Problem Description

You need to count how many distinct "ideal arrays" of length n can be formed, where each element is between 1 and maxValue.

An array is considered "ideal" if it satisfies two conditions:

1. Every element must be a value from 1 to maxValue (inclusive)
2. Each element (except the first) must be divisible by the element immediately before it - meaning arr[i] is divisible by arr[i-1] for all positions i > 0

For example, [1, 2, 6] is an ideal array because 2 is divisible by 1, and 6 is divisible by 2. Similarly, [3, 6, 6] is ideal because 6 is divisible by 3, and 6 is divisible by 6.

The key insight is that in an ideal array, the values form a non-decreasing sequence where each element is a multiple of the previous one. This means if we start with value a, the next value must be a*k for some integer k ≥ 1.

The solution uses dynamic programming where f[i][j] represents the number of distinct sequences ending with value i and having exactly j distinct values. Since we have an array of length n but only j distinct values, we need to distribute these j values across n positions. This is done using combinations - specifically C(n-1, j-1), which represents choosing j-1 positions from n-1 possible positions to place the transitions between different values.

The algorithm:

1. Preprocesses combination values using Pascal's triangle
2. Computes f[i][j] by considering all multiples of each value
3. Combines the results by multiplying the number of distinct sequences with the ways to distribute them across the array length

Since the answer can be very large, return it modulo 10^9 + 7.

Intuition

The key observation is that an ideal array has a special structure: each element must be a multiple of the previous one. This means if we look at the sequence of distinct values in the array, they form a divisibility chain like [a, a*k₁, a*k₁*k₂, ...].

Let's think about this problem in two parts:

1. What distinct values can appear in the array? Since each value must divide the next, we're essentially building chains of divisors. For example, starting from 2, we could have chains like [2, 4, 8] or [2, 6, 12].

2. How can we arrange these distinct values in an array of length n? Since the array must be non-decreasing (due to the divisibility constraint), we can have repetitions. For instance, with distinct values [2, 4] and array length 4, we could have [2, 2, 4, 4] or [2, 4, 4, 4].

This leads us to think about the problem as:

• First, find all possible chains of distinct values where each divides the next
• Then, figure out how to distribute these values across n positions

For the first part, we can use dynamic programming. Let f[i][j] represent the number of ways to form a chain of exactly j distinct values ending with value i. We can build longer chains by considering all multiples of i: if we have a chain ending at i with j values, we can extend it to any multiple k*i (where k*i ≤ maxValue) to get a chain with j+1 values.

For the second part, imagine we have j distinct values to place in n positions. Since the array must be non-decreasing, this is equivalent to dividing n positions into j groups (where each group contains copies of the same value). This is a classic "stars and bars" problem - we need to place j-1 dividers among n-1 possible positions, giving us C(n-1, j-1) ways.

The maximum number of distinct values in any chain is limited because each value must be at least double the previous one (smallest multiplier is 2). So the longest possible chain has length approximately log₂(maxValue), which is why we only need to consider up to about 15 distinct values.

By combining these two insights - counting the chains of distinct values and then distributing them across the array - we arrive at the final answer: sum over all possible ending values i and chain lengths j of f[i][j] * C(n-1, j-1).

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

The implementation consists of three main parts: preprocessing combinations, computing the DP table for distinct value chains, and calculating the final answer.

Step 1: Precompute Combinations using Pascal's Triangle

We build a 2D array c[i][j] to store C(i, j) values using the recurrence relation:

• c[i][j] = c[i-1][j] + c[i-1][j-1]
• Base case: c[i][0] = 1 for all i

c = [[0] * 16 for _ in range(n)]
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

We only need combinations up to 16 columns because the maximum chain length is bounded by log₂(maxValue) + 1 ≈ 15.

Step 2: Build DP Table for Distinct Value Chains

Initialize f[i][j] where:

• i represents the ending value (1 to maxValue)
• j represents the number of distinct values in the chain
• f[i][j] = number of ways to form a chain of j distinct values ending at i

Base case: f[i][1] = 1 for all i (single element chains).

f = [[0] * 16 for _ in range(maxValue + 1)]
for i in range(1, maxValue + 1):
    f[i][1] = 1

For each chain length j from 1 to 14, and each value i, we extend the chain to all its multiples:

for j in range(1, 15):
    for i in range(1, maxValue + 1):
        k = 2
        while k * i <= maxValue:
            f[k * i][j + 1] = (f[k * i][j + 1] + f[i][j]) % mod
            k += 1

This builds chains by considering: if we have a chain of length j ending at i, we can extend it to k*i (for k ≥ 2) to create a chain of length j+1.

