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2426. Number of Pairs Satisfying Inequality

Difficulty Topics

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

Solution

number-of-pairs-satisfying-inequality.py
from sortedcontainers import SortedList

class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        N = len(nums1)
        sl = SortedList()
        res = 0

        # nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff
        # nums1[i] - nums2[i] <= diff + nums1[j] - nums2[j]

        for j in range(N):
            curr = diff + nums1[j] - nums2[j]

            index = sl.bisect_right(curr)
            res += index

            sl.add(nums1[j] - nums2[j])

        return res