0561MediumArrayHash Table

Triple Readings That Split Evenly

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Trains the technique from

LeetCode 2964Number of Divisible Triplet Sums

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A monitoring rig stores one reading per channel in the array nums, and a reporting tool batches readings into groups whose total has to divide evenly among d shifts.

Count the position triples (i, j, k) with i < j < k for which nums[i] + nums[j] + nums[k] is a multiple of d. Two triples are the same triple when they use the same three positions, so each set of three positions is counted once even if the readings there are equal.

Return that count.

Examples

Example 1

Input: nums = [9, 4, 7, 8, 6, 12], d = 5
Output: 3

Three position triples qualify: (0, 1, 2) totals 20, (0, 1, 5) totals 25 and (2, 4, 5) totals 25, and each of those is a multiple of 5.

Example 2

Input: nums = [2, 2, 2, 2], d = 3
Output: 4

Every triple of positions totals 6, which is a multiple of 3, and there are 4 ways to choose three of the four positions.

Example 3

Input: nums = [6, 9], d = 4
Output: 0

Fewer than three readings exist, so no triple can be formed.

Constraints

1 <= nums.length <= 1000
1 <= nums[i] <= 10^9
1 <= d <= 10^9
With at most 1000 readings the count never exceeds 166167000.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def divisible_triplet_count(nums: list[int], d: int) -> int:

Java

public int divisibleTripletCount(int[] nums, int d)

The intended approach

Time O(n^2) lookups, one per pair of middle and last position.Space O(min(n, d)) for the residue tally.

Walk the middle position `j` from left to right, keeping a hash map that tallies the residues modulo `d` of every reading strictly before `j`. For each later position `k`, the earlier reading has to carry the residue that brings the three-way total to zero modulo `d`, so one map lookup adds all matching choices of `i` at once. Building the map incrementally is what keeps the position order `i < j < k` intact.

Pythonentry point: divisible_triplet_count

def divisible_triplet_count(nums, d):
    # O(n^2) time, O(min(n, d)) space: fix the middle and the last position, then
    # look up how many earlier readings carry the residue that completes the sum.
    n = len(nums)
    total = 0
    seen = {}
    for j in range(1, n):
        r = nums[j - 1] % d
        seen[r] = seen.get(r, 0) + 1
        base = nums[j] % d
        for k in range(j + 1, n):
            need = (-(base + nums[k])) % d
            total += seen.get(need, 0)
    return total

The obvious version

Three nested loops over every increasing position triple, adding the three readings and testing the remainder directly.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_divisible_triplet_count(nums, d):
    n = len(nums)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if (nums[i] + nums[j] + nums[k]) % d == 0:
                    total += 1
    return total

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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