Step 3: Calculate Final Answer

For each possible ending value i and chain length j, multiply:

• f[i][j]: number of distinct value chains
• c[n-1][j-1]: ways to distribute j distinct values across n positions

ans = 0
for i in range(1, maxValue + 1):
    for j in range(1, 16):
        ans = (ans + f[i][j] * c[-1][j - 1]) % mod

The formula c[n-1][j-1] represents choosing j-1 positions from n-1 possible spots to place dividers, effectively distributing j distinct values into n array positions while maintaining non-decreasing order.

Time Complexity: O(maxValue × log² maxValue) - For each value up to maxValue, we consider chains up to length log maxValue, and for each chain, we check multiples up to maxValue.

Space Complexity: O(n × log maxValue + maxValue × log maxValue) - For storing the combination table and DP table.

Example Walkthrough

Let's walk through a small example with n = 3 and maxValue = 5 to illustrate the solution approach.

Step 1: Precompute Combinations

We need combinations C(i,j) for distributing distinct values. For n=3, we build:

c[0][0] = 1
c[1][0] = 1, c[1][1] = 1
c[2][0] = 1, c[2][1] = 2, c[2][2] = 1

So C(2,0) = 1, C(2,1) = 2, C(2,2) = 1.

Step 2: Build DP Table for Distinct Value Chains

Initialize f[i][1] = 1 for all values 1 to 5 (single-element chains):

• f[1][1] = 1, f[2][1] = 1, f[3][1] = 1, f[4][1] = 1, f[5][1] = 1

Now build chains of length 2 by considering multiples:

• From value 1: multiples are 2,3,4,5 → f[2][2] = 1, f[3][2] = 1, f[4][2] = 1, f[5][2] = 1
• From value 2: multiple is 4 → f[4][2] += 1 (now f[4][2] = 2)

After processing all values for length 2:

• f[2][2] = 1 (chain: 1→2)
• f[3][2] = 1 (chain: 1→3)
• f[4][2] = 2 (chains: 1→4 and 2→4)
• f[5][2] = 1 (chain: 1→5)

Build chains of length 3:

• From f[2][2] = 1: multiple is 4 → f[4][3] = 1 (chain: 1→2→4)

Step 3: Calculate Final Answer

Now we combine the chains with ways to distribute them:

For chains of length 1 (need to fill 3 positions with 1 distinct value):

• Ways to distribute: C(2,0) = 1
• Contribution: (f[1][1] + f[2][1] + f[3][1] + f[4][1] + f[5][1]) × 1 = 5 × 1 = 5
• Examples: [1,1,1], [2,2,2], [3,3,3], [4,4,4], [5,5,5]

For chains of length 2 (need to fill 3 positions with 2 distinct values):

• Ways to distribute: C(2,1) = 2 (either [a,a,b] or [a,b,b])
• Contribution: (f[2][2] + f[3][2] + f[4][2] + f[5][2]) × 2 = (1+1+2+1) × 2 = 10
• Examples for chain 1→2: [1,1,2] and [1,2,2]
• Examples for chain 2→4: [2,2,4] and [2,4,4]

For chains of length 3 (need to fill 3 positions with 3 distinct values):

• Ways to distribute: C(2,2) = 1 (only [a,b,c])
• Contribution: f[4][3] × 1 = 1 × 1 = 1
• Example: [1,2,4]

Total: 5 + 10 + 1 = 16 distinct ideal arrays

Some of the 16 arrays are:

• Length-1 chains: [1,1,1], [2,2,2], [3,3,3], [4,4,4], [5,5,5]
• Length-2 chains: [1,1,2], [1,2,2], [1,1,3], [1,3,3], [1,1,4], [1,4,4], [2,2,4], [2,4,4], [1,1,5], [1,5,5]
• Length-3 chain: [1,2,4]

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × 16 + maxValue × 16 × maxValue + maxValue × 16) which simplifies to O(n + maxValue²)

The time complexity breaks down into three main parts:

1. Building the combination table c: The nested loops run for n iterations and up to 16 iterations, giving O(n × 16) = O(n).

