How do you solve Count the Number of Ideal Arrays on LeetCode?

Count the Number of Ideal Arrays (LeetCode #2338) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * maxValue log(maxValue)) time complexity and O(maxValue) space complexity. Key insight: Ideal arrays are non-decreasing.

What is the time complexity of Count the Number of Ideal Arrays?

The optimal solution for Count the Number of Ideal Arrays runs in O(n * maxValue log(maxValue)) time and uses O(maxValue) space. The complexity arises from iterating through each length and each value, while the inner loop runs in a logarithmic manner due to the divisibility checks.

What is the brute force solution for Count the Number of Ideal Arrays?

The brute force approach for Count the Number of Ideal Arrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach generates all possible arrays of length n with values from 1 to maxValue and checks if they satisfy the ideal array conditions. This is simple but inefficient due to the large number of combinations. The optimal Optimal Solution approach improves this to O(n * maxValue log(maxValue)).

What algorithm or data structure is used to solve Count the Number of Ideal Arrays?

Count the Number of Ideal Arrays uses the following concepts: Math, Dynamic Programming, Combinatorics, Number Theory. The recommended patterns to study are: Dynamic Programming, Number Theory.

What are the interview tips for Count the Number of Ideal Arrays?

Understand the problem constraints thoroughly. Think about dynamic programming as a way to optimize counting problems. Practice similar problems to get familiar with patterns.

What are common mistakes when solving Count the Number of Ideal Arrays?

Not considering the modulo operation. Failing to optimize the counting process.

#2338

Count the Number of Ideal Arrays

Hard

MathDynamic ProgrammingCombinatoricsNumber TheoryDynamic ProgrammingNumber Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * maxValue log(maxValue))
Space	O(1)	O(maxValue)

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Intuition

Time O(n * maxValue log(maxValue))Space O(maxValue)

The optimal solution uses dynamic programming to count the number of ideal arrays without generating them explicitly. By leveraging the properties of divisibility, we can efficiently compute the counts based on previously computed values.

⚙️

Algorithm

3 steps

1Step 1: Create an array dp where dp[i] represents the number of ideal arrays of length n that end with the value i.
2Step 2: For each value j from 1 to maxValue, calculate dp[j] based on the sum of dp[k] for all k that divide j.
3Step 3: The result is the sum of dp[j] for all j from 1 to maxValue.

solution.py13 lines

1# Full working Python code
2def countIdealArrays(n, maxValue):
3    MOD = 10**9 + 7
4    dp = [0] * (maxValue + 1)
5    for i in range(1, maxValue + 1):
6        dp[i] = 1  # Each number can form an ideal array of length 1
7    for _ in range(1, n):
8        new_dp = [0] * (maxValue + 1)
9        for j in range(1, maxValue + 1):
10            for k in range(j, maxValue + 1, j):
11                new_dp[k] = (new_dp[k] + dp[j]) % MOD
12        dp = new_dp
13    return sum(dp) % MOD

ℹ

Complexity note: The complexity arises from iterating through each length and each value, while the inner loop runs in a logarithmic manner due to the divisibility checks.

1Ideal arrays are non-decreasing.
2Divisibility creates a tree-like structure of valid sequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.