<p>Given an integer array <code>nums</code>, return the sum of <code>floor(nums[i] / nums[j])</code> for all pairs of indices <code>0 &lt;= i, j &lt; nums.length</code> in the array. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>The <code>floor()</code> function returns the integer part of the division.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,5,9]
<strong>Output:</strong> 10
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 49
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

from collections import Counter  
from typing import List
class Solution:  
    def sumOfFlooredPairs(self, nums: List[int]) -> int:   
        result, var1 = 0, max(nums) + 1  
        count = Counter(nums) 
        prefix_sum = [0] * var1  
        for i in range(1, var1):  
            prefix_sum[i] = prefix_sum[i - 1] + count[i]   
        contribution = lambda num: sum(count[num] * (prefix_sum[-1] - prefix_sum[i - 1]) for i in range(num, var1, num))  
        for idx, num in enumerate(count):  
            result += contribution(num)
        return result % (10**9 + 7)
36400