1159MediumArrayTwo PointersBinary SearchSorting

Triples Totalling Under the Limit

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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 tray holds the readings readings.

Return how many triples of positions i < j < k have readings[i] + readings[j] + readings[k] strictly below limit.

Examples

Example 1

Input: readings = [0, 0, 0], limit = 1
Output: 1

The tray has one triple and it totals nothing, which is below one.

Example 2

Input: readings = [1, 1, 1], limit = 3
Output: 0

The one triple totals 3, which is not strictly below 3.

Example 3

Input: readings = [-100, -100, -100], limit = -100
Output: 1

The three readings total -300, comfortably below the limit.

Constraints

0 <= readings.length <= 3500
-100 <= readings[i] <= 100
-100 <= limit <= 100
the answer is at most 10^9

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 three_sum_smaller(readings: list[int], limit: int) -> int:

Java

public int threeSumSmaller(int[] readings, int limit)

The intended approach

Time O(n^2) after ordering the tray.Space O(n) for the ordered copy.

Which positions a triple comes from never matters, only the three readings, so the tray can be put in order first. That turns the counting into a sweep rather than a search. Fix the smallest of the three and set one marker just after it and another at the far end. If the three readings the markers pick out fall below the limit, then so does every reading between the two markers standing in for the far one, because they are all no larger. So the whole run between the markers can be counted in one stroke and the near marker stepped forward. If the three do not fall below the limit, the far marker has to come in, since nothing smaller than it will help while the near marker stays put. Each fixed smallest reading costs one pass of the two markers, so the whole count is the tray squared rather than cubed.

Pythonentry point: three_sum_smaller

def three_sum_smaller(readings, limit):
    order = sorted(readings)
    n = len(order)
    total = 0
    for i in range(n - 2):
        lo = i + 1
        hi = n - 1
        while lo < hi:
            if order[i] + order[lo] + order[hi] < limit:
                total += hi - lo
                lo += 1
            else:
                hi -= 1
    return total

The obvious version

Take every triple of positions in increasing order, add the three readings and count the ones falling below the limit. Costs the tray cubed.

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_three_sum_smaller(readings, limit):
    n = len(readings)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if readings[i] + readings[j] + readings[k] < limit:
                    total += 1
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 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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