Pinduoduo·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate

Apr 2026

Summary

Pinduoduo software engineer interview with a classic matrix generation problem. Pretty standard coding round, the spiral order part is where most people trip up if they haven't seen it before.

Questions Asked (1)

Given two integers m and n, generate an m x n matrix filled with values 1 through m*n placed in spiral order, starting from the top-left and moving right, then down, then left, then up, spiraling inward until all cells are filled.

Algorithms & Data Structures

Author's notes

The boundary tracking is what gets you.

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AI HintsAI Generated

Suggested Approach

Use a layer-by-layer simulation approach, maintaining four boundaries (top, bottom, left, right) that shrink as each spiral layer is filled. Iterate through the matrix in the order right, down, left, up, updating boundaries after each direction, and fill cells with an incrementing counter from 1 to m*n.

Pro tip: Clarify edge cases upfront (e.g., single row/column, m or n equal to 1) and mention that the algorithm runs in O(m*n) time and O(1) extra space (excluding output), which is optimal. Also, consider discussing an alternative recursive approach to show depth, but emphasize the iterative boundary method for its simplicity and efficiency.

1. Initialize boundaries and counter

Set top=0, bottom=m-1, left=0, right=n-1, and initialize a counter val=1. Create an m x n matrix to fill.

2. Traverse right along top row

Fill from left to right along the top boundary, then increment top. If top > bottom, break.

3. Traverse down along right column

Fill from top to bottom along the right boundary, then decrement right. If left > right, break.

4. Traverse left along bottom row

Fill from right to left along the bottom boundary, then decrement bottom. If top > bottom, break.

5. Traverse up along left column

Fill from bottom to top along the left boundary, then increment left. If left > right, break. Repeat steps 2-5 until all cells are filled.

Key Points to Mention

Boundary variables (top, bottom, left, right) to track the current layer.
Direction order: right, down, left, up, with boundary updates after each direction.
Termination condition: stop when boundaries cross (top > bottom or left > right).
Time complexity O(m*n) and space complexity O(1) extra space (excluding output).
Edge cases: single row, single column, and m or n equal to 1.
Potential alternative: recursive layer-by-layer approach, but iterative is preferred for clarity and efficiency.

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