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Decagon·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Decagon software engineer interview, got a classic grid DP problem. Nothing unexpected, pretty standard coding round.

Questions Asked (1)

Q1

A robot starts at the top-left of an m x n grid and can only move right or down. How many unique paths exist to reach the bottom-right corner?

Algorithms & Data Structures
Author's notes

Classic DP, I've seen this a dozen times.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain the dynamic programming approach where the number of paths to each cell is the sum of paths from the cell above and the cell to the left. Derive the recurrence relation and discuss how to optimize space, possibly mentioning the combinatorial solution.

Pro tip: Mention that the problem can be solved in O(min(m,n)) space by using a 1D array, and note that the combinatorial formula C(m+n-2, m-1) provides a direct mathematical solution, showing depth beyond basic DP.

1. Clarify the problem

Confirm that the grid is m x n, moves are only right or down, and the goal is to count unique paths from (0,0) to (m-1,n-1). Ask about edge cases like m=1 or n=1.

2. Define the DP state

Let dp[i][j] be the number of unique paths to cell (i,j). Base case: dp[0][0] = 1. For first row and first column, there is only one path.

3. Derive recurrence relation

For other cells, dp[i][j] = dp[i-1][j] + dp[i][j-1] because you can arrive from above or from the left.

4. Optimize space

Use a 1D array of size n (or m) to compute row by row, updating dp[j] += dp[j-1] for each row, reducing space complexity to O(min(m,n)).

5. Discuss alternative and complexity

Mention the combinatorial solution: total moves = (m-1) downs + (n-1) rights, so paths = C(m+n-2, m-1). Time complexity O(m*n) for DP, O(min(m,n)) space; combinatorial O(min(m,n)) time.

Key Points to Mention

  • Dynamic programming approach with recurrence dp[i][j] = dp[i-1][j] + dp[i][j-1]
  • Base cases: first row and first column have exactly 1 path
  • Space optimization using a 1D array to achieve O(min(m,n)) space
  • Combinatorial formula: C(m+n-2, m-1) or C(m+n-2, n-1)
  • Time and space complexity analysis: O(m*n) time, O(min(m,n)) space for optimized DP
  • Edge cases: m=1 or n=1, and large grids where combinatorial might overflow (use modulo if needed)

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