Use a depth-first search (DFS) traversal, passing the current accumulated number down the recursion. At each leaf, add the accumulated number to a running total. This approach is intuitive and efficient, with O(n) time and O(h) space.
Pro tip: Clarify edge cases upfront, such as an empty tree or a tree with a single node, and mention that you can optimize space by using an iterative approach with a stack if recursion depth is a concern.
Confirm that each root-to-leaf path forms a number by concatenating digits, and the goal is to sum all such numbers. Ask clarifying questions about input constraints and edge cases.
Select DFS (preorder) to naturally build numbers along paths. Explain that BFS is also possible but requires storing partial numbers in the queue.
Create a helper function that takes a node and the current number formed so far. Update the number as current * 10 + node.val, and if it's a leaf, return that number; otherwise, recurse on children and sum results.
If the node is null, return 0. If the node is a leaf, return the current number. This ensures correct summation.
State that time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack, where h is tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem: if the array is unsorted, sorting it will reorder elements, so the original indices are lost. Therefore, the problem likely means to find all indices in the original array where the target appears, or to return the indices in the sorted array. Discuss both interpretations and choose one based on interviewer feedback. Then, propose an efficient solution: for original indices, a simple linear scan suffices; for sorted indices, sort the array and use binary search to find the range of target values, then return all indices in that range.
Pro tip: Always clarify ambiguities before coding; interviewers value candidates who identify edge cases and ask questions. Also, consider mentioning that if the array is already sorted, binary search is optimal, but if not, sorting first may be unnecessary overhead.
Ask whether the array is sorted or unsorted, and whether the indices should refer to the original array or the sorted array. Confirm if duplicates exist and if the target may not be present.
For original indices: linear scan O(n). For sorted indices: sort O(n log n) then binary search O(log n) to find range, then collect indices. Compare trade-offs.
If original indices: iterate through array, collect indices where element equals target. If sorted indices: sort array (if needed), use binary search to find first and last occurrence of target, then return all indices in between.
State time and space complexity for each approach. For linear scan: O(n) time, O(1) extra space (excluding output). For sort+binary search: O(n log n) time, O(n) space if sorting a copy, or O(1) if in-place and output not counted.
Consider empty array, target not present, all elements equal to target, and large arrays. Discuss how to handle duplicates efficiently.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use the BST property to traverse from the root, keeping track of the closest value seen so far. At each node, compare the node's value with the target, update the closest if needed, and move left or right based on whether the target is smaller or larger. This yields O(h) time and O(1) space.
Pro tip: Clarify whether the tree is balanced or not, and discuss how the algorithm adapts to skewed trees. Also, mention that you can stop early if you find an exact match, which can improve average performance.
Ask if the tree is a standard BST with unique values, and confirm that 'closest' means minimizing absolute difference. Also, check if the tree can be empty or if the target can be any integer.
Explain that you'll traverse from the root, maintaining the closest value found so far. At each step, compare the current node's value with the target and update the closest if the absolute difference is smaller.
If the target is less than the current node's value, move to the left child; if greater, move to the right child. If equal, return immediately as it's the closest possible.
State that the time complexity is O(h) where h is the height of the tree, and space complexity is O(1) for iterative traversal. Mention that in the worst case (skewed tree), h = n.
Discuss what to do if the tree is empty (return null or throw exception), and how to handle multiple nodes with the same closest difference (any is acceptable).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Stack-based approach: track indices of unmatched open parens, flag unmatched closes as you go, then remove whatever's left.
Use a stack to track indices of unmatched opening parentheses and a set to track indices of unmatched closing parentheses. After one pass, remove characters at those indices to produce a valid string. Alternatively, use a counter-based two-pass approach to identify the minimum removals without extra space.
Pro tip: Clarify whether you need to return any valid string or all possible valid strings; the latter requires backtracking and is more complex. Also, mention that the problem guarantees a unique answer if you remove the minimum number, but multiple valid strings may exist.
Confirm that you need to remove the minimum number of parentheses to make the string valid, and that the string contains only parentheses and letters. Ask if you need to return the resulting string or just the count.
Decide between stack-based (easy to implement, O(n) space) and counter-based (O(1) space) approaches. Explain the trade-offs.
For stack: iterate, push indices of '(', pop for ')', mark unmatched. For counter: first pass left-to-right to remove excess ')', second pass right-to-left to remove excess '('.
Build the output string by skipping characters at indices marked for removal. Ensure the result is valid and has minimum removals.
Test with edge cases: empty string, all parentheses, nested, multiple invalid. Analyze time and space complexity.
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: immutable array, multiple range sum queries, and efficiency requirements. Then propose a prefix sum array that precomputes cumulative sums, enabling O(1) query time with O(n) preprocessing and O(n) space. Explain the trade-offs and mention alternative approaches like segment trees or Fenwick trees, but emphasize that prefix sums are optimal for immutability.
Pro tip: Mention that for immutable arrays, prefix sums are the optimal solution because they achieve constant-time queries with linear preprocessing, and discuss how you would handle edge cases like empty ranges or large sums (e.g., using 64-bit integers).
Ask about the array size, number of queries, whether the array is truly immutable, and if updates are needed. Confirm that queries are frequent and that O(1) query time is desired.
Describe building a prefix sum array where prefix[i] = sum of elements from index 0 to i-1. Then a range sum from i to j is prefix[j+1] - prefix[i].
State that preprocessing takes O(n) time and space, and each query takes O(1) time. Compare with segment trees (O(n) build, O(log n) query) and explain why prefix sums are better for immutable arrays.
Cover empty ranges, large sums (use 64-bit integers), and how to handle negative numbers. Mention that the prefix array can be built in a single pass.
Summarize the solution and optionally mention how to handle updates if the array were mutable (e.g., using a Fenwick tree).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize this as a dynamic programming problem where the minimum cost to reach step i is the cost of step i plus the minimum of the costs to reach steps i-1 and i-2. Define the base cases carefully: you can start at step 0 or step 1 with their respective costs, and the top is beyond the last step. Then implement an iterative solution with O(n) time and O(1) space, explaining each step clearly.
Pro tip: Clarify the problem constraints upfront (e.g., whether the top is beyond the last step, and if costs are non-negative) and mention edge cases like empty or single-step staircases. This shows attention to detail and prevents incorrect assumptions.
Confirm the input format, what 'top' means (usually beyond the last step), and any constraints (e.g., cost values, staircase size). Ask if you can start at step 0 or step 1.
Let dp[i] be the minimum cost to reach step i. Then dp[i] = cost[i] + min(dp[i-1], dp[i-2]) for i >= 2, with base cases dp[0] = cost[0] and dp[1] = cost[1].
Since only the last two values are needed, use two variables to track the minimum costs for the previous two steps, reducing space complexity to O(1).
Write clean code (e.g., in Python) and walk through a small example like cost = [10, 15, 20] to verify the logic. Handle edge cases such as empty array or single step.
State that the time complexity is O(n) and space complexity is O(1). Mention 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.