2. Computing the dynamic programming table f:

   • The outer loop runs for 15 iterations (j from 1 to 14)
   • The middle loop runs for maxValue iterations (i from 1 to maxValue)
   • The inner while loop runs approximately maxValue/i times for each i
   • The total iterations for the inner loop across all values of i is approximately maxValue × (1/1 + 1/2 + 1/3 + ... + 1/maxValue) ≈ maxValue × log(maxValue)
   • However, since we're iterating through all multiples, the total work is O(15 × maxValue × maxValue) = O(maxValue²)

3. Computing the final answer: Two nested loops running maxValue × 16 times, giving O(maxValue × 16) = O(maxValue).

Overall: O(n + maxValue² + maxValue) = O(n + maxValue²)

Space Complexity: O(n × 16 + maxValue × 16) which simplifies to O(n + maxValue)

The space complexity comes from:

• Array c of size n × 16 = O(n)
• Array f of size (maxValue + 1) × 16 = O(maxValue)

Overall: O(n + maxValue)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Understanding of Array Distribution

Pitfall: A common mistake is misunderstanding how to distribute j distinct values across n positions. Many incorrectly assume we need C(n, j) combinations, which would be choosing j positions out of n for the distinct values.

Why it's wrong: The array must be non-decreasing, and we're not choosing which positions get which values. Instead, we're determining where transitions between different values occur.

Correct approach: Use C(n-1, j-1) which represents choosing j-1 transition points from n-1 possible positions between elements. This divides the array into j segments, each containing copies of the same value.

Example: For array length n=5 with j=3 distinct values [a, b, c]:

• We need 2 transition points (from a to b, and from b to c)
• These 2 points can be chosen from 4 possible positions (between elements)
• Result: C(4, 2) = 6 ways to arrange like [a,a,b,b,c], [a,b,b,b,c], etc.

2. Off-by-One Errors in DP Transition

Pitfall: Starting the multiplier k from 1 instead of 2 when extending chains:

# Wrong
k = 1  # This would include the same value, not creating a strictly increasing sequence
while k * i <= maxValue:
    f[k * i][j + 1] = (f[k * i][j + 1] + f[i][j]) % mod
    k += 1

Why it's wrong: Starting from k=1 would mean f[i][j+1] includes transitions from f[i][j], creating sequences with repeated consecutive values that aren't strictly increasing in terms of distinct values.

Correct approach: Start k from 2 to ensure each new value is strictly greater than the previous:

k = 2  # Ensures k*i > i, maintaining strictly increasing distinct values
while k * i <= maxValue:
    f[k * i][j + 1] = (f[k * i][j + 1] + f[i][j]) % mod
    k += 1

3. Insufficient Array Bounds for Chain Length

Pitfall: Using a fixed small size like 10 or 12 for the maximum chain length without proper calculation:

# Potentially wrong
MAX_LENGTH = 10  # Too small for some inputs

Why it's wrong: The maximum chain length depends on maxValue. For example, the chain [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024] has 11 distinct values when maxValue ≥ 1024.

Correct approach: Calculate based on the logarithm of maxValue. Since each value must be at least double the previous one in the worst case, the maximum chain length is approximately log₂(maxValue) + 1:

MAX_LENGTH = 16  # Safe for maxValue up to 10^4
# Or dynamically: MAX_LENGTH = min(16, int(math.log2(maxValue)) + 2)

4. Forgetting Modulo Operations

Pitfall: Not applying modulo consistently, especially in the combination calculation:

# Wrong - can cause integer overflow
combinations[i][j] = combinations[i - 1][j] + combinations[i - 1][j - 1]

Correct approach: Apply modulo after every arithmetic operation:

combinations[i][j] = (combinations[i - 1][j] + combinations[i - 1][j - 1]) % MOD

5. Mishandling the Final Answer Calculation

Pitfall: Using wrong indices for the combination lookup:

# Wrong
ans = (ans + f[i][j] * c[n][j]) % mod  # Wrong indices

Correct approach: Use c[n-1][j-1] to properly distribute j distinct values across n positions:

ans = (ans + f[i][j] * c[n-1][j-1]) % mod