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Applovin·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Applovin SWE interview with a grid path counting problem. Pretty standard dynamic programming territory but they pushed into some interesting follow-up territory around 4-directional movement that tripped me up a bit.

Questions Asked (1)

Q1

Given an R x C grid with a start cell, an end cell, and some blocked cells, count the number of distinct paths from start to end. Movement is restricted to right and down only.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Got through the core solution fine.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (grid size, number of blocked cells) and confirm that paths are counted modulo a large prime if needed. Then present a dynamic programming solution where dp[i][j] represents the number of paths to cell (i,j), computed as the sum of paths from the top and left neighbors, with blocked cells having 0 paths. Finally, discuss time and space complexity and possible optimizations like using a 1D array.

Pro tip: Mention that if the grid is very large but blocked cells are sparse, you can use combinatorics with inclusion-exclusion or coordinate compression to avoid O(R*C) time. Also, always ask about modulo to prevent integer overflow.

1. Clarify constraints and edge cases

Ask about grid dimensions, number of blocked cells, whether start or end can be blocked, and if the answer should be modulo a prime. Confirm movement is only right and down.

2. Define DP state and recurrence

Let dp[i][j] be the number of paths from start to (i,j). Initialize dp[start] = 1. For each cell, if blocked, dp[i][j] = 0; else dp[i][j] = dp[i-1][j] + dp[i][j-1].

3. Implement and optimize space

Iterate row by row, using a 1D array to store the previous row's values. Update dp[j] = dp[j] + dp[j-1] for each row, skipping blocked cells.

4. Analyze complexity and trade-offs

Time complexity is O(R*C), space O(C). Discuss alternative approaches like combinatorics for sparse obstacles, and when they might be preferable.

5. Test with examples and edge cases

Walk through a small grid with obstacles, verify the count, and check cases where start or end is blocked, or no path exists.

Key Points to Mention

  • Dynamic programming recurrence: dp[i][j] = dp[i-1][j] + dp[i][j-1]
  • Handling blocked cells by setting dp to 0
  • Space optimization using a 1D array
  • Time and space complexity: O(R*C) time, O(C) space
  • Modulo arithmetic to prevent overflow
  • Edge cases: start/end blocked, no path, large grid

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