Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs.
A pair (i, j) is fair if:
0 <= i < j < n, and
lower <= nums[i] + nums[j] <= upper
Example 1:
Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6
Output: 6
Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5).
Example 2:
Input: nums = [1,7,9,2,5], lower = 11, upper = 11
Output: 1
Explanation: There is a single fair pair: (2,3).
Constraints:
1 <= nums.length <= 105
nums.length == n
-109 <= nums[i] <= 109
-109 <= lower <= upper <= 109
Solutions
Solution 1: Sorting + Binary Search
First, we sort the array nums in ascending order. Then, for each nums[i], we use binary search to find the lower bound j of nums[j], i.e., the first index that satisfies nums[j] >= lower - nums[i]. Then, we use binary search again to find the lower bound k of nums[k], i.e., the first index that satisfies nums[k] >= upper - nums[i] + 1. Therefore, [j, k) is the index range for nums[j] that satisfies lower <= nums[i] + nums[j] <= upper. The count of these indices corresponding to nums[j] is k - j, and we can add this to the answer. Note that $j > i$.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array nums.