← Molocoads Interview Insights

Molocoads·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Interviewed for a software engineer role at Molocoads and got hit with a pretty involved array problem that had a follow-up I was not expecting. The core question was manageable but the circular variant threw me off a bit.

Questions Asked (2)

Q1

Given an integer array, for each index i, find the maximum length of a contiguous subarray that contains i and whose maximum element is exactly nums[i].

Algorithms & Data Structures
Author's notes

The key insight is that you're basically looking for how far left and right you can walk from i before hitting something strictly larger.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

For each index i, the maximum subarray containing i with maximum exactly nums[i] is bounded by the nearest greater elements to the left and right. Use a monotonic stack to compute these boundaries in O(n) time, then the length is right_boundary - left_boundary - 1.

Pro tip: Clarify how to handle duplicates: if there are equal elements, define boundaries to avoid double-counting, e.g., use strict greater on one side and greater-or-equal on the other.

1. Understand the problem

Restate the problem: for each i, find the longest contiguous subarray that includes i and whose maximum is exactly nums[i]. Note that any element greater than nums[i] cannot be in the subarray.

2. Identify boundaries

The subarray must be bounded by the nearest elements greater than nums[i] on the left and right. If no such element exists, the boundary is -1 or n.

3. Compute boundaries efficiently

Use a monotonic decreasing stack to find the previous greater element and next greater element for each index in O(n) time.

4. Calculate lengths

For each i, the maximum length is (next_greater_index - prev_greater_index - 1). Collect these lengths into the result array.

5. Handle duplicates

If there are equal elements, decide on a consistent rule (e.g., previous greater-or-equal and next strictly greater) to ensure each subarray is counted exactly once.

Key Points to Mention

  • Monotonic stack for finding previous and next greater elements in O(n).
  • Time complexity O(n) and space complexity O(n).
  • Handling duplicates to avoid overcounting or undercounting.
  • Edge cases: all elements equal, strictly increasing/decreasing arrays.
  • The subarray must include index i, so the length is at least 1.
  • Proof of correctness: any subarray with maximum nums[i] cannot include a greater element, so it must lie within the boundaries.

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

Q2

Follow-up: how would you modify your solution if the array is circular, meaning the subarray can wrap around from the end back to the beginning?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I fumbled.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

First, clarify the original problem and solution (e.g., maximum subarray sum). Then, explain that for a circular array, the maximum subarray can either be non-wrapping (original solution) or wrapping (total sum minus the minimum subarray sum). Finally, handle edge cases like all negative numbers.

Pro tip: Mention that the wrapping case is equivalent to finding the minimum subarray sum and subtracting it from the total sum, but be careful when all numbers are negative—then the maximum subarray is the non-wrapping one.

1. Restate the original problem and solution

Briefly summarize the original problem (e.g., maximum subarray sum) and the approach used (e.g., Kadane's algorithm). This sets the context for the modification.

2. Identify the two cases for circular arrays

Explain that the maximum subarray can either not wrap around (handled by the original solution) or wrap around (which means it consists of a suffix and a prefix).

3. Derive the wrapping case solution

Show that the wrapping case is equivalent to total sum minus the minimum subarray sum (non-circular). This is because the elements not in the maximum wrapping subarray form a contiguous minimum subarray.

4. Handle edge cases

If all numbers are negative, the minimum subarray sum equals the total sum, leading to an empty subarray for the wrapping case. In that case, return the non-wrapping maximum (which is the least negative number).

5. Combine results and analyze complexity

The final answer is the maximum of the non-wrapping and wrapping cases. Time complexity remains O(n) with two passes (one for max subarray, one for min subarray), and space O(1).

Key Points to Mention

  • Kadane's algorithm for maximum subarray sum
  • Kadane's algorithm for minimum subarray sum
  • Total sum of the array
  • Edge case: all negative numbers
  • Time and space complexity analysis
  • Proof that wrapping case = total sum - min subarray sum

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