Sum of Floored Pairs — LeetCode 1862 Python Solution

The problem

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.

Example

Input

nums = [2,5,9]

Output

10

Explanation

floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0

Python solution

class Solution:
    def sumOfFlooredPairs(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        cnt = Counter(nums)
        mx = max(nums)
        s = [0] * (mx + 1)
        for i in range(1, mx + 1):
            s[i] = s[i - 1] + cnt[i]
        ans = 0
        for y in range(1, mx + 1):
            if cnt[y]:
                d = 1
                while d * y <= mx:
                    ans += cnt[y] * d * (s[min(mx, d * y + y - 1)] - s[d * y - 1])
                    ans %= mod
                    d += 1
        return ans

Complexity

Pattern: Prefix Sum

Precompute running totals once so any range query becomes a single subtraction. LeetCode 1862. Sum of Floored Pairs is filed here because LeetCode tags it Prefix Sum, which is the vocabulary this hub collects.

The prefix sum guide has the Python template for the pattern and the 157 LeetCode problems that use it.

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