Hard
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:
arr[i]
is a value from 1
to maxValue
, for 0 <= i < n
.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 109 + 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 <= 104
1 <= maxValue <= 104
import java.util.Arrays;
public 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);
}
}