2426. Number of Pairs Satisfying Inequality

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 $O(n \times \log n)$.

Python3JavaC++Go

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

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;
    }
}

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;
    }
};

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)
}

2426. Number of Pairs Satisfying Inequality

Description

Solutions

Solution 1: Binary Indexed Tree

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