LeetCode #3577 — MEDIUM

Count the Number of Computer Unlocking Permutations

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[1,2,3]

Example 2

[3,3,3,4,4,4]

Core Insight

What unlocks the optimal approach

Ensure that the element at index 0 has the unique minimum complexity (no other element can match its value).
Fix index 0 as the first in the unlocking order.
The remaining indices from <code>1</code> to <code>n - 1</code> can then be arranged arbitrarily, yielding <code>factorial(n - 1)</code> possible orders.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3577: Count the Number of Computer Unlocking Permutations
class Solution {
    public int countPermutations(int[] complexity) {
        final int mod = (int) 1e9 + 7;
        long ans = 1;
        for (int i = 1; i < complexity.length; ++i) {
            if (complexity[i] <= complexity[0]) {
                return 0;
            }
            ans = ans * i % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #3577: Count the Number of Computer Unlocking Permutations
func countPermutations(complexity []int) int {
	mod := int64(1e9 + 7)
	ans := int64(1)
	for i := 1; i < len(complexity); i++ {
		if complexity[i] <= complexity[0] {
			return 0
		}
		ans = ans * int64(i) % mod
	}
	return int(ans)
}

# Accepted solution for LeetCode #3577: Count the Number of Computer Unlocking Permutations
class Solution:
    def countPermutations(self, complexity: List[int]) -> int:
        mod = 10**9 + 7
        ans = 1
        for i in range(1, len(complexity)):
            if complexity[i] <= complexity[0]:
                return 0
            ans = ans * i % mod
        return ans

// Accepted solution for LeetCode #3577: Count the Number of Computer Unlocking Permutations
impl Solution {
    pub fn count_permutations(complexity: Vec<i32>) -> i32 {
        const MOD: i64 = 1_000_000_007;
        let mut ans = 1i64;
        for i in 1..complexity.len() {
            if complexity[i] <= complexity[0] {
                return 0;
            }
            ans = ans * i as i64 % MOD;
        }
        ans as i32
    }
}

// Accepted solution for LeetCode #3577: Count the Number of Computer Unlocking Permutations
function countPermutations(complexity: number[]): number {
    const mod = 1e9 + 7;
    let ans = 1;
    for (let i = 1; i < complexity.length; i++) {
        if (complexity[i] <= complexity[0]) {
            return 0;
        }
        ans = (ans * i) % mod;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.