← Verkada Inc. Interview Insights
Start by clarifying the problem constraints and edge cases, then explain the optimal O(n) Kadane's algorithm. Walk through a small example to demonstrate understanding, and finally discuss potential optimizations or variations.
Pro tip: Mention that Kadane's algorithm can be adapted to return the subarray itself, and highlight its linear time complexity as optimal for this problem. Also, briefly note how you would handle all-negative arrays.
Ask about input size, possible values (negative, zero, positive), and whether the subarray must be non-empty. Confirm the expected return type.
Briefly mention the O(n^2) or O(n^3) solution to show you understand the problem, then transition to the optimal approach.
Describe the dynamic programming approach: maintain current sum and max sum, resetting current sum to 0 when it becomes negative. Emphasize O(n) time and O(1) space.
Use a small array (e.g., [-2,1,-3,4,-1,2,1,-5,4]) to illustrate how the algorithm works step by step, showing updates to current and max sums.
Mention how to return the subarray indices, handle all-negative arrays, and note that this is optimal for the problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints (e.g., unlimited coin supply, positive denominations, target >= 0). Then explain that this is a classic dynamic programming problem where you build up the minimum coins for each amount from 0 to target, using the recurrence dp[i] = min(dp[i - coin] + 1) for each coin. Finally, discuss time/space complexity and possible optimizations or edge cases.
Pro tip: Mention that while greedy works for some coin systems (like US coins), it fails for arbitrary denominations, so DP is the safe general solution. Also, note that you can optimize space to O(target) and that BFS can be an alternative if you think of it as a shortest-path problem.
Ask about coin denominations (positive integers?), target amount (non-negative?), and whether coins can be reused unlimited times. Discuss edge cases like target = 0 (return 0) or impossible amounts (return -1).
Let dp[i] be the minimum coins to make amount i. Initialize dp[0] = 0 and dp[i] = infinity for i > 0. For each amount i from 1 to target, and for each coin c <= i, update dp[i] = min(dp[i], dp[i - c] + 1).
After filling the DP table, if dp[target] is still infinity, return -1; otherwise return dp[target]. Mention that you can use a large number like target+1 as infinity to avoid overflow.
Time complexity is O(target * number of coins), space is O(target). You can reduce space to O(target) (already minimal) and note that BFS can solve it in O(target * coins) as well but may be more intuitive for shortest path.
Walk through a small example (e.g., coins = [1,2,5], target = 11) to verify. Compare with greedy approach and explain why greedy fails for some denominations (e.g., coins = [1,3,4], target = 6).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.