# 2338. Count the Number of Ideal Arrays

## Description

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 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

## Solutions

• class Solution {
private int[][] f;
private int[][] c;
private int n;
private int m;
private static final int MOD = (int) 1e9 + 7;

public int idealArrays(int n, int maxValue) {
this.n = n;
this.m = maxValue;
this.f = new int[maxValue + 1][16];
for (int[] row : f) {
Arrays.fill(row, -1);
}
c = new int[n][16];
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= i && j < 16; ++j) {
c[i][j] = j == 0 ? 1 : (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
}
}
int ans = 0;
for (int i = 1; i <= m; ++i) {
ans = (ans + dfs(i, 1)) % MOD;
}
return ans;
}

private int dfs(int i, int cnt) {
if (f[i][cnt] != -1) {
return f[i][cnt];
}
int res = c[n - 1][cnt - 1];
if (cnt < n) {
for (int k = 2; k * i <= m; ++k) {
res = (res + dfs(k * i, cnt + 1)) % MOD;
}
}
f[i][cnt] = res;
return res;
}
}

• class Solution {
public:
int m, n;
const int mod = 1e9 + 7;
vector<vector<int>> f;
vector<vector<int>> c;

int idealArrays(int n, int maxValue) {
this->m = maxValue;
this->n = n;
f.assign(maxValue + 1, vector<int>(16, -1));
c.assign(n, vector<int>(16, 0));
for (int i = 0; i < n; ++i)
for (int j = 0; j <= i && j < 16; ++j)
c[i][j] = !j ? 1 : (c[i - 1][j] + c[i - 1][j - 1]) % mod;
int ans = 0;
for (int i = 1; i <= m; ++i) ans = (ans + dfs(i, 1)) % mod;
return ans;
}

int dfs(int i, int cnt) {
if (f[i][cnt] != -1) return f[i][cnt];
int res = c[n - 1][cnt - 1];
if (cnt < n)
for (int k = 2; k * i <= m; ++k)
res = (res + dfs(k * i, cnt + 1)) % mod;
f[i][cnt] = res;
return res;
}
};

• class Solution:
def idealArrays(self, n: int, maxValue: int) -> int:
@cache
def dfs(i, cnt):
res = c[-1][cnt - 1]
if cnt < n:
k = 2
while k * i <= maxValue:
res = (res + dfs(k * i, cnt + 1)) % mod
k += 1
return res

c = [[0] * 16 for _ in range(n)]
mod = 10**9 + 7
for i in range(n):
for j in range(min(16, i + 1)):
c[i][j] = 1 if j == 0 else (c[i - 1][j] + c[i - 1][j - 1]) % mod
ans = 0
for i in range(1, maxValue + 1):
ans = (ans + dfs(i, 1)) % mod
return ans


• func idealArrays(n int, maxValue int) int {
mod := int(1e9) + 7
m := maxValue
c := make([][]int, n)
f := make([][]int, m+1)
for i := range c {
c[i] = make([]int, 16)
}
for i := range f {
f[i] = make([]int, 16)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(int, int) int
dfs = func(i, cnt int) int {
if f[i][cnt] != -1 {
return f[i][cnt]
}
res := c[n-1][cnt-1]
if cnt < n {
for k := 2; k*i <= m; k++ {
res = (res + dfs(k*i, cnt+1)) % mod
}
}
f[i][cnt] = res
return res
}
for i := 0; i < n; i++ {
for j := 0; j <= i && j < 16; j++ {
if j == 0 {
c[i][j] = 1
} else {
c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod
}
}
}
ans := 0
for i := 1; i <= m; i++ {
ans = (ans + dfs(i, 1)) % mod
}
return ans
}