2338. Count the Number of Ideal Arrays

HardMathDynamic ProgrammingCombinatoricsNumber Theory

Problem Description

You're tasked with finding the number of distinct "ideal" integer arrays of a certain length n, each containing integers that range from 1 to maxValue. An array qualifies as "ideal" if it meets two conditions:

Every element arr[i] takes on a value between 1 and maxValue (inclusive).
For every i from 1 to n-1, the element arr[i] is divisible by the element before it arr[i - 1].

The result should be returned as the count of these distinct arrays modulo 10^9 + 7.

Intuition

The problem requires counting configurations under certain constraints, which is a classic dynamic programming challenge. The idea is to use a memoized recursive function (dfs in the given code) to explore different array configurations, starting from each potential value [1, maxValue] as the first element. The dynamic programming function serves to count the distinct ways to build up the ideal array by incrementally choosing subsequent elements that follow the divisibility rule.

Here's the approach followed in the provided solution:

Initialize a combinatorics table c that will be used later to calculate combinations efficiently (c[n][k] will represent the number of combinations to pick k items from n items).
The dfs function is recursively used to explore all possible valid next elements for a given current element i and a count cnt of how many elements have been chosen so far. If a number k*i is still within maxValue, it can become the next element in the ideal array.
The number of ways to extend the array from the current element i and cnt is the sum of ways to extend all valid next elements (k*i, cnt+1).
It's important to keep in mind that the answer could be very large so the modulo operation is performed at every step to ensure the numbers stay within the bounds.
We iterate over all possible starting elements from 1 to maxValue and use dfs to count all distinct arrays starting from that element.

Through this recursive exploration that carefully counts combinations while respecting the constraints, we can amass the total number of distinct ideal arrays.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

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What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8

1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12

1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

[30, 20, 10][100, 33, 30][10, 12, 20]

I don’t know

Solution Approach

The solution uses dynamic programming, recursion with memoization, combinatorics (for counting combinations), and modulus operations to prevent integer overflow. Here's a walkthrough of the implementation details:

Dynamic Programming and Memoization

A two-dimensional array called c is prepared in advance to store binomial coefficients which are used for calculating combinations. This uses the well-known formula C(n, k) = C(n-1, k) + C(n-1, k-1) to populate the array efficiently.
The dfs function is memoized using the @cache decorator (assuming Python 3.9+), which means that it remembers the results of previous computations and doesn't re-calculate them, leading to significant time savings.

Recursive Function `dfs`

The dfs function takes two arguments: the current value i and the count cnt of elements in the array so far. This function calculates the number of ways to build an ideal array starting with the value i and having cnt elements already.
The base case is that when cnt reaches n, no more elements can be added, and so c[n-1][cnt-1] is returned, which represents the number of ways to arrange cnt elements in positions up to n-1.
For each possible next multiple k * i (k >= 2), the function recursively calculates the number of ideal arrays that can be built by adding this next multiple, incrementing cnt by 1 for each recursive call.
The modulo mod = 10**9 + 7 is applied at every addition to keep the number manageable as per the problem statement.

Combination Preprocessing

A loop is used to calculate and cache binomial coefficients in advance for the combinatorial part of the computation. This is essential for efficiently calculating the number of combinations in the recursion.

Summation and Modulo Application

A loop iterates from 1 to maxValue, calling dfs(i, 1) for each i. The function returns the number of distinct ideal arrays starting with the value i.
The results are summed up (with modulo applied) to get the final answer.

Final Result

The answer variable ans is initialized to 0. It accumulates the number of ideal arrays calculated by the dfs function for every possible starting element i.
The final answer is then returned, which represents the total number of distinct ideal arrays of length n with values up to maxValue.

By applying dynamic programming with recursive memoization and combinatorial mathematics, the solution efficiently calculates the total number of configurations that meet the problem's requirements.

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Example Walkthrough

Let's illustrate the solution approach using an example where the length of the array n is 3, and the maximum value maxValue is 4. We will walk through some steps to understand how the algorithm works in this simple case.

