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Cooler Gauges Downstream

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Trains the technique from

LeetCode 315Count of Smaller Numbers After Self

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.

Temperature gauges are fixed along a cooling pipe in the order the coolant flows through them. readings[i] is the temperature in degrees at the gauge in position i, and a gauge lies downstream of another when its position is larger.

For each gauge, an engineer wants to know how many of the gauges downstream of it read strictly cooler than it does. A gauge reading exactly the same temperature does not count, and gauges upstream are ignored.

Return an array cooler of the same length as readings, where cooler[i] is that count for the gauge in position i.

Examples

Example 1

Input: readings = [9, 4, 12, 2]
Output: [2, 1, 1, 0]

Downstream of the 9 sit 4, 12 and 2, of which 4 and 2 are cooler, giving 2. Downstream of the 4 sit 12 and 2, of which only 2 is cooler, giving 1. Downstream of the 12 sits 2 alone, giving 1, and the last gauge has nothing downstream, giving 0.

Example 2

Input: readings = [6, 6, 6, 6]
Output: [0, 0, 0, 0]

Every gauge reads the same temperature, and an equal reading is not cooler, so no gauge counts any of the ones downstream of it.

Example 3

Input: readings = [-5, -9, -5, -20, 0]
Output: [2, 1, 1, 0, 0]

The first gauge reads -5 and finds -9 and -20 cooler downstream, giving 2, while the -5 in the middle finds only -20, giving 1. The gauge reading -20 has just the 0 downstream, which is warmer, so its count is 0.

Constraints

1 <= readings.length <= 10^5
-10^4 <= readings[i] <= 10^4

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 cooler_downstream(readings: list[int]) -> list[int]:

Java

public List<Integer> coolerDownstream(int[] readings)

The intended approach

Time O(n log V) for n gauges over a temperature range of V = 2 * 10^4 + 1 values: each gauge costs one prefix query and one update.Space O(V) for the tally, plus the returned array.

Sweep the pipe from the last gauge back to the first, keeping a Fenwick tally of the temperatures already swept, which are exactly the ones downstream of the current gauge. The temperatures come from a fixed range, so a reading is used directly as a tally index after shifting it up by 10001 to keep every index positive. Answering one gauge is a prefix query one index below its own shifted reading, which excludes equal readings, and the gauge is then added to the tally before moving upstream.

Pythonentry point: cooler_downstream

def cooler_downstream(readings):
    # a Fenwick tally over shifted readings, filled from the last gauge backwards
    SHIFT = 10001            # -10^4 maps to 1, so no index is ever zero
    span = 20002
    tally = [0] * (span + 1)
    out = [0] * len(readings)
    for i in range(len(readings) - 1, -1, -1):
        slot = readings[i] + SHIFT - 1
        seen = 0
        while slot > 0:
            seen += tally[slot]
            slot -= slot & (-slot)
        out[i] = seen
        slot = readings[i] + SHIFT
        while slot <= span:
            tally[slot] += 1
            slot += slot & (-slot)
    return out

The obvious version

For each gauge, walk every later position and count the readings strictly below it. Quadratic, but it is the rule written out literally.

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_cooler_downstream(readings):
    out = []
    for i in range(len(readings)):
        seen = 0
        for j in range(i + 1, len(readings)):
            if readings[j] < readings[i]:
                seen += 1
        out.append(seen)
    return out

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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