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Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Salesforce SWE interview with a greedy heap problem that looks straightforward until you realize the optimal strategy isn't obvious at all. One question, but it had some depth to it.

Questions Asked (1)

Q1

You have an array of building heights, a supply of bricks, and a fixed number of ladders. Moving to a taller building costs either bricks equal to the height difference or one ladder. Find the furthest building index you can reach using both resources optimally.

Algorithms & Data Structures
Author's notes

My first instinct was to use ladders on the biggest jumps, which is kind of right but you don't know which jumps are biggest until you've already passed them.

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AI HintsAI Generated

Suggested Approach

Use a greedy strategy with a min-heap to track the largest height differences encountered. Allocate bricks to smaller gaps and ladders to the largest gaps, ensuring optimal resource usage. Iterate through the array, and when resources are insufficient, return the current index.

Pro tip: Clarify that ladders should be used for the largest height differences to conserve bricks, and mention that the heap approach achieves O(n log L) time where L is the number of ladders.

1. Understand the problem and constraints

Restate the problem: given an array of building heights, bricks, and ladders, find the furthest index reachable. Note that moving to a taller building requires either bricks equal to the height difference or one ladder.

2. Choose the right data structure

Use a min-heap to keep track of the largest height differences encountered so far. This allows efficient replacement of a ladder with bricks when a larger gap appears.

3. Iterate and manage resources

Traverse the array from left to right. For each positive height difference, push it onto the min-heap. If the heap size exceeds the number of ladders, use bricks to cover the smallest difference (pop from heap). If bricks become negative, return the current index.

4. Return the result

If the loop completes, return the last index of the array, indicating all buildings are reachable.

Key Points to Mention

  • Greedy approach: use ladders for the largest height differences to minimize brick usage.
  • Min-heap to efficiently track and replace the smallest difference when ladders are exceeded.
  • Time complexity: O(n log L) where n is the number of buildings and L is the number of ladders.
  • Space complexity: O(L) for the heap.
  • Edge cases: no height differences, insufficient resources at the start, and large height differences.
  • Optimality: prove that using ladders on the largest gaps is always optimal.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.