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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1801_1900.s1862_sum_of_floored_pairs;

// #Hard #Array #Math #Binary_Search #Prefix_Sum
// #2022_05_08_Time_115_ms_(70.91%)_Space_57.2_MB_(81.82%)

import java.util.Arrays;

/**
 * 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 <= 10⁵
 * *   1 <= nums[i] <= 10⁵
**/
public class Solution {
    public int sumOfFlooredPairs(int[] nums) {
        long mod = 1000000007;
        Arrays.sort(nums);
        int max = nums[nums.length - 1];
        int[] counts = new int[max + 1];
        long[] qnts = new long[max + 1];
        for (int k : nums) {
            counts[k]++;
        }
        for (int i = 1; i < max + 1; i++) {
            if (counts[i] == 0) {
                continue;
            }
            int j = i;
            while (j <= max) {
                qnts[j] += counts[i];
                j = j + i;
            }
        }
        for (int i = 1; i < max + 1; i++) {
            qnts[i] = (qnts[i] + qnts[i - 1]) % mod;
        }
        long sum = 0;
        for (int k : nums) {
            sum = (sum + qnts[k]) % mod;
        }
        return (int) sum;
    }
}