0562MediumArrayTwo PointersBinary SearchGreedySorting

Bracing Strut Triples

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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 workshop has a pile of steel struts; nums[i] is the length of the strut in slot i, and a length of 0 marks a slot holding an offcut that is too short to use. The struts arrive in no particular order.

Three struts can be bolted into a rigid triangular brace exactly when each of their lengths is strictly less than the sum of the other two, which is what gives the brace a positive area. A flat, collapsed arrangement does not count.

Count the triples of slots (i, j, k) with i < j < k whose three struts form a rigid brace. Struts of equal length in different slots are different struts, so each triple of slots is counted separately.

Examples

Example 1

Input: nums = [6, 3, 4, 6]
Output: 4

All four triples of slots qualify: 6, 3, 4 has every length under the sum of the other two, and so do 6, 3, 6 and 6, 4, 6 and 3, 4, 6.

Example 2

Input: nums = [2, 3, 5]
Output: 0

The only triple has 2 + 3 equal to 5, so the arrangement is flat rather than rigid and nothing is counted.

Example 3

Input: nums = [0, 1, 1]
Output: 0

Slot 0 holds an offcut of length 0, and 0 + 1 does not exceed 1, so the only triple fails.

Constraints

1 <= nums.length <= 1000
0 <= nums[i] <= 1000
With at most 1000 struts 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 triangle_number(nums: list[int]) -> int:

Java

public int triangleNumber(int[] nums)

The intended approach

Time O(n^2): the sort is O(n log n) and each of the n sweeps is linear.Space O(n) for the sorted copy of the lengths.

Sort the lengths, which lets each candidate be described by the position of its longest strut: once the lengths are ascending, only the pair of shorter struts has to beat the longest one, since the other two comparisons hold automatically. Fix that longest position, then sweep a low and a high pointer over the shorter prefix: when the pair clears the longest strut, every position between the two pointers also clears it with the high strut, so the whole span is counted at once and the high pointer steps down; otherwise the low pointer steps up.

Pythonentry point: triangle_number

def triangle_number(nums):
    # O(n^2) time, O(n) space: with the lengths in ascending order the longest
    # side of a candidate is fixed by its position, so a two-pointer sweep counts
    # every partner pair for it in linear time.
    arr = sorted(nums)
    n = len(arr)
    total = 0
    for top in range(n - 1, 1, -1):
        lo = 0
        hi = top - 1
        while lo < hi:
            if arr[lo] + arr[hi] > arr[top]:
                total += hi - lo
                hi -= 1
            else:
                lo += 1
    return total

The obvious version

Three nested loops over every increasing slot triple, testing all three inequalities on the untouched input order.

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_triangle_number(nums):
    n = len(nums)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                a = nums[i]
                b = nums[j]
                c = nums[k]
                if a + b > c and a + c > b and b + c > a:
                    total += 1
    return total

How this problem was checked

Reference solution executed against 13 test cases — 13 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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