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J.P. Morgan·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

J.P. Morgan software engineer coding round with two algorithm problems. The grid traversal one was way harder than the anagram warmup and I don't think I handled the complexity analysis cleanly.

Questions Asked (2)

Q1

Given an array of strings, group together all strings that are anagrams of each other. Return the grouped lists and the total number of groups. Analyze the time and space complexity of your solution.

Algorithms & Data Structures
Author's notes

Classic problem, felt fine.

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AI HintsAI Generated

Suggested Approach

Use a hash map to group strings by a canonical key that is identical for all anagrams, such as the sorted string or a character count signature. Iterate through the array, compute the key for each string, and append the string to the corresponding group. Finally, return the groups and the number of groups.

Pro tip: Mention that sorting each string costs O(k log k) per string, but using a character count key can reduce it to O(k) per string, which is more efficient for large alphabets or long strings. Also, clarify that the total time complexity is O(n * k log k) with sorting and O(n * k) with counting, where n is the number of strings and k is the maximum length.

1. Clarify and Define

Confirm the definition of anagrams and edge cases (e.g., empty strings, different lengths, case sensitivity). Ask if the output should preserve any order.

2. Choose a Canonical Key

Decide on a key that uniquely identifies anagrams: sorted string or character count signature. Discuss trade-offs (time vs. space).

3. Group with Hash Map

Iterate through the array, compute the key for each string, and use a hash map to map key to a list of anagrams. Append the string to the corresponding list.

4. Return Results

Extract the grouped lists from the hash map and return them along with the total number of groups (size of the map).

5. Analyze Complexity

Analyze time and space complexity: O(n * k log k) time with sorting, O(n * k) with counting; O(n * k) space for storing all strings and keys.

Key Points to Mention

  • Hash map usage for grouping
  • Canonical key: sorted string vs. character count
  • Time complexity: O(n * k log k) with sorting, O(n * k) with counting
  • Space complexity: O(n * k) for storing strings and keys
  • Edge cases: empty strings, single string, all anagrams, no anagrams
  • Trade-offs between sorting and counting approaches

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

Q2

Given a 2D grid of characters and a target string, count the total number of distinct paths through the grid that spell out the string. You can move to any orthogonally adjacent cell and cannot reuse a cell within a single path. Return -1 if no valid path exists. Discuss an efficient approach and its complexity.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This one wrecked me a bit.

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AI HintsAI Generated

Suggested Approach

Model the problem as a DFS/backtracking search from each cell matching the first character, exploring all 4-directional moves while tracking visited cells to avoid reuse. Optimize with memoization on (row, col, index) if revisiting states is allowed, but note that the no-reuse constraint complicates memoization; discuss trade-offs. Return the total count or -1 if zero.

Pro tip: Clarify the constraints upfront (grid size, string length, whether paths can revisit cells across different paths) and mention that the no-reuse rule makes the problem NP-hard in general, so an exact solution may be exponential; propose pruning or approximation if the input is large.

1. Clarify requirements and constraints

Ask about grid dimensions, string length, character set, and whether paths can share cells across different paths. Confirm that 'distinct paths' means different sequences of cell coordinates.

2. Design a backtracking DFS solution

For each cell matching the first character, recursively explore all 4-directional neighbors that match the next character, marking cells as visited and unmarking on backtrack. Count complete matches.

3. Analyze time and space complexity

Time: O(4^L * M*N) worst-case, where L is string length and M*N is grid size; space: O(L) for recursion stack plus visited set. Discuss pruning (e.g., early termination if remaining characters cannot be matched).

4. Discuss optimizations and trade-offs

Mention memoization is tricky due to visited-state; consider bidirectional search or meet-in-the-middle for long strings. If paths can reuse cells, memoization on (r,c,index) reduces to O(M*N*L).

5. Handle edge cases and return value

If no path found, return -1. Handle empty string (return 0 or 1 depending on definition), single-cell grid, and strings longer than total cells.

Key Points to Mention

  • Backtracking with visited set to enforce no cell reuse within a path
  • Time complexity exponential in string length due to branching factor 4
  • Space complexity O(L) for recursion depth and visited tracking
  • Pruning techniques: early exit if remaining characters cannot be matched
  • Trade-off: memoization not directly applicable with visited constraint; alternative DP if reuse allowed
  • Edge cases: empty string, no path (return -1), grid smaller than string length

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