dark alpha capital·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Interviewed at Dark Alpha Capital for a Software Engineer role and got hit with a bitmask DP problem that felt like it came straight out of a competitive programming contest. The question was essentially LC 1066 territory, assign workers to resources minimizing total cost, and they wanted a full end-to-end implementation plus a discussion of the formulation and complexity.

Questions Asked (1)

Implement a bitmask DP solution for an assignment problem where N workers must each be assigned to a unique resource to minimize total cost. Walk through the state definition, transitions, and time complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This one took me a minute to even set up the state correctly.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by defining the DP state as the minimum cost to assign the first k workers to a subset of resources, where k is the number of set bits in the mask. Then describe the transition: for each unassigned resource, add its cost with the current worker and take the minimum. Finally, analyze the time complexity as O(N * 2^N) and space as O(2^N).

Pro tip: Mention that you can optimize space by using a 1D DP array of size 2^N, and that the number of set bits in the mask implicitly gives the current worker index, avoiding an extra dimension.

1. Define the state

Let dp[mask] be the minimum total cost to assign workers 0..k-1 to the resources indicated by the set bits in mask, where k = popcount(mask).

2. Base case and initialization

Initialize dp[0] = 0 and all other dp values to infinity. Iterate over masks in increasing order.

3. Transition

For each mask, let i = popcount(mask). For each resource j not in mask, update dp[mask | (1<<j)] = min(dp[mask | (1<<j)], dp[mask] + cost[i][j]).

4. Answer extraction

The answer is dp[(1<<N)-1], representing all workers assigned to all resources.

5. Complexity analysis

There are 2^N states, and for each state we iterate over up to N resources, giving O(N * 2^N) time. Space is O(2^N) for the DP array.

Key Points to Mention

State definition: dp[mask] = min cost to assign first popcount(mask) workers to resources in mask.
Transition: assign next worker to an unset bit in mask.
Time complexity: O(N * 2^N) because each state processes up to N transitions.
Space complexity: O(2^N) for the DP array.
Optimization: use 1D array and popcount to determine current worker.
Comparison to Hungarian algorithm: bitmask DP is simpler but exponential, suitable for small N (N ≤ 20).

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