First, we've initialized a combinatorics table c for calculating combinations. This table would look something like this for n=3, though the actual table might have more entries for efficiency in real problems:

	0	1	2	3
0	1	0	0	0
1	1	1	0	0
2	1	2	1	0
3	1	3	3	1

Now, let's examine how we might use the dfs function to compute the number of ideal arrays using memoization and recursion.

Starting with i=1, we call dfs(1, 1). This means that we are looking to build an array that starts with 1 and currently has one element.
Since i=1 and cnt=1, we look at the next multiples of i, which are 2*1, 3*1, and 4*1. Each of these can be the next element in the array because they are all divisible by 1.
We then make recursive calls to dfs for each of these multiples, incrementing cnt each time. So we get dfs(2, 2), dfs(3, 2), and dfs(4, 2).
When dfs is called with cnt=3, we reach the base case, and we count the number of configurations using our combinatorics table c. In this case, it returns c[2][2] which is 1 because there's only one way to arrange two elements in two positions.
Next, we would move on to i=2 and do the same process, but this time the possible multiples are 4*2 because multiples like 6*2, 8*2 exceed maxValue.
Similarly for i=3 and i=4, we notice that there are no valid multiples within the maxValue for these starting points, so the recursive calls would immediately hit the base case and return c[n-1][cnt-1].

Once all possible starting i values are considered, we would sum up all the computed ways, applying the modulo as necessary, to obtain the total number of distinct ideal arrays.

In our small example, let's say the dfs calls return the following configurations:

For i=1, we get 3 arrays [1,1,1], [1,2,2], [1,4,4] (assuming dfs for i=3 returns 0 since it has no multiples within maxValue).
For i=2, we get 1 array [2,2,2].
For i=3, we get 0 arrays as explained.
For i=4, we also get 0 arrays.

Adding these up, we get a total of 4 distinct "ideal" integer arrays for n=3 and maxValue=4.

Each of the sums is calculated modulo 10^9 + 7 to prevent overflow and ensure the calculation remains within the required bounds. So, our final answer for this example would be 4 % (10^9 + 7), which is still 4.

Solution Implementation

1# The provided code is already in Python 3 syntax, but I've made changes
2# to improve readability, standardized naming, and added comments for better understanding.
3
4from functools import lru_cache  # Importing lru_cache for caching
5
6class Solution:
7    def idealArrays(self, n: int, maxValue: int) -> int:
8        # Using the lru_cache decorator to memoize the results of the dfs function
9        @lru_cache(maxsize=None)
10        def dfs(value, length):
11            """ 
12            Performs a depth-first search starting with a given value
13            and length of the sequence, and returns the number of ideal arrays.
14            """
15            # Initialize result with the combination C(n-1, length-1)
16            result = comb_table[-1][length - 1]
17            # If we haven't reached the desired length, continue building the sequence
18            if length < n:
19                multiple = 2
20                # Iterate through the possible multiples of 'value' within the maxValue
21                while multiple * value <= maxValue:
22                    # Recursively call dfs and update the result
23                    result = (result + dfs(multiple * value, length + 1)) % MODULE
24                    multiple += 1
25            return result
26      
27        # Comb table creation using dynamic programming to calculate combinations
28        comb_table = [[0] * 16 for _ in range(n)]
29        MODULE = 10**9 + 7  # Module for the problem, to prevent integer overflow
30        for i in range(n):
31            for j in range(min(16, i + 1)):
32                # Base case: There is one way to choose 0 items
33                if j == 0:
34                    comb_table[i][j] = 1
35                else:
36                    # Use the recurrence relation C(n, k) = C(n-1, k) + C(n-1, k-1)
37                    comb_table[i][j] = (comb_table[i - 1][j] + comb_table[i - 1][j - 1]) % MODULE
38      
39        # Initial answer is set to zero
40        answer = 0
41        # Iterate through all possible starting values
42        for i in range(1, maxValue + 1):
43            # Update the answer by using the dfs function
44            answer = (answer + dfs(i, 1)) % MODULE
45      
46        return answer  # Return the final answer
47

