Start by clarifying the problem: singly linked list, iterative and/or recursive, return new head. Then present the iterative solution with three pointers (prev, curr, next), walking through the pointer manipulation. If time permits, also present the recursive solution, explaining the base case and recursive step.
Pro tip: Always draw the linked list and pointer movements on a whiteboard or paper before coding; it prevents off-by-one errors and demonstrates systematic thinking. Also, mention edge cases like empty list or single node upfront.
Confirm whether to implement iteratively, recursively, or both. Ask about input constraints (e.g., empty list, single node) and expected return (new head).
Describe using three pointers: prev (initially null), curr (head), and next. Iterate while curr is not null, reversing the link and advancing pointers.
Trace the algorithm on a small list (e.g., 1->2->3) to show how pointers change and why it works. This builds confidence and catches errors.
Write clean, bug-free code for the iterative solution. If asked, also implement the recursive version, explaining the base case and recursive call.
State time and space complexity for both approaches. Discuss trade-offs (iterative uses O(1) space, recursive uses O(n) stack space). Suggest test cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one actually tripped me up more than I expected.
First, identify the layer (ring) of the spiral that contains the given coordinate by computing the maximum of the absolute values of x and y. Then, determine the side of the square ring the point lies on and calculate the offset from the ring's starting corner to compute the exact value using arithmetic formulas.
Pro tip: Emphasize that the solution is O(1) by deriving a closed-form formula, and mention that you would validate it with a few test cases, including edge cases like the origin and points on the axes.
Compute the layer number k as the maximum of |x| and |y|. This identifies which square ring the point belongs to.
The largest value in layer k is (2k+1)^2, and the smallest is (2k-1)^2 + 1. Use these to establish the base value for the layer.
Determine which side of the square the point lies on (right, top, left, or bottom) by checking the coordinates relative to the layer boundaries. Compute the offset along that side from the starting corner.
Use the base value and offset to calculate the final value with a simple arithmetic expression, ensuring the direction of the spiral (counterclockwise outward) is correctly accounted for.
Test the formula with known points such as (0,0)=1, (1,0)=2, (1,1)=3, etc., to confirm correctness and handle any special cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem by tracking how many times each bulb is toggled, which equals the number of divisors of its position. Recognize that only perfect squares have an odd number of divisors, so those bulbs remain on. Conclude that the bulbs at positions 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are on.
Pro tip: Explain the reasoning clearly and connect it to the mathematical property of divisors; this shows you can derive a solution rather than just recall it. Also, mention that this is a classic problem often used to test pattern recognition and mathematical insight.
Clarify that each person k toggles bulbs at positions that are multiples of k, meaning bulb i is toggled once for each divisor of i.
For bulb i, the number of toggles equals the number of divisors of i. Initially off, so it ends on if toggled an odd number of times.
Most numbers have an even number of divisors because they pair up, except perfect squares where one divisor is repeated (the square root).
Thus, only bulbs at perfect square positions (1, 4, 9, ..., 100) are toggled an odd number of times and remain on.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Copy the next node's value into the current node, then skip over the next node.
Explain that since we don't have access to the previous node, we can't delete the given node directly. Instead, copy the data from the next node into the current node, then delete the next node. This achieves O(1) time complexity by effectively overwriting the current node with the next node's data and bypassing the next node.
Pro tip: Mention the edge case where the node to delete is the last node: in that case, this approach fails because there is no next node. In a real interview, discuss how you would handle it (e.g., mark it as dummy or assume it's not the last node).
Confirm that the node to delete is not the last node and that the list is singly linked. Also, ensure that the node is guaranteed to be in the list.
Describe the approach: copy the data from the next node into the current node, then update the current node's next pointer to skip the next node.
Discuss what happens if the node is the last node: the approach doesn't work directly. Mention possible solutions or assumptions.
State that the time complexity is O(1) and space complexity is O(1), as we only modify pointers and copy data.
Write the code: node.data = node.next.data; node.next = node.next.next; (and optionally free the next node in languages with manual memory management).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The key insight is that you never want the slow people (5 and 6) to make a return trip.
Start by restating the problem and clarifying constraints. Then, explain the two main strategies for bridge crossing puzzles: the 'fastest two shuttle' method and the 'two slowest together' method. Finally, apply the optimal strategy step-by-step to achieve a total time under 13 minutes, verifying each move.
Pro tip: Demonstrate algorithmic thinking by comparing the two strategies and showing why the chosen one is optimal. This shows you can analyze trade-offs, which is crucial for software engineering.
