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, andlower <= 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 <= 105nums.length == n-109 <= nums[i] <= 109-109 <= lower <= upper <= 109
Solution 1: Sorting + Binary Search
C++
class Solution { public: long long countFairPairs(vector<int>& nums, int lower, int upper) { long long ans = 0; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size(); ++i) { auto j = lower_bound(nums.begin() + i + 1, nums.end(), lower - nums[i]); auto k = lower_bound(nums.begin() + i + 1, nums.end(), upper - nums[i] + 1); ans += k - j; } return ans; } };
JAVA
class Solution { public long countFairPairs(int[] nums, int lower, int upper) { Arrays.sort(nums); long ans = 0; int n = nums.length; for (int i = 0; i < n; ++i) { int j = search(nums, lower - nums[i], i + 1); int k = search(nums, upper - nums[i] + 1, i + 1); ans += k - j; } return ans; } private int search(int[] nums, int x, int left) { int right = nums.length; while (left < right) { int mid = (left + right) >> 1; if (nums[mid] >= x) { right = mid; } else { left = mid + 1; } } return left; } }
PYTHON
from typing import List from bisect import bisect_left class Solution: def countFairPairs(self, nums: List[int], lower: int, upper: int) -> int: nums.sort() ans = 0 for i in range(len(nums)): j = bisect_left(nums, lower - nums[i], i + 1) k = bisect_left(nums, upper - nums[i] + 1, i + 1) ans += k - j return ans
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