← Squarepoint Capital Interview Insights
This was the one I was least prepared for going in.
Clarify the definition of drawdown as the decline from a running peak to a subsequent trough, ensuring the returned value is negative. Then implement a single-pass algorithm that tracks the running maximum and the maximum drawdown, recording the start and end indices when a new maximum drawdown is found. Finally, analyze the time and space complexity, noting the O(n) time and O(1) space.
Pro tip: Explicitly state that the drawdown value is negative (e.g., -0.25 for a 25% decline) and confirm the index convention (0-based vs 1-based) to avoid off-by-one errors. Also, mention that the start index is the index of the peak, not the start of the decline.
Confirm that drawdown is the percentage decline from a peak to a trough, returned as a negative number. Clarify whether indices are 0-based or 1-based and whether the start index refers to the peak or the beginning of the decline.
Use a single pass through the series, maintaining the current peak value and its index. For each point, compute the drawdown from the current peak; if it's more negative than the maximum drawdown seen so far, update the maximum drawdown and record the peak index as start and current index as end.
Write clean code with clear variable names (e.g., max_drawdown, peak_value, peak_index). Handle edge cases such as empty series or series with no drawdown (return 0 or None with appropriate indices).
Validate the function with simple cases (e.g., [1, 2, 1, 3] -> max drawdown -0.5 from index 1 to 2) and edge cases (monotonically increasing, monotonically decreasing, constant).
State that the algorithm runs in O(n) time because it makes a single pass, and uses O(1) extra space since only a few variables are needed regardless of input size.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a single pass through the array, tracking the minimum price seen so far and the maximum profit achievable by selling at the current price. Update the minimum and profit at each step, ensuring O(n) time and O(1) space.
Pro tip: Clarify that you're assuming the array represents daily prices and that you can buy and sell on the same day if profitable, but typically the sell must be after the buy. Also, mention that returning 0 handles the case of no profit.
Confirm that the array is non-empty, prices are positive, and that a single buy-sell transaction is allowed. Discuss edge cases like decreasing prices or single element.
Initialize min_price to the first element and max_profit to 0. These will track the lowest price seen and the best profit so far.
For each price, update min_price to the minimum of current min_price and the price. Then compute potential profit as price - min_price and update max_profit if it's larger.
After the loop, return max_profit. If no profit was possible, max_profit remains 0.
State that the algorithm runs in O(n) time because it makes a single pass, and O(1) space because it uses only a few variables.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem and edge cases, then explain the dynamic programming approach using a 1D array where dp[i] represents the minimum coins to make amount i. Initialize dp[0]=0 and others to infinity, iterate through amounts and coins to update dp[i] = min(dp[i], dp[i-coin]+1). Finally, discuss time and space complexity and potential optimizations.
Pro tip: Mention that while DP gives the optimal solution, for large targets or certain coin systems, greedy might work but is not always correct; showing awareness of trade-offs impresses interviewers. Also, consider using BFS for unweighted graph interpretation if appropriate.
Ask about constraints: coin denominations positive? target non-negative? Can target be 0? What if no combination exists? This shows thoroughness.
Let dp[i] be the minimum coins to make amount i. Recurrence: dp[i] = min(dp[i - coin] + 1) for all coins ≤ i. Base case: dp[0] = 0, others infinity.
Iterate i from 1 to target, and for each coin, update dp[i] if i >= coin. Return dp[target] if finite else -1.
Time complexity O(target * number of coins), space O(target). Mention that this is pseudo-polynomial.
Mention space optimization (only need dp array), or using BFS for unweighted graph, or greedy with caveats.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.