The Live Coding Round

By the time you reach the live coding round, you have already cleared an assessment that graded itself. This round puts a person on the other side of the screen. It is a single problem, usually inside an hour, run on a shared editor (CoderPad is common) or, if you would rather, your own IDE with your screen shared. Autocomplete and IntelliSense are fine, and you can code in whatever language you are fastest in. The interviewer is there to watch you work, and by candidate reports they lean collaborative: go quiet for a minute on a stuck step and a hint tends to arrive rather than silence.

The problem itself sits around the easy to lower-medium range and leans practical over tricky. There is no exotic data structure to recall. Working code that actually runs, with the time and space complexity stated out loud, is the baseline everyone is expected to hit. A pass comes from the work around the algorithm. Sketch the approach before you type, name the edge cases in the prompt, and test your own solution against them before the interviewer has to ask.

How the round runs

Expect one problem for roughly an hour. You get the prompt, you talk it through, you write real code, and you run it. It is live and non-adversarial, so the interviewer will often nudge you past a stuck point or point at a case you missed. Treat it as a conversation rather than a silent exam. The named problems that recur are small and concrete: the highest-frequency substring of a fixed length, merging several sorted lists, and a sort with the complexity spelled out. None is a research problem. All three reward the same habits.

What it rewards

Three things carry the signal, and none of them is raw cleverness. First, communication: say the approach and its cost before you commit to it, so the interviewer can steer early instead of watching you finish the wrong thing. Second, testing: run your code, and run it against the awkward inputs, not just the happy path. Third, edge cases: the empty input, the single element, the tie, the boundary. State the complexity in both time and space when you land the solution, and be ready to say what you would change if an input got large. Do that consistently and a modest algorithm reads as a strong hire.

Asked at HubSpot

These three are the ones that come up. Build each one until you can talk through the approach, code it cleanly, name its edge cases, and state its complexity without pausing. That fluency, more than any trick, is what the round is measuring.

1

Most Frequent Substring: Fixed-Window Counting

EasyFrequently asked

Problem

Given a string text and an integer k, return the length-k substring that appears most often, counting overlapping occurrences. If several substrings tie for the highest count, return the lexicographically smallest one. If text is shorter than k, return the empty string.

For example, in "abcabc" with k = 3 the windows are abc, bca, cab, abc; "abc" appears twice and wins.

Input: text = "abcabc", k = 3

Output: "abc"

The four windows are abc, bca, cab, abc. "abc" appears twice, every other window once.

Input: text = "abab", k = 1

Output: "a"

"a" and "b" each appear twice; the tie breaks to the lexicographically smaller "a".

How this uses the pattern

Slide a fixed window of length k across the string, count each window in a hash map, then pick the largest count with a deterministic tie-break.

Most Frequent Substring: Fixed-Window Counting

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Test Cases

Input: ["abcabc", 3]Expected: "abc"

Input: ["aaaa", 2]Expected: "aa"

Input: ["abcde", 2]Expected: "ab"

Input: ["abab", 1]Expected: "a"

Input: ["ab", 3]Expected: ""

Need a hint? (3 available)

The HubSpot angle

HubSpot runs this as a talk-through, not a speed run. The interviewer wants to hear you name the window mechanic and state the cost as you write it. Slicing each window is O(n·k), and a rolling hash gets you to O(n). Picking the simpler one on purpose is a fine answer as long as you say why. Bring up the short-string case and the tie before you are asked, because "which one do you return when two tie?" is the follow-up that catches people who never defined it.

Take it further

A natural extension is to return the top N substrings by frequency rather than just the winner, ordered by count and then lexicographically. The counting is identical; what changes is that you rank the map instead of scanning it once for a single max.

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2

Merge k Sorted Lists: Min-Heap Merge

MediumCommonly asked

Problem

You are given lists, a list of already-sorted integer lists. Merge them into one list sorted in non-decreasing order and return it. Any of the inner lists may be empty, and lists itself may be empty.

This one is often framed on linked lists in the room, where each list is a chain of nodes rather than an array. The merge technique is identical; only the traversal boilerplate changes.

Input: lists = [[1,4,5],[1,3,4],[2,6]]

Output: [1,1,2,3,4,4,5,6]

The three sorted lists interleave into a single sorted run.

Input: lists = []

Output: []

No lists means nothing to merge.

How this uses the pattern

Push the front of each list into a min-heap, then repeatedly pop the smallest and push the next element from the list it came from.

Merge k Sorted Lists: Min-Heap Merge

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Test Cases

Input: [[[1,4,5],[1,3,4],[2,6]]]Expected: [1,1,2,3,4,4,5,6]

Input: [[]]Expected: []

Input: [[[]]]Expected: []

Input: [[[1],[0]]]Expected: [0,1]

Input: [[[-3,-1,2],[-2,0],[5]]]Expected: [-3,-2,-1,0,2,5]

Need a hint? (3 available)

The HubSpot angle

The interviewer is happy to hand you the ListNode boilerplate, because the boilerplate is not the point. The signal is whether you reach for a heap and can say why it beats gluing everything together and sorting. A heap of size k gives O(N log k), against O(N log N) for the naive sort. Say that out loud. Then test the empty-list and single-list cases first, since a rushed heap version tends to throw on exactly those.

Take it further

A common variation asks you to do it without a heap. Divide and conquer: pair the lists up and merge each pair, halving the number of lists on every pass. It reaches the same O(N log k) with O(1) extra structure, and it is the answer to "what if you cannot use a priority queue?"

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3

Top K Frequent Words: Rank the Full List

EasySometimes asked

Problem

Given words, a list of strings, return the distinct words ranked by how often they appear, most frequent first. When two words appear equally often, order them by ascending lexicographic order. Return an empty list when words is empty.

Input: words = ["i","love","leetcode","i","love","coding"]

Output: ["i","love","coding","leetcode"]

"i" and "love" appear twice, and "i" sorts before "love". "coding" and "leetcode" appear once, and "coding" sorts before "leetcode".

Input: words = ["b","a","b","c","a","a"]

Output: ["a","b","c"]

Counts are a=3, b=2, c=1, already in descending order.

How this uses the pattern

Count the words, then sort the distinct words by a compound key: count descending, then word ascending.

Top K Frequent Words: Rank the Full List

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Test Cases

Input: [["i","love","leetcode","i","love","coding"]]Expected: ["i","love","coding","leetcode"]

Input: [["b","a","b","c","a","a"]]Expected: ["a","b","c"]

Input: [[]]Expected: []

Input: [["z","z","a"]]Expected: ["z","a"]

Input: [["m","n","m","n"]]Expected: ["m","n"]

Need a hint? (3 available)

The HubSpot angle

The algorithm here is easy on purpose. The round tests the sort more than the counting. HubSpot wants the comparator said out loud, descending by count and then ascending by word, and the complexity that comes with it, O(m log m) over m distinct words plus O(n) to count them. If you say "sort by frequency" and stop, the follow-up is immediate: what breaks a tie? Put the answer in the comparator, and it never has to be a conversation.

Take it further

The classic tightening is to return only the top K instead of the full ranking. A heap of size K gets the answer in O(n + m log K), which beats a full sort when K is small relative to the number of distinct words.

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