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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 $nums1[i]-nums2[i]\le nums1[j]-nums2[j]+diff$. 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 $nums[i]\le nums[j]+diff$.

We can enumerate $j$ from small to large, find out how many numbers before it satisfy $nums[i]\le nums[j]+diff$, 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 $nums[i]\le nums[j]+diff$ in $O(\log n)$ time.

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

• Java
• C++
• Python
• Go

• 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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