1import java.util.Arrays;
2
3class Solution {
4    private int[][] memoization;
5    private int[][] combinationMatrix;
6    private int numElements;
7    private int maxValue;
8    private static final int MOD = (int) 1e9 + 7;
9
10    public int idealArrays(int n, int maxValue) {
11        this.numElements = n;
12        this.maxValue = maxValue;
13
14        // Initialize memoization array with -1 to indicate uncalculated states
15        this.memoization = new int[maxValue + 1][16];
16        for (int[] row : memoization) {
17            Arrays.fill(row, -1);
18        }
19
20        // Pre-calculate the combination values (n choose k) up to n this combinationMatrix will be used in dfs
21        combinationMatrix = new int[n][16];
22        for (int i = 0; i < n; ++i) {
23            for (int j = 0; j <= i && j < 16; ++j) {
24                combinationMatrix[i][j] = (j == 0 ? 1 : (combinationMatrix[i - 1][j] + combinationMatrix[i - 1][j - 1]) % MOD);
25            }
26        }
27
28        int answer = 0;
29
30        // Calculate the total number of ideal arrays for all starting values
31        for (int i = 1; i <= maxValue; ++i) {
32            answer = (answer + dfs(i, 1)) % MOD;
33        }
34
35        return answer;
36    }
37
38    // Depth First Search to find the number of ideal arrays starting with 'startValue'
39    // and having 'length' number of distinct integers sorted in strictly increasing order.
40    private int dfs(int startValue, int length) {
41        // If previously computed, retrieve the result from memoization table
42        if (memoization[startValue][length] != -1) {
43            return memoization[startValue][length];
44        }
45
46        // Base value is 'n choose length-1'
47        int count = combinationMatrix[numElements - 1][length - 1];
48
49        // If length is less than the number of elements, continue to find all possible arrays
50        if (length < numElements) {
51            for (int k = 2; startValue * k <= maxValue; ++k) {
52                count = (count + dfs(startValue * k, length + 1)) % MOD;
53            }
54        }
55
56        // Store the result in the memoization table before returning
57        memoization[startValue][length] = count;
58        return count;
59    }
60}
61

1class Solution {
2public:
3    int maxValue, sequenceLength;
4    const int MOD = 1e9 + 7; // Modulus for the problem to handle large numbers
5
6    // f = memoization table for results of dfs function with index i and count cnt
7    // c = combinations table to store number of combinations to choose j elements from a set of size i
8    vector<vector<int>> f;
9    vector<vector<int>> c;
10
11    // Main function that starts the process to count ideal arrays
12    int idealArrays(int n, int maxValue) {
13        this->maxValue = maxValue;
14        this->sequenceLength = n;
15        // Initialize memoization table with -1 to represent uncalculated states
16        f.assign(maxValue + 1, vector<int>(16, -1));
17        // Initialize combinations table with 0 values
18        c.assign(n, vector<int>(16, 0));
19      
20        // Precalculate combinations using Pascal's triangle relationship
21        for (int i = 0; i < n; ++i) {
22            for (int j = 0; j <= i && j < 16; ++j) {
23                c[i][j] = j == 0 ? 1 : (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
24            }
25        }
26
27        int ans = 0;
28        // Sum the results of dfs for each number starting from 1 to maxValue
29        for (int i = 1; i <= maxValue; ++i) {
30            ans = (ans + dfs(i, 1)) % MOD;
31        }
32        return ans;
33    }
34
35    // Depth-first search function that calculates the number of ideal arrays 
36    // ending with the integer 'i', having 'cnt' unique integers
37    int dfs(int i, int cnt) {
38        // If the result has already been calculated, return it to avoid recalculation
39        if (f[i][cnt] != -1) return f[i][cnt];
40
41        // Base result is the number of sequences that can be formed with 'cnt' copies of 'i'
42        int res = c[sequenceLength - 1][cnt - 1];
43
44        // If we can add more unique numbers to the array
45        if (cnt < sequenceLength) {
46            // Try to extend the sequence by adding multiples of 'i'
47            for (int k = 2; i * k <= maxValue; ++k) {
48                res = (res + dfs(i * k, cnt + 1)) % MOD;
49            }
50        }
51
52        // Store the calculated result in the memoization table
53        f[i][cnt] = res;
54        return res;
55    }
56};
57

1const MOD = 1e9 + 7; // Modulus for the problem to handle large numbers
2
3let maxValue: number; // The maximum value that array elements can take
4let sequenceLength: number; // The length of the sequence
5let f: number[][]; // Memoization table for results of dfs function with index i and count cnt
6let c: number[][]; // Combinations table to store number of combinations
7
8// Function to initialize the necessary variables and tables
9function initializeSolution(n: number, mValue: number): void {
10  maxValue = mValue; // Store the provided maximum value
11  sequenceLength = n; // Store the provided sequence length
12  f = Array.from({ length: maxValue + 1 }, () => Array(16).fill(-1)); // Initialize memoization table with -1
13  c = Array.from({ length: n }, () => Array(16).fill(0)); // Initialize combinations table with 0
14
15  // Precalculate combinations using Pascal's triangle relationship
16  for (let i = 0; i < n; ++i) {
17    for (let j = 0; j <= i && j < 16; ++j) {
18      c[i][j] = j === 0 ? 1 : (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
19    }
20  }
21}
22
23// Main function to calculate the count of ideal arrays
24function idealArrays(n: number, mValue: number): number {
25  initializeSolution(n, mValue);
26
27  let ans = 0;
28  // Sum the results of dfs for each number starting from 1 to maxValue
29  for (let i = 1; i <= maxValue; ++i) {
30    ans = (ans + dfs(i, 1)) % MOD;
31  }
32  return ans;
33}
34
35// Depth-first search function to calculate the count of ideal arrays
36// ending with the integer 'i', having 'cnt' unique integers
37function dfs(i: number, cnt: number): number {
38  // Check the memoization table to avoid recalculating
39  if (f[i][cnt] !== -1) return f[i][cnt];
40
41  // The base result is the count of sequences that can be formed with 'cnt' copies of 'i'
42  let res = c[sequenceLength - 1][cnt - 1];
43
44  // If more unique numbers can be added to the array
45  if (cnt < sequenceLength) {
46    // Extend the sequence by adding multiples of 'i'
47    for (let k = 2; i * k <= maxValue; ++k) {
48      res = (res + dfs(i * k, cnt + 1)) % MOD;
49    }
50  }
51
52  // Store the calculated result in the memoization table
53  f[i][cnt] = res;
54  return res;
55}
56

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Time and Space Complexity

The given Python code represents a solution for computing the number of ideal arrays of length n with elements not exceeding maxValue. It uses dynamic programming with memoization (caching previous results) and combinatorial mathematics.

Time Complexity

The time complexity of this solution depends on the depth of the recursion and the number of iterations. Let's break down the time complexity:

The dfs function computes the number of ways to build an array starting with the element i and having cnt elements so far. It recursively calls itself while the multiples of i are less than or equal to maxValue. The memoization (@cache) dramatically reduces the number of computations by storing already computed values of dfs(i, cnt).
The recursion depth is at most log(maxValue) because we are considering the multiples of i until maxValue. Since the multiples are found by multiplying by k (k >= 2), the number of recursive calls for dfs is limited by the logarithmic scale of maxValue.
Inside every recursive call, a constant-time operation of combination calculation (+, %) is performed. For each i, the recursion explores log(maxValue) levels.
The outer loop runs from 1 to maxValue.

So, roughly, the time complexity can be estimated as O(maxValue * log(maxValue)).

Space Complexity

The space complexity includes the space taken by the recursive call stack and the space used to store combinations:

The cache used for memoization will have at most maxValue * n entries because each unique pair of (i, cnt) will be stored only once.
The space for storing combinations is O(n * 16) since we have limited the combination array to 16 elements wide.
The call stack's maximum depth would be log(maxValue) due to the nature of the recursion.

The dominant term in the space complexity is the cache size i.e. O(maxValue * n), so the overall space complexity can be considered as O(maxValue * n).

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

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2338. Count the Number of Ideal Arrays

Problem Description

Intuition

Solution Approach

Dynamic Programming and Memoization

Recursive Function dfs

Combination Preprocessing

Summation and Modulo Application

Final Result

Example Walkthrough

Solution Implementation

Recursive Function `dfs`