0344HardArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetTreap

Reservoir Swing Bands

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 reservoir logs one net level change per hour in nums, where a negative entry means the level dropped that hour and a positive entry means it rose.

A stretch is any run of one or more consecutive hours, written nums[i .. j] with i <= j. Its swing is the sum of the entries it covers. Two stretches are different when their hour ranges differ, even if their swings agree.

Given the inclusive band lower to upper, return how many stretches have a swing of at least lower and at most upper.

Examples

Example 1

Input: nums = [3,-4,2,5,-6], lower = -1, upper = 4
Output: 8

Eight stretches land inside the band: `nums[0..0]`, `nums[0..1]`, `nums[0..2]`, `nums[0..4]`, `nums[1..3]`, `nums[2..2]`, `nums[2..4]` and `nums[3..4]`, with swings of 3, -1, 1, 0, 3, 2, 1 and -1. Every one of those swings sits between -1 and 4 inclusive.

Example 2

Input: nums = [8,-8], lower = 0, upper = 0
Output: 1

Only `nums[0..1]` sums to 0. The single-hour stretches swing by 8 and -8, both outside the band.

Example 3

Input: nums = [6], lower = 6, upper = 9
Output: 1

The one stretch available swings by 6, and 6 lies inside the band because both ends of the band count.

Constraints

1 <= nums.length <= 10^5
-2^31 <= nums[i] <= 2^31 - 1
-10^5 <= lower <= upper <= 10^5
The count always fits in a signed 32-bit integer.

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_range_sum(nums: list[int], lower: int, upper: int) -> int:

Java

public int countRangeSum(int[] nums, int lower, int upper)

The intended approach

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

Build the running totals `pre[0..n]`, where the swing of `nums[i..j]` is `pre[j + 1] - pre[i]`. Counting stretches then means counting index pairs `a < b` whose running totals differ by something inside the band. Merge passes of increasing width sort the running totals; before each merge, both runs are already sorted, so two moving pointers over the left run give the count of partners for every entry of the right run. Each pair of indices is split by exactly one merge, so nothing is counted twice.

Pythonentry point: count_range_sum

def count_range_sum(nums, lower, upper):
    # A stretch from i to j totals pre[j + 1] - pre[i], so the task becomes:
    # count index pairs a < b in the prefix table whose difference lands in the band.
    # Sorting the table bottom-up with merge passes lets every pair be counted once,
    # in the single pass where a sits in the left run and b in the right run.
    # Time O(n log n).  Space O(n) for the prefix table and the merge buffer.
    n = len(nums)
    pre = [0] * (n + 1)
    running = 0
    for i in range(n):
        running += nums[i]
        pre[i + 1] = running
    size = n + 1
    buf = [0] * size
    total = 0
    width = 1
    while width < size:
        start = 0
        while start < size:
            mid = min(start + width, size)
            end = min(start + 2 * width, size)
            if mid < end:
                lo = start
                hi = start
                for b in range(mid, end):
                    x = pre[b]
                    while lo < mid and pre[lo] < x - upper:
                        lo += 1
                    while hi < mid and pre[hi] <= x - lower:
                        hi += 1
                    total += hi - lo
                i = start
                j = mid
                k = start
                while i < mid and j < end:
                    if pre[i] <= pre[j]:
                        buf[k] = pre[i]
                        i += 1
                    else:
                        buf[k] = pre[j]
                        j += 1
                    k += 1
                while i < mid:
                    buf[k] = pre[i]
                    i += 1
                    k += 1
                while j < end:
                    buf[k] = pre[j]
                    j += 1
                    k += 1
                pre[start:end] = buf[start:end]
            start += 2 * width
        width *= 2
    return total

The obvious version

Fix a starting hour, extend the stretch one hour at a time while keeping the running swing, and test the band at every step. Quadratic in the number of hours.

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_band_stretches(nums, lower, upper):
    n = len(nums)
    found = 0
    for i in range(n):
        total = 0
        for j in range(i, n):
            total += nums[j]
            if lower <= total <= upper:
                found += 1
    return found

How this problem was checked

Reference solution executed against 10 test cases — 10 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