sherlocks-array-merging-algorithm

Watson gave Sherlock a collection of arrays V. Here each V_i is an array of variable length. It is guaranteed that if you merge the arrays into one single array, you'll get an array, M, of n distinct integers in the range [1, n].

Watson asks Sherlock to merge V into a sorted array. Sherlock is new to coding, but he accepts the challenge and writes the following algorithm:

M ← [] (an empty array).
k ← number of arrays in the collection V.
While there is at least one non-empty array in V:
- T ← [] (an empty array) and i ← 1.
- While i ≤ k :
  - If V_i is not empty:
    - Remove the first element of V_i and push it to T.
  - i ← i + 1.
- While T is not empty:
  - Remove the minimum element of T and push it to M.
Return T as the output.

Let's see an example. Let V be { [3,5], [1], [2,4] }. The image below demonstrates how Sherlock will do the merging according to the algorithm:

Sherlock isn't sure if his algorithm is correct or not. He ran Watson's input, V, through his pseudocode algorithm to produce an output, M, that contains an array of n integers. However, Watson forgot the contents of V and only has Sherlock's M with him! Can you help Watson reverse-engineer M to get the original contents of V?

Given m, find the number of different ways to create collection V such that it produces m when given to Sherlock's algorithm as input. As this number can be quite large, print it modulo 10^9+7.

Notes:

Two collections of arrays are different if one of the following is true:
- Their sizes are different.
- Their sizes are the same but at least one array is present in one collection but not in the other.
Two arrays, A and B, are different if one of the following is true:
- Their sizes are different.
- Their sizes are the same, but there exists an index i such that a_i ≠ b_i.

Input Format :

The first line contains an integer, n, denoting the size of array M. The second line contains n space-separated integers describing the respective values of m_0, m_1, ..., m_{n-1}.

Constraints :

$$ 1 \le n \le 1200 $$

$$ 1 \le m_i \le n $$

Output Format :

Print the number of different ways to create collection V, modulo 10^9 + 7.

Sample Input 0:	Sample Output 0:
`3 1 2 3`	`4`

Explanation 0 :

There are four distinct possible collections:

1. V = {[1,2,3]}
1. V = {[1], [2], [3]}
1. V = {[1,3], [2]}
1. v = {[1], [2,3]}

Thus, we print the result of 4 mod (10^9+7) = 4 as our answer.

Sample Input 1:	Sample Output 1:
`2 2 1`	`1`

Explanation 1 :

The only distinct possible collection is V = {[2, 1]}, so we print the result of 1mod(10^9+7) = 1 as our answer.

Solution :

#include <iostream>
#include <cstring>
using namespace std;

const int MAX_N = 1205;
const int MOD = 1e9 + 7;

int n;
int m[MAX_N]; // Input array M
int right_bound[MAX_N]; // Right boundaries of increasing sequences

int factorial[MAX_N]; // Factorial array
int inverse_factorial[MAX_N]; // Inverse factorial array

int dp[MAX_N][MAX_N]; // Dynamic programming array

// Recursive function to compute the number of ways to create collection V
int countWays(int last, int pos) {
    // Base case: if we reached the end of array M, return the inverse of last
    if (pos > n) return inverse_factorial[last];

    // If the value is already computed, return it
    if (dp[last][pos] != -1) return dp[last][pos];

    long long res = 0;
    // Iterate over the possible next positions to split the array
    for (int i = pos; i <= min(right_bound[pos], last + pos - 1); ++i) {
        // Compute the number of ways recursively and update the result
        res += (((1LL * factorial[i - pos + 1] * countWays(i - pos + 1, i + 1)) % MOD) * inverse_factorial[last - (i - pos + 1)]) % MOD;
    }
    // Store the result in dp array and return it
    return dp[last][pos] = res % MOD;
}

int main() {
    cin >> n;
    // Input array M
    for (int i = 1; i <= n; ++i) {
        cin >> m[i];
    }
    // Calculate right boundaries of increasing sequences
    right_bound[n] = n;
    for (int i = n - 1; i >= 1; --i) {
        if (m[i + 1] > m[i]) right_bound[i] = right_bound[i + 1];
        else right_bound[i] = i;
    }
    // Precompute factorial array
    factorial[0] = 1;
    for (int i = 1; i <= n; ++i) {
        factorial[i] = (1LL * factorial[i - 1] * i) % MOD;
    }
    // Precompute inverse factorial array using modulo inverse
    inverse_factorial[0] = 1;
    inverse_factorial[1] = 1;
    for (int i = 2; i <= n; ++i) {
        inverse_factorial[i] = ((-1LL * (MOD / i) * inverse_factorial[MOD % i]) % MOD + MOD) % MOD;
    }
    for (int i = 2; i <= n; ++i) {
        inverse_factorial[i] = (1LL * inverse_factorial[i - 1] * inverse_factorial[i]) % MOD;
    }
    // Initialize dp array with -1
    memset(dp, -1, sizeof(dp));

    int ans = 0;
    // Compute the answer
    for (int i = 1; i <= right_bound[1]; ++i) {
        ans += (1LL * factorial[i] * countWays(i, i + 1)) % MOD;
        if (ans >= MOD) ans -= MOD;
    }
    cout << ans << endl; // Output the answer
    return 0;
}

Code Walkthrough :

This is a problem from Dynamic Programming.

Header Files and Namespaces:

#include <iostream>
#include <cstring>
using namespace std;

These lines include necessary header files for input-output operations and string manipulation. The using namespace std; directive allows using standard namespace functions and objects without prefixing them with std::.

