0964MediumArrayTwo PointersBinary SearchSorting

Pairs Whose Total Falls in Range

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Trains the technique from

LeetCode 2563Count the Number of Fair Pairs

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.

Readings are given as nums, along with bounds lower and upper.

A pair of positions i before j is in range when nums[i] + nums[j] is at least lower and at most upper.

Return how many pairs are in range.

Examples

Example 1

Input: nums = [14, 3, 27, 9, 41], lower = 20, upper = 40
Output: 3

Sorted, the readings run 3, 9, 14, 27, 41. The pairs totalling between 20 and 40 are 3 and 27, 9 and 27, 14 and 27, and 3 and 41... of those the last totals 44, so it does not count, leaving the three that do.

Example 2

Input: nums = [0, 0, 0, 0], lower = 0, upper = 0
Output: 6

Every pair totals nothing, and there are six pairs among four readings.

Example 3

Input: nums = [1, 2, 3], lower = 100, upper = 200
Output: 0

No pair totals anywhere near the range.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
-10^9 <= lower <= upper <= 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 count_fair_pairs(nums: list[int], lower: int, upper: int) -> int:

Java

public long countFairPairs(int[] nums, int lower, int upper)

The intended approach

Time O(n log n) for n readings, the sort dominating the two linear walks.Space O(n) for the sorted readings.

Which positions the two readings came from never matters, only their values, so the readings may be sorted. Then counting the pairs whose total lands in a range is the difference of two easier counts: pairs totalling at most `upper`, less pairs totalling at most `lower - 1`. Counting the pairs totalling at most some bound over a sorted list is one walk from both ends. Start with a pointer at each end; if the two readings under them total at most the bound, then every reading from the left pointer up to just before the right one pairs with the right pointer's reading and also stays inside, so that many pairs are counted at once and the right pointer moves in. Otherwise the right pointer's reading is too large for anything, so it moves in with nothing counted.

Pythonentry point: count_fair_pairs

def count_fair_pairs(nums, lower, upper):
    order = sorted(nums)
    n = len(order)

    def at_most(bound):
        total = 0
        left = 0
        right = n - 1
        while left < right:
            if order[left] + order[right] <= bound:
                total += right - left
                left += 1
            else:
                right -= 1
        return total

    return at_most(upper) - at_most(lower - 1)

The obvious version

Try every pair of positions and count those whose total lands in the range. Quadratic in the reading count and it never sorts anything.

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

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

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