error in Range Boundaries

The Pitfall: Confusing whether the end index is inclusive or exclusive, leading to incorrect boundary marking.

Common Mistake:

# Thinking end_idx is exclusive and marking wrong boundary
difference_array[start_idx] += increment
if end_idx < length:  # ❌ Wrong condition
    difference_array[end_idx] -= increment  # ❌ Wrong index

Remember: The problem states end_idx is inclusive, so we need to mark the boundary at end_idx + 1 to stop the increment effect after position end_idx.

4. Not Returning the Accumulated Result

The Pitfall: Returning the difference array itself instead of its prefix sum.

Incorrect:

return difference_array  # ❌ Returns differences, not final values

Correct:

return list(accumulate(difference_array))  # ✅ Returns cumulative sum