README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2563.Count%20the%20Number%20of%20Fair%20Pairs/README_EN.md	1720	Weekly Contest 332 Q2	Array Two Pointers Binary Search Sorting

2563. Count the Number of Fair Pairs

中文文档

Description

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.

Python3

class Solution:
    def countFairPairs(self, nums: List[int], lower: int, upper: int) -> int:
        nums.sort()
        ans = 0
        for i, x in enumerate(nums):
            j = bisect_left(nums, lower - x, lo=i + 1)
            k = bisect_left(nums, upper - x + 1, lo=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;
    }
}

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;
    }
};

Go

func countFairPairs(nums []int, lower int, upper int) (ans int64) {
	sort.Ints(nums)
	for i, x := range nums {
		j := sort.Search(len(nums), func(h int) bool { return h > i && nums[h] >= lower-x })
		k := sort.Search(len(nums), func(h int) bool { return h > i && nums[h] >= upper-x+1 })
		ans += int64(k - j)
	}
	return
}

TypeScript

function countFairPairs(nums: number[], lower: number, upper: number): number {
    const search = (x: number, l: number): number => {
        let r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };

    nums.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0; i < nums.length; ++i) {
        const j = search(lower - nums[i], i + 1);
        const k = search(upper - nums[i] + 1, i + 1);
        ans += k - j;
    }
    return ans;
}