2563. Count the Number of Fair Pairs

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

class Solution {
public:
long long lower_bound(vector<int>& nums, int low, int high, int element) {
while (low <= high) {
int mid = low + ((high - low) / 2);
if (nums[mid] >= element) {
high = mid - 1;
} else
low = mid + 1;
}
return low;
}
long long countFairPairs(vector<int>& nums, int lower, int upper) {
sort(nums.begin(), nums.end());
long long ans = 0;
for (int i = 0; i < nums.size(); i++) {
// Assume we have picked nums[i] as the first pair element.

// `low` indicates the number of possible pairs with sum < lower.
int low =
lower_bound(nums, i + 1, nums.size() - 1, lower - nums[i]);

// `high` indicates the number of possible pairs with sum <= upper.
int high =
lower_bound(nums, i + 1, nums.size() - 1, upper - nums[i] + 1);

// Their difference gives the number of elements with sum in the
// given range.
ans += 1LL * (high - low);
}
return ans;
}
};