LeetCode in Kotlin

3577. Count the Number of Computer Unlocking Permutations

Medium

You are given an array complexity of length n.

There are n locked computers in a room with labels from 0 to n - 1, each with its own unique password. The password of the computer i has a complexity complexity[i].

The password for the computer labeled 0 is already decrypted and serves as the root. All other computers must be unlocked using it or another previously unlocked computer, following this information:

• You can decrypt the password for the computer i using the password for computer j, where j is any integer less than i with a lower complexity. (i.e. j < i and complexity[j] < complexity[i])
• To decrypt the password for computer i, you must have already unlocked a computer j such that j < i and complexity[j] < complexity[i].

Find the number of permutations of [0, 1, 2, ..., (n - 1)] that represent a valid order in which the computers can be unlocked, starting from computer 0 as the only initially unlocked one.

Since the answer may be large, return it modulo 10⁹ + 7.

Note that the password for the computer with label 0 is decrypted, and not the computer with the first position in the permutation.

Example 1:

Input: complexity = [1,2,3]

Output: 2

Explanation:

The valid permutations are:

• [0, 1, 2]
  • Unlock computer 0 first with root password.
  • Unlock computer 1 with password of computer 0 since complexity[0] < complexity[1].
  • Unlock computer 2 with password of computer 1 since complexity[1] < complexity[2].
• [0, 2, 1]
  • Unlock computer 0 first with root password.
  • Unlock computer 2 with password of computer 0 since complexity[0] < complexity[2].
  • Unlock computer 1 with password of computer 0 since complexity[0] < complexity[1].

Example 2:

Input: complexity = [3,3,3,4,4,4]

Output: 0

Explanation:

There are no possible permutations which can unlock all computers.

Constraints:

• 2 <= complexity.length <= 105
• 1 <= complexity[i] <= 109

Solution

class Solution {
    fun countPermutations(complexity: IntArray): Int {
        val n = complexity.size
        for (i in 1..<n) {
            if (complexity[i] <= complexity[0]) {
                return 0
            }
        }
        var ans: Long = 1
        for (x in 2..<n) {
            ans = (ans * x) % MOD
        }
        return ans.toInt()
    }

    companion object {
        private const val MOD = 1000000007
    }
}

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