Constants and Arrays:

const int MAX_N = 1205;
const int MOD = 1e9 + 7;

int n;
int m[MAX_N]; // Input array M
int right_bound[MAX_N]; // Right boundaries of increasing sequences

int factorial[MAX_N]; // Factorial array
int inverse_factorial[MAX_N]; // Inverse factorial array

int dp[MAX_N][MAX_N]; // Dynamic programming array

MAX_N defines the maximum size of arrays.
MOD is the modulo value.
m stores the input array M.
right_bound stores the right boundaries of increasing sequences.
factorial stores factorials.
inverse_factorial stores inverse factorials.
dp is a dynamic programming array.

Recursive Function `countWays`:

int countWays(int last, int pos) {
    if (pos > n) return inverse_factorial[last];
    if (dp[last][pos] != -1) return dp[last][pos];

    long long res = 0;
    for (int i = pos; i <= min(right_bound[pos], last + pos - 1); ++i) {
        res += (((1LL * factorial[i - pos + 1] * countWays(i - pos + 1, i + 1)) % MOD) * inverse_factorial[last - (i - pos + 1)]) % MOD;
    }
    return dp[last][pos] = res % MOD;
}

Parameters: last is the length of the last segment considered, pos is the current position in the array m.
Base Case: If pos exceeds n, return the inverse factorial of last (no more elements to consider).
Memoization: If the result for dp[last][pos] is already computed, return it.
Recursive Calculation: Iterate over possible next positions (i) to split the array. For each position i, the function makes a recursive call to countWays with the new segment length i - pos + 1 and the new starting position i + 1.
- Calculate the number of ways recursively.
- Multiply with factorial and inverse factorial for proper combination counting, ensuring results are modulo MOD.
  - factorial[i - pos + 1]: Factorial of the length of the new segment.
  - countWays(i - pos + 1, i + 1): Recursive call for the next segment.
  - inverse_factorial[last - (i - pos + 1)]: Inverse factorial of the remaining length after considering the new segment.
Memoize and Return Result: Store the result in dp and return it.

Main Function:

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> m[i];
    }

    // Calculate right boundaries of increasing sequences
    right_bound[n] = n;
    for (int i = n - 1; i >= 1; --i) {
        if (m[i + 1] > m[i]) right_bound[i] = right_bound[i + 1];
        else right_bound[i] = i;
    }

    // Precompute factorial array
    factorial[0] = 1;
    for (int i = 1; i <= n; ++i) {
        factorial[i] = (1LL * factorial[i - 1] * i) % MOD;
    }

    // Precompute inverse factorial array using modulo inverse
    inverse_factorial[0] = 1;
    inverse_factorial[1] = 1;
    for (int i = 2; i <= n; ++i) {
        inverse_factorial[i] = ((-1LL * (MOD / i) * inverse_factorial[MOD % i]) % MOD + MOD) % MOD;
    }
    for (int i = 2; i <= n; ++i) {
        inverse_factorial[i] = (1LL * inverse_factorial[i - 1] * inverse_factorial[i]) % MOD;
    }

    memset(dp, -1, sizeof(dp)); // Initialize dp array with -1

    int ans = 0;
    // Compute the answer
    for (int i = 1; i <= right_bound[1]; ++i) {
        ans += (1LL * factorial[i] * countWays(i, i + 1)) % MOD;
        if (ans >= MOD) ans -= MOD;
    }
    cout << ans << endl; // Output the answer
    return 0;
}

`Input Reading`:

Read the integer n (size of array m).
Read the elements of the array m.

`Right Boundaries Calculation`:

Initialize right_bound[n] to n.
Traverse the array from the end to the beginning:
- If the next element is greater, propagate the right boundary.
- Otherwise, set the right boundary to the current index.

`Factorials and Inverse Factorials Precomputation`:

Compute the factorials modulo MOD.
Compute the modular inverses of the factorials using the property:
- inverse_factorial[i] = ((-1 * (MOD / i) * inverse_factorial[MOD % i]) % MOD + MOD) % MOD.
  
  Inverse Factorial using Fermat’s Little Theorem.
Ensure results are modulo MOD.

`Initialize dp Array`:

Set all elements of dp to -1 using memset.

`Compute the Answer`:

Iterate over possible initial segment lengths.
Use the countWays function to compute the number of valid ways for each segment length.
Sum the results, ensuring the sum is modulo MOD.

`Output the Result`:

Print the computed answer.

The countWays function calculates the number of ways to split the array starting from a given position into increasing subsequences. It uses:

Recursion: To explore all possible ways to split the array.
Memoization: To store already computed results for efficiency.
Factorials and Inverse Factorials: To correctly handle combinations under modular arithmetic constraints.

This approach ensures that all possible partitions are considered efficiently, and the results are computed accurately within the constraints of the problem.

This code efficiently calculates the number of different ways to create collection V, satisfying the conditions specified in the problem statement.

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sherlocks-array-merging-algorithm.MD

sherlocks-array-merging-algorithm.MD

sherlocks-array-merging-algorithm

Solution :

Code Walkthrough :

Header Files and Namespaces:

Constants and Arrays:

Recursive Function `countWays`:

Main Function:

`Input Reading`:

`Right Boundaries Calculation`:

`Factorials and Inverse Factorials Precomputation`:

`Initialize dp Array`:

`Compute the Answer`:

`Output the Result`:

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sherlocks-array-merging-algorithm

Solution :

Code Walkthrough :

Header Files and Namespaces:

Constants and Arrays:

Recursive Function countWays:

Main Function:

Input Reading:

Right Boundaries Calculation:

Factorials and Inverse Factorials Precomputation:

Initialize dp Array:

Compute the Answer:

Output the Result:

Recursive Function `countWays`:

`Input Reading`:

`Right Boundaries Calculation`:

`Factorials and Inverse Factorials Precomputation`:

`Initialize dp Array`:

`Compute the Answer`:

`Output the Result`: