Count the Number of Computer Unlocking Permutations

1. Preliminary Check: Iterate from i = 1 to n-1. For each i, check if there exists at least one j < i such that complexity[j] < complexity[i]. If this condition fails for any i, no computer i can ever be unlocked, so no full permutation is possible. Return 0.
2. Calculate Prerequisites: For each computer k from 0 to n-1, we need to find the minimum complexity among its prerequisites. Let's denote this as m[k]. A prerequisite j for k satisfies j < k and complexity[j] < complexity[k]. So, m[k] = min({complexity[j] | j < k and complexity[j] < complexity[k]}). If k has no prerequisites, we can consider m[k] to be infinity. This step takes O(n^2) time.
3. Iterative Calculation: The core idea is that the total number of valid permutations can be built iteratively. Let f(i) be the number of valid permutations for the first i computers {0, ..., i-1}. The number of permutations for {0, ..., i} can be found by determining the number of valid positions to insert computer i into any valid permutation of {0, ..., i-1}. It can be shown that this number of insertion spots is i - |K_i|, where K_i is a specific subset of computers {0, ..., i-1}.
4. Identify K_i: The set K_i consists of computers k < i that do not depend on any of i's prerequisites. Formally, K_i = {k < i | complexity[k] >= complexity[i] and m[k] >= complexity[i]}. We calculate the size of this set, |K_i|, for each i from 1 to n-1.
5. Calculate Total Permutations: The final answer is the product of the number of choices at each step. Initialize ans = 1. Iterate i from 1 to n-1. In each step, calculate |K_i| by iterating k from 0 to i-1. The number of ways to place computer i is i - |K_i|. Update the answer: ans = (ans * (i - |K_i|)) % MOD. The total time complexity for this part is O(n^2).
6. Return Result: Return the final calculated ans.

The algorithm proceeds as follows:

First, we perform a basic check. For any computer i > 0 to be unlockable, it must have at least one prerequisite j < i with complexity[j] < complexity[i]. We can iterate through each i from 1 to n-1 and check if such a j exists. If not, we return 0. This check takes O(n^2).

Next, we define m[k] as the minimum complexity among all prerequisites of computer k. We can compute m[k] for all k from 0 to n-1 with a nested loop, which also takes O(n^2).

The main logic calculates the result ans as the product ans = product_{i=1 to n-1} (i - |K_i|). The set K_i is defined as {k < i | complexity[k] >= complexity[i] and m[k] >= complexity[i]}. We can compute |K_i| for each i by iterating k from 0 to i-1 and checking the conditions. This nested loop structure results in an O(n^2) complexity for the main calculation.

1class Solution {2    public int countPermutations(int[] complexity) {3        int n = complexity.length;4        long MOD = 1_000_000_007;5 6        // Preliminary check: O(n^2)7        for (int i = 1; i < n; i++) {8            boolean hasPrereq = false;9            for (int j = 0; j < i; j++) {10                if (complexity[j] < complexity[i]) {11                    hasPrereq = true;12                    break;13                }14            }15            if (!hasPrereq) {16                return 0;17            }18        }19 20        // Precompute m[k]: min complexity of prerequisites for k: O(n^2)21        long[] m = new long[n];22        for (int k = 0; k < n; k++) {23            m[k] = Long.MAX_VALUE;24            for (int j = 0; j < k; j++) {25                if (complexity[j] < complexity[k]) {26                    m[k] = Math.min(m[k], complexity[j]);27                }28            }29        }30 31        long ans = 1;32        // Main calculation: O(n^2)33        for (int i = 1; i < n; i++) {34            int countKi = 0;35            for (int k = 0; k < i; k++) {36                if (complexity[k] >= complexity[i] && m[k] >= complexity[i]) {37                    countKi++;38                }39            }40            long choices = i - countKi;41            ans = (ans * choices) % MOD;42        }43 44        return (int) ans;45    }46}

1. Coordinate Compression: The complexities can be large, but only their relative order matters. We gather all unique complexity values, sort them, and map each to a smaller integer rank. This takes O(N log N).
2. Optimized Prerequisite Calculation: We can calculate size[i] (number of prerequisites for i) and m[i] (minimum complexity of prerequisites for i) for all i efficiently. We process computers sorted by their complexity. We use a data structure like a Segment Tree or Fenwick Tree (BIT) over the computer indices [0, ..., n-1]. When processing computer i, we query the data structure for information about computers j < i that have already been processed (i.e., have smaller complexity). This allows computing all size[i] and m[i] values in O(N log N) time. During this step, we also perform the preliminary check: if size[i] == 0 for any i > 0, return 0.
3. Optimized |K_i| Calculation: The main bottleneck is calculating |K_i| = |{k < i | complexity[k] >= complexity[i] and m[k] >= complexity[i]}| for all i. This is a series of 2D range counting problems. We can solve this efficiently in O(N log N) using an offline approach with a Fenwick Tree (BIT).
   • Create events for points and queries. Points are (k, complexity[k], m[k]). Queries are (i, complexity[i]).
   • We need to count points k such that k < i, complexity[k] >= complexity[i], and m[k] >= complexity[i].
   • We can process all points and queries offline. Let's create events based on the complexity values. We group points and queries by their complexity C. We iterate through C in descending order.
   • We use a BIT on the indices [0, ..., n-1]. When processing complexity C, we first add all points k with complexity[k] == C to another data structure (e.g., lists grouped by m[k]). Then, we add points k with m[k] == C to our BIT at index k. Finally, for all queries i with complexity[i] == C, we query the BIT for the sum up to i-1. This gives us the count |K_i|.
4. Final Calculation: With all |K_i| values computed, we calculate the final answer ans = product_{i=1 to n-1} (i - |K_i|) in O(N) time.

The overall formula remains ans = product_{i=1 to n-1} (i - |K_i|). We optimize the computation of m_k and |K_i|.

1. Computing m_k and size_k in O(N log N): We first coordinate compress the complexity values. Then, we iterate through computers k sorted by complexity[k]. We use a segment tree over indices j < k. When processing k, all computers j processed so far have complexity[j] <= complexity[k]. We query the segment tree for min(complexity[j]) and count(j) over the range [0, k-1] to find m_k and size_k. After processing all computers with the same complexity as k, we update the segment tree with their information. This computes all m_k and size_k in O(N log N).

2. Computing |K_i| for all i in O(N log N): This is the most complex part. We need to find |{k < i | complexity[k] >= complexity[i] and m[k] >= complexity[i]}| for each i. This is a set of 3D counting queries. We can solve this with a sweep-line algorithm.

• Create a list of all complexity and m values, then coordinate compress them.
• Create two lists of events: points and queries.
  • points: For each k from 0 to n-1, add (complexity[k], m[k], k).
  • queries: For each i from 1 to n-1, add (complexity[i], i).
• Sort both points and queries by complexity in descending order.
• Use a Fenwick Tree (BIT) on indices [0, ..., n-1].
• Use two pointers, one for points (p_ptr) and one for queries (q_ptr). Iterate through queries.
• For each query (C, i):
  • Advance p_ptr to add all points (c_p, m_p, k_p) where c_p >= C to a temporary structure, grouped by m_p.
  • This is still complex. A better offline approach is to combine points and queries and sort by the main coordinate.
  • Let's sweep on the index t from 0 to n-1. We need a 2D data structure to handle queries on (complexity, m). A persistent segment tree can solve this in O(N log N). At each step t, we update the persistent segment tree with the point (complexity[t], m[t]) and for any query i=t, we use the tree's state to answer the 2D range query.

1// The O(N log N) solution is highly complex to implement from scratch.2// The following is a conceptual representation of the logic.3// A full implementation would require helper classes for Fenwick Tree/Segment Tree,4// coordinate compression, and careful offline event handling.5 6class Solution {7    public int countPermutations(int[] complexity) {8        // High-level logic for an O(N log N) solution9        int n = complexity.length;10        long MOD = 1_000_000_007;11 12        // 1. Coordinate compress complexity values. O(N log N)13 14        // 2. Compute size[i] and m[i] for all i in O(N log N)15        //    using a Fenwick/Segment tree and processing computers sorted by complexity.16        //    If any size[i] == 0 for i > 0, return 0.17        long[] m = new long[n]; // Assume m is computed18        // ... computation of m ...19 20        // 3. Compute |K_i| for all i in O(N log N)21        //    This is a 3D counting problem: count k < i s.t. c[k] >= c[i] and m[k] >= c[i].22        //    Can be solved with a sweep-line algorithm or a persistent segment tree.23        //    Let's assume `countsKi` array is populated with |K_i| values.24        int[] countsKi = new int[n]; // Assume this is computed25        // ... computation of countsKi ...26 27        // 4. Final calculation28        long ans = 1;29        for (int i = 1; i < n; i++) {30            long choices = i - countsKi[i];31            ans = (ans * choices) % MOD;32        }33 34        return (int) ans;35    }36}

Note: A full, correct O(N log N) implementation is very advanced. The provided snippet illustrates the high-level structure, as the details of the data structures are extensive.