#2338

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count the Number of Ideal Arrays

This problem involves counting the number of ideal arrays of a given length under certain conditions using state transition dynamic programming.

math dynamic programming combinatorics number theory

Problem Statement

You are given two integers, n and maxValue, describing the length and maximum value of an ideal array, respectively. The array must satisfy certain conditions to be considered ideal.

Return the number of distinct ideal arrays of length n. Since the result can be large, return it modulo 10^9 + 7.

Examples

Example 1

Input: n = 2, maxValue = 5

Output: 10

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

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

Solution Approach

State Transition Dynamic Programming

The key to solving this problem is recognizing it as a state transition problem. Using dynamic programming, maintain the count of ideal arrays as you iterate through each possible value. The transition from one state to another is based on whether the value in the array repeats or increases.

Modulo Operation to Manage Large Numbers

Since the answer can be very large, all calculations must be done modulo 10^9 + 7. This is crucial to avoid overflow and ensure the result fits within the problem constraints.

Optimizing Space Complexity

The problem involves handling large arrays, and thus, it is important to optimize both time and space complexity. Using dynamic programming with reduced space complexity ensures the solution is efficient even for the upper limits of n and maxValue.

Complexity Analysis

Metric	Value
Time	O((n+\omega(m))\cdot\omega(m)+m\omega(m))
Space	O((n+\log(m))\cdot\log(m))

The time complexity is O((n + ω(m)) * ω(m) + m * ω(m)), where ω(m) is the maximum number of unique elements in the array. The space complexity is O((n + log(m)) * log(m)), optimized to handle large inputs efficiently.

What Interviewers Usually Probe

Candidates should demonstrate a clear understanding of dynamic programming and state transitions.
Look for efficient handling of large numbers and the use of modulo operations.
Candidates should optimize both time and space complexity while working with dynamic arrays.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle the modulo operation can lead to incorrect results for large numbers.
Not optimizing the dynamic programming approach can result in time or space inefficiencies.
Overcomplicating the state transition process when the solution can be simplified by recognizing the non-decreasing property of the array.

Follow-up variants

The problem could be modified by changing the allowed range for values in the array, requiring adjustments to the dynamic programming approach.
Instead of returning the count modulo 10^9 + 7, the problem could ask for the sum of all possible ideal arrays.
The problem could involve a different type of array structure, such as allowing for strictly increasing arrays or arrays with no repetition.

FAQ

What is the state transition dynamic programming approach for this problem?

In this problem, you use dynamic programming to manage the transition between different states of the array, ensuring that the array remains non-decreasing as it grows.

How do I handle large numbers in this problem?

You must take the result modulo 10^9 + 7 to prevent overflow and keep the numbers manageable.

What is the time complexity of this solution?

The time complexity is O((n + ω(m)) * ω(m) + m * ω(m)), ensuring the solution can handle large inputs efficiently.

What are the main pitfalls to avoid in this problem?

Make sure to correctly apply the modulo operation, optimize the dynamic programming approach, and avoid overcomplicating the state transition process.

Can the approach be optimized further for space?

Yes, the space complexity can be optimized by reducing the memory used for storing intermediate states, which is handled in the solution.

terminal

Solution

Solution 1: Dynamic Programming

Let $f[i][j]$ represent the number of sequences ending with $i$ and consisting of $j$ distinct elements. The initial value is $f[i][1] = 1$.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

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

Solution 2

#### Python3

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

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

Continue Practicing

#1735 count ways to make array with product #2400 number of ways to reach a position after exactly k steps #2597 the number of beautiful subsets