README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2426.Number%20of%20Pairs%20Satisfying%20Inequality/README_EN.md	2030	Biweekly Contest 88 Q4	Binary Indexed Tree Segment Tree Array Binary Search Divide and Conquer Ordered Set Merge Sort

2426. Number of Pairs Satisfying Inequality

中文文档

Description

You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that:

• 0 <= i < j <= n - 1 and
• nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

 

Example 1:

Input: nums1 = [3,2,5], nums2 = [2,2,1], diff = 1
Output: 3
Explanation:
There are 3 pairs that satisfy the conditions:
1. i = 0, j = 1: 3 - 2 <= 2 - 2 + 1. Since i < j and 1 <= 1, this pair satisfies the conditions.
2. i = 0, j = 2: 3 - 5 <= 2 - 1 + 1. Since i < j and -2 <= 2, this pair satisfies the conditions.
3. i = 1, j = 2: 2 - 5 <= 2 - 1 + 1. Since i < j and -3 <= 2, this pair satisfies the conditions.
Therefore, we return 3.

Example 2:

Input: nums1 = [3,-1], nums2 = [-2,2], diff = -1
Output: 0
Explanation:
Since there does not exist any pair that satisfies the conditions, we return 0.

 

Constraints:

• n == nums1.length == nums2.length
• 2 <= n <= 105
• -104 <= nums1[i], nums2[i] <= 104
• -104 <= diff <= 104

Solutions

Solution 1: Binary Indexed Tree

We can transform the inequality in the problem to $nums1[i] - nums2[i] \leq nums1[j] - nums2[j] + diff$. Therefore, if we calculate the difference between the corresponding elements of the two arrays and get another array $nums$, the problem is transformed into finding the number of pairs in $nums$ that satisfy $nums[i] \leq nums[j] + diff$.

We can enumerate $j$ from small to large, find out how many numbers before it satisfy $nums[i] \leq nums[j] + diff$, and thus calculate the number of pairs. We can use a binary indexed tree to maintain the prefix sum, so we can find out how many numbers before it satisfy $nums[i] \leq nums[j] + diff$ in $O(\log n)$ time.

The time complexity is $O(n \times \log n)$.

Python3

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        tree = BinaryIndexedTree(10**5)
        ans = 0
        for a, b in zip(nums1, nums2):
            v = a - b
            ans += tree.query(v + diff + 40000)
            tree.update(v + 40000, 1)
        return ans

Java

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public static final int lowbit(int x) {
        return x & -x;
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }
}

class Solution {
    public long numberOfPairs(int[] nums1, int[] nums2, int diff) {
        BinaryIndexedTree tree = new BinaryIndexedTree(100000);
        long ans = 0;
        for (int i = 0; i < nums1.length; ++i) {
            int v = nums1[i] - nums2[i];
            ans += tree.query(v + diff + 40000);
            tree.update(v + 40000, 1);
        }
        return ans;
    }
}

C++

class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) {}

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    long long numberOfPairs(vector<int>& nums1, vector<int>& nums2, int diff) {
        BinaryIndexedTree* tree = new BinaryIndexedTree(1e5);
        long long ans = 0;
        for (int i = 0; i < nums1.size(); ++i) {
            int v = nums1[i] - nums2[i];
            ans += tree->query(v + diff + 40000);
            tree->update(v + 40000, 1);
        }
        return ans;
    }
};

Go

type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= this.lowbit(x)
	}
	return s
}

func numberOfPairs(nums1 []int, nums2 []int, diff int) int64 {
	tree := newBinaryIndexedTree(100000)
	ans := 0
	for i := range nums1 {
		v := nums1[i] - nums2[i]
		ans += tree.query(v + diff + 40000)
		tree.update(v+40000, 1)
	}
	return int64(ans)
}