0760MediumArrayHash TableTwo PointersSortingCounting

Weight Triples Hitting The Target

Tracked in this browser only

Trains the technique from

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 depot belt has weighed every parcel in a batch, giving loads, where loads[i] is the weight of the parcel at position i in kilograms.

A dispatcher wants to know how many ways three parcels can be pulled off the belt so their weights add up to exactly target kilograms. Count triples of positions (i, j, k) with i < j < k and loads[i] + loads[j] + loads[k] == target. Two triples are different when the positions differ, even if the three weights are identical.

The count can be enormous, so return it modulo 1000000007.

Examples

Example 1

Input: loads = [2, 2, 5, 3, 1, 9], target = 10
Output: 2

Positions (0, 2, 3) weigh 2, 5 and 3, which add to 10, and positions (1, 2, 3) weigh the same three amounts. No other choice of three positions adds to 10.

Example 2

Input: loads = [50, 50, 50, 50, 100, 0], target = 150
Output: 8

Any three of the four parcels weighing 50 kg add to 150 kg, which is four triples. Taking the 0 kg parcel, the 100 kg parcel and one of the four 50 kg parcels also adds to 150 kg, which is four more.

Example 3

Input: loads = [4, 5, 6], target = 15
Output: 1

There is only one choice of three positions on this belt, and 4 + 5 + 6 is 15, so it qualifies.

Constraints

3 <= loads.length <= 3000
0 <= loads[i] <= 100
0 <= target <= 300

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 count_weight_triples(loads: list[int], target: int) -> int:

Java

public int countWeightTriples(int[] loads, int target)

The intended approach

Time O(n + W^2) for n parcels and W = 101 weights.Space O(W) for the tally.

Only 101 different weights exist, so tally how many parcels carry each one and then loop over weight pairs `x <= y`, deriving `z = target - x - y` and keeping it only when `y <= z <= 100`. Turn each qualifying weight triple into a number of position triples by case: all three weights equal contributes `C(c, 3)` from that one tally, two equal contributes `C(c, 2)` times the third tally, and three different weights contributes the plain product. Reduce modulo `1000000007` as the running total grows.

Pythonentry point: count_weight_triples

def count_weight_triples(loads, target):
    # Tally how many parcels carry each weight, then walk weight triples
    # x <= y <= z once and turn each into a count of index triples.
    # Time O(W^2) for W = 101 distinct weights, plus O(n) to tally. Space O(W).
    MOD = 1000000007
    tally = [0] * 101
    for w in loads:
        tally[w] += 1
    total = 0
    for x in range(101):
        if tally[x] == 0:
            continue
        for y in range(x, 101):
            if tally[y] == 0:
                continue
            z = target - x - y
            if z < y or z > 100 or tally[z] == 0:
                continue
            cx = tally[x]
            if x == y == z:
                total += cx * (cx - 1) * (cx - 2) // 6
            elif x == y:
                total += cx * (cx - 1) // 2 * tally[z]
            elif y == z:
                cy = tally[y]
                total += cx * (cy * (cy - 1) // 2)
            else:
                total += cx * tally[y] * tally[z]
            total %= MOD
    return total % MOD

The obvious version

Three nested loops over positions `i < j < k`, adding one whenever the three weights hit the target. Cubic in the number of parcels, so only usable on short belts.

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_count_weight_triples(loads, target):
    MOD = 1000000007
    n = len(loads)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if loads[i] + loads[j] + loads[k] == target:
                    total += 1
    return total % MOD

How this problem was checked

Reference solution executed against 11 test cases — 11 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in