LeetCode in Kotlin
1862. Sum of Floored Pairs
Hard
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 <= 1051 <= nums[i] <= 105
Solution
class Solution {
fun sumOfFlooredPairs(nums: IntArray): Int {
val mod: Long = 1000000007
nums.sort()
val max = nums[nums.size - 1]
val counts = IntArray(max + 1)
val qnts = LongArray(max + 1)
for (k in nums) {
counts[k]++
}
for (i in 1 until max + 1) {
if (counts[i] == 0) {
continue
}
var j = i
while (j <= max) {
qnts[j] += counts[i].toLong()
j = j + i
}
}
for (i in 1 until max + 1) {
qnts[i] = (qnts[i] + qnts[i - 1]) % mod
}
var sum: Long = 0
for (k in nums) {
sum = (sum + qnts[k]) % mod
}
return sum.toInt()
}
}