0164HardArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetTreap

Steep Decline Pairs

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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 transmitter logs one signed power level per sample into levels, in the order the samples were taken. Levels are calibration offsets, so they may be negative, zero or positive.

An analyst wants to know how often the signal fell away sharply between two samples, no matter how far apart those samples sat. Sample i and sample j make a steep decline when i < j and levels[i] > 2 * levels[j], using ordinary signed arithmetic. The comparison is strict: a level that is exactly double the later one does not qualify.

Count the steep declines in the log and return that count. Note that a pair is judged only by its two levels and by which sample came first; the samples in between are irrelevant.

Examples

Example 1

Input: levels = [6, 1, 4, 2]
Output: 2

The pair of samples holding 6 then 1 qualifies because 6 beats 2, and 6 then 2 qualifies because 6 beats 4. The pair holding 4 then 2 fails, since 4 is exactly double and not more.

Example 2

Input: levels = [-1, -5, 3]
Output: 1

Only -1 followed later by -5 qualifies, because doubling -5 gives -10 and -1 sits above it.

Example 3

Input: levels = [10, 5]
Output: 0

Doubling the later level gives exactly the earlier one, and the rule needs a strict excess, so nothing is counted.

Constraints

1 <= levels.length <= 5 * 10^4
-2^31 <= levels[i] <= 2^31 - 1
The count is at most 10^10.

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 steep_declines(levels: list[int]) -> int:

Java

public int steepDeclines(int[] levels)

The intended approach

Time O(n log n)Space O(n)

Split the log in half. Every steep decline is either entirely inside one half or crosses the split, so recurse on both halves first, which also leaves each half sorted. Reordering a half is harmless because the split itself already records which sample came first, and a crossing pair only needs that. With both halves sorted the crossing pairs fall out in linear time: sweep the left half upward and advance a single pointer over the right half while the left level still beats double the right level, then add how far that pointer travelled. Merging the two sorted halves afterwards hands the caller's level up the recursion ready for its own count, so the whole thing is a merge sort that returns a count alongside the sorted data.

Pythonentry point: steep_declines

def steep_declines(levels):
    # Time O(n log n): a merge sort where each merge also does one linear
    # two-pointer sweep to count the declines crossing the split.
    # Space O(n): a working copy of the log plus a merge buffer.
    work = list(levels)
    buffer = [0] * len(work)

    def sort_and_count(lo, hi):
        if hi - lo <= 1:
            return 0
        mid = (lo + hi) // 2
        found = sort_and_count(lo, mid) + sort_and_count(mid, hi)
        right = mid
        for left in range(lo, mid):
            while right < hi and work[left] > 2 * work[right]:
                right += 1
            found += right - mid
        a = lo
        b = mid
        out = lo
        while a < mid and b < hi:
            if work[a] <= work[b]:
                buffer[out] = work[a]
                a += 1
            else:
                buffer[out] = work[b]
                b += 1
            out += 1
        while a < mid:
            buffer[out] = work[a]
            a += 1
            out += 1
        while b < hi:
            buffer[out] = work[b]
            b += 1
            out += 1
        work[lo:hi] = buffer[lo:hi]
        return found

    return sort_and_count(0, len(work))

The obvious version

Test every ordered pair of samples directly with two nested loops and tally the ones that satisfy the rule.

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_steep_declines(levels):
    total = 0
    n = len(levels)
    for i in range(n):
        for j in range(i + 1, n):
            if levels[i] > 2 * levels[j]:
                total += 1
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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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