Free AI LeetCode Communication Coach — 3Sum

The problem

Given an integer array nums, return all unique triplets that sum to zero.

Example: nums = [-1,0,1,2,-1,-4] → [[-1,-1,2],[-1,0,1]]

Sort + two pointers + duplicate skip.

nums.sort()
res = []
for i in range(len(nums)):
    if i and nums[i] == nums[i-1]:
        continue
    left, right = i + 1, len(nums) - 1
    while left < right:
        total = nums[i] + nums[left] + nums[right]
        if total == 0:
            res.append([nums[i], nums[left], nums[right]])
            left += 1
            while left < right and nums[left] == nums[left-1]:
                left += 1
        elif total < 0:
            left += 1
        else:
            right -= 1
return res

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “I sort, fix one index, then two pointers.”
• “I skip duplicates on i, left, and right.”

Practise out loud

1. 1. Restate

   I need every unique triplet that adds to zero.

2. 2. Approach

   I sort the array. For each index i I run two pointers on the rest looking for a pair that sums to minus nums[i]. I skip duplicates.

3. 3. Example

   On −1,0,1,2,−1,−4 after sort, when i is −1 I find −1,2 and 0,1. I skip the next −1 so I do not repeat.

4. 4. Time and space

   Time is O(n²). Extra space is O(1) besides the output.

5. 5. Edge cases

   I would check arrays shorter than three and the all-zeros case.

Common mistakes when explaining 3Sum

• Forgetting to say how you skip duplicates.
• Claiming O(n) when the two-pointer loop is O(n²).