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

  • 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;
        }
    };
    
  • 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
    
    
  • 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)
    }
    

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