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.

Input & Output

Example 1 — Basic Case

$ Input: n = 2, maxValue = 3

› Output: 5

💡 Note: The 5 ideal arrays are: [1,1], [1,2], [1,3], [2,2], [3,3]. Each array satisfies: all elements ≤ 3, and each element divides the next.

Example 2 — Longer Array

$ Input: n = 3, maxValue = 2

› Output: 4

💡 Note: The 4 ideal arrays are: [1,1,1], [1,1,2], [1,2,2], [2,2,2]. All arrays have length 3 with elements ≤ 2.

Example 3 — Single Element

$ Input: n = 1, maxValue = 4

› Output: 4

💡 Note: With length 1, any single value is ideal: [1], [2], [3], [4]. No divisibility constraint applies.

Constraints

2 ≤ n ≤ 10⁴
1 ≤ maxValue ≤ 10⁴

Visualization

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Asked in

G Google 15 f Meta 8 a Amazon 5

This problem requires counting ideal arrays where each element divides the next. The key insight is that ideal arrays correspond to non-decreasing sequences of divisor relationships. The optimal approach uses prime factorization and combinatorics to count arrangements without generating them. Time: O(maxValue log log maxValue), Space: O(maxValue)

Common Approaches

✓ Brute Force Recursion

⏱️ Time: O(maxValue^n) Space: O(n)

Try every possible value at each position and recursively check if it forms a valid ideal array. This approach explores all combinations but is extremely inefficient.

Combinatorics with Prime Factorization

⏱️ Time: O(maxValue log log maxValue + maxValue × π(maxValue) × n) Space: O(maxValue)

Each ideal array corresponds to a non-decreasing sequence of prime factor multiplicities. Use combinatorics to count the number of ways to distribute prime factors across array positions.

Dynamic Programming

⏱️ Time: O(n × maxValue²) Space: O(n × maxValue)

Build a 2D DP table where dp[i][v] represents the number of ideal arrays of length i ending with value v. For each position and value, sum up all arrays that can extend to this value.

Brute Force Recursion — Algorithm Steps

Start with first position, try all values 1 to maxValue
For each subsequent position, try all multiples of current value
Count arrays that satisfy all conditions

Visualization

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Step-by-Step Walkthrough

Start

Try each value 1 to maxValue as first element

Expand

For each element, try all its multiples as next element

Count

Count valid arrays of length n

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD 1000000007

int n, maxValue;

long long backtrack(int pos, int lastVal) {
    if (pos == n) {
        return 1;
    }
    
    long long count = 0;
    int curr = lastVal;
    while (curr <= maxValue) {
        count = (count + backtrack(pos + 1, curr)) % MOD;
        curr += lastVal;
    }
    return count;
}

long long solution(int n_val, int maxVal) {
    n = n_val;
    maxValue = maxVal;
    
    long long result = 0;
    for (int start = 1; start <= maxValue; start++) {
        result = (result + backtrack(1, start)) % MOD;
    }
    
    return result;
}

int main() {
    int n_input, maxValue_input;
    
    scanf("%d", &n_input);
    scanf("%d", &maxValue_input);
    
    printf("%lld\n", solution(n_input, maxValue_input));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(maxValue^n)

At each position we try up to maxValue possibilities, with n positions

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth is n

⚡ Linearithmic Space

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Count the Number of Ideal Arrays - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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