Restate the problem: four people with crossing times 1, 2, 5, 6 minutes, one torch, at most two cross at a time, pair moves at slower pace. Goal: total time < 13 minutes.
Recognize two common strategies: (A) fastest two shuttle the torch to bring others over, (B) two slowest cross together while fastest return. Compare their total times.
Use strategy B: 1 and 2 cross (2 min), 1 returns (1 min), 5 and 6 cross (6 min), 2 returns (2 min), 1 and 2 cross (2 min). Total = 13 minutes. This is the known optimal solution for times 1, 2, 5, 6.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Went with the min/max bounds approach rather than in-order traversal.
Use a recursive approach that passes down the allowable range (min, max) for each node, ensuring every node's value falls within its valid interval. Alternatively, perform an in-order traversal and verify that the sequence is strictly increasing. Clearly explain the chosen method, its time and space complexity, and handle edge cases like duplicate values.
Pro tip: Mention that a common mistake is only comparing a node with its immediate children; instead, emphasize the need to enforce global constraints via ranges or in-order traversal. Also, discuss how to handle duplicates based on the problem's definition (usually strict inequality).
Ask whether duplicate values are allowed and confirm the definition of a valid BST (e.g., left subtree < node < right subtree).
Decide between the range-based recursive method or the in-order traversal method, and explain why one might be preferred.
Describe the steps of your chosen approach, including how you initialize and update the bounds or how you track the previous node during traversal.
State the time complexity (O(n)) and space complexity (O(h) for recursion stack or O(n) for iterative in-order with stack).
Discuss edge cases such as an empty tree, a single node, duplicate values, and skewed trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sort by start time, then sweep through merging whenever the current interval's start is within the previous one's end.
Start by clarifying the problem constraints (e.g., whether intervals are sorted, inclusive/exclusive endpoints, and expected output format). Then propose a sort-based approach: sort intervals by start time, iterate through them, and merge overlapping intervals by comparing the current interval's start with the previous merged interval's end. Finally, analyze time and space complexity and discuss edge cases.
Pro tip: Mention that sorting is the key to achieving O(n log n) time, and that without sorting, the problem would require O(n^2) comparisons. Also, proactively discuss how you would handle edge cases like empty input or intervals that touch at endpoints.
Ask about input format, whether intervals are sorted, endpoint inclusivity, and expected output. Confirm if intervals are given as pairs [start, end] and if merging touching intervals (e.g., [1,2] and [2,3]) is required.
Propose sorting intervals by start time, then merging in a single pass. Explain why this yields O(n log n) time due to sorting, and O(n) space for the output.
Describe iterating through sorted intervals: if the current interval's start <= last merged interval's end, update the end to max of both ends; otherwise, add the last merged interval to the result and start a new one.
State time and space complexity. Discuss edge cases: empty list, single interval, all overlapping, none overlapping, and intervals with same start times.
Walk through a concrete example, such as [[1,3],[2,6],[8,10],[15,18]] -> [[1,6],[8,10],[15,18]], to verify correctness and demonstrate understanding.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The sub-O(n log n) approach uses a bucket/pigeonhole argument.
First, clarify the problem and constraints, then propose a sorting-based O(n log n) solution as a baseline. Next, introduce the pigeonhole principle to achieve O(n) time by bucketing elements into n-1 intervals, and finally discuss trade-offs and edge cases.
Pro tip: Mention that the O(n) solution uses O(n) extra space, so if memory is constrained, sorting might be preferable. Also, handle duplicates and negative numbers gracefully.
Confirm that the array is unsorted, elements can be any integers (including negatives and duplicates), and we need the maximum difference between consecutive elements in the sorted order.
Sort the array and scan adjacent elements to find the maximum difference. This takes O(n log n) time and O(1) extra space (if sorting in-place).
Use the pigeonhole principle: if there are n elements, the maximum gap is at least ceil((max-min)/(n-1)). Create n-1 buckets of that size, distribute elements, and track min/max per bucket. The maximum gap is between the max of one bucket and the min of the next non-empty bucket.
The bucket approach runs in O(n) time and uses O(n) extra space. Compare with sorting: O(n log n) time, O(1) extra space. Discuss when each is preferable.
Consider arrays with fewer than 2 elements, all elements equal, or large ranges. Ensure the algorithm handles duplicates correctly (they fall into the same bucket, so gaps within buckets are zero).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.