README.md

1862. Sum of Floored Pairs

Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0 <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 109 + 7.

The floor() function returns the integer part of the division.

Example 1:

Input: nums = [2,5,9]
Output: 10
Explanation:
floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
floor(5 / 2) = 2
floor(9 / 2) = 4
floor(9 / 5) = 1
We calculate the floor of the division for every pair of indices in the array then sum them up.

Example 2:

Input: nums = [7,7,7,7,7,7,7]
Output: 49

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

Solutions (Python)

1. Solution

class Solution:
    def sumOfFlooredPairs(self, nums: List[int]) -> int:
        prevsum = 0
        ret = 0

        nums.sort()

        for j in range(len(nums)):
            if j > 0 and nums[j] == nums[j - 1]:
                ret = (ret + prevsum) % 1000000007
                continue

            i = bisect.bisect(nums, nums[j] - 1)
            prevsum = 0

            for x in range(1, nums[-1] // nums[j] + 1):
                k = bisect.bisect(nums, nums[j] * (x + 1) - 1, lo=i)
                prevsum = (prevsum + x * (k - i)) % 1000000007
                i = k

            ret = (ret + prevsum) % 1000000007

        return ret