2338. Count the Number of Ideal Arrays

Problem

You are given two integers n and maxValue, which are used to describe an ideal array.

A 0-indexed integer array arr of length n is considered ideal if the following conditions hold:

• Every arr[i] is a value from 1 to maxValue, for 0 <= i < n.

• Every arr[i] is divisible by arr[i - 1], for 0 < i < n.

Return **the number of *distinct* ideal arrays of length **n. Since the answer may be very large, return it modulo 10^9 + 7.

  Example 1:

Input: n = 2, maxValue = 5
Output: 10
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (5 arrays): [1,1], [1,2], [1,3], [1,4], [1,5]
- Arrays starting with the value 2 (2 arrays): [2,2], [2,4]
- Arrays starting with the value 3 (1 array): [3,3]
- Arrays starting with the value 4 (1 array): [4,4]
- Arrays starting with the value 5 (1 array): [5,5]
There are a total of 5 + 2 + 1 + 1 + 1 = 10 distinct ideal arrays.

Example 2:

Input: n = 5, maxValue = 3
Output: 11
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (9 arrays): 
   - With no other distinct values (1 array): [1,1,1,1,1] 
   - With 2nd distinct value 2 (4 arrays): [1,1,1,1,2], [1,1,1,2,2], [1,1,2,2,2], [1,2,2,2,2]
   - With 2nd distinct value 3 (4 arrays): [1,1,1,1,3], [1,1,1,3,3], [1,1,3,3,3], [1,3,3,3,3]
- Arrays starting with the value 2 (1 array): [2,2,2,2,2]
- Arrays starting with the value 3 (1 array): [3,3,3,3,3]
There are a total of 9 + 1 + 1 = 11 distinct ideal arrays.

  Constraints:

• 2 <= n <= 10^4

• 1 <= maxValue <= 10^4

Solution

class Solution {
    public int idealArrays(int n, int maxValue) {
        int mod = (int) (1e9 + 7);
        int maxDistinct = (int) (Math.log(maxValue) / Math.log(2)) + 1;
        int[][] dp = new int[maxDistinct + 1][maxValue + 1];
        Arrays.fill(dp[1], 1);
        dp[1][0] = 0;
        for (int i = 2; i <= maxDistinct; i++) {
            for (int j = 1; j <= maxValue; j++) {
                for (int k = 2; j * k <= maxValue && dp[i - 1][j * k] != 0; k++) {
                    dp[i][j] += dp[i - 1][j * k];
                }
            }
        }
        int[] sum = new int[maxDistinct + 1];
        for (int i = 1; i <= maxDistinct; i++) {
            sum[i] = Arrays.stream(dp[i]).sum();
        }
        long[] invs = new long[Math.min(n, maxDistinct) + 1];
        invs[1] = 1;
        for (int i = 2; i < invs.length; i++) {
            invs[i] = mod - mod / i * invs[mod % i] % mod;
        }
        long result = maxValue;
        long comb = (long) n - 1;
        for (int i = 2; i <= maxDistinct && i <= n; i++) {
            result += (sum[i] * comb) % mod;
            comb *= n - i;
            comb %= mod;
            comb *= invs[i];
            comb %= mod;
        }
        return (int) (result % mod);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).