The Triplet Hunt

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L5
Asked in
Ripple

Problem

You are reconciling a ledger: nums is a list of integer adjustments and k is a target balance. Return every unique trio of values that together sum to k, each trio listed from its smallest value to its largest. The same three values never appear twice, and the trios come back in ascending order.

Every path that works gets a seat at the table.

Example 1

Input:

nums = [-4,-1,-1,0,1,2], k = 0

Output:

[[-1,-1,2],[-1,0,1]]

Example 2

Input:

nums = [1,2,3,4,5], k = 9

Output:

[[1,3,5],[2,3,4]]

Worked solution and explanation

What this really is

Strip the ledger framing and this is 3-sum: find every unique value-trio that hits a target. Everyone reaches for the triple loop; what separates candidates is killing duplicate trios without leaning on a set of tuples. Sort first, fix the smallest element, then sweep the rest from both ends. The trap is the dedup: skip equal values at the fixed index AND after each hit, or you emit [-1, 0, 1] twice and fail on any input with repeats.

Break down the requirements

Step 1: Sort the input

Sorting enables two-pointer search and makes duplicate detection trivial.

Step 2: Outer loop fixes one element; inner loop uses two pointers

For each i, search [i+1, n-1] for pairs summing to k - nums[i].

Step 3: Skip duplicates at all levels

Advance past repeated values after recording each triplet.

The solution

Sorted array with fix-one two-pointer search
def the_triplet_hunt(nums: list[int], k: int) -> list[list[int]]:
    nums.sort()
    n = len(nums)
    triplets = []
    for i in range(n - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        left, right = i + 1, n - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == k:
                triplets.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < k:
                left += 1
            else:
                right -= 1
    return triplets

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