LeetCode Entry

3577. Count the Number of Computer Unlocking Permutations

10.12.2025 medium 2025 kotlin rust

Permutations to unlock all c[j] right and bigger than c[i]

3577. Count the Number of Computer Unlocking Permutations medium blog post substack youtube

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Problem TLDR

Permutations to unlock all c[j] right and bigger than c[i] #medium #combinatorics

Intuition

    // 123456
    // 1 then take all values bigger as second position
    // 1 p[23456]
    // 12 p[3456]
    // 13 p[2456], the number is n!
    // what if we have duplicates?
    // 1223456
    // we are not allowed to sort
    // 1654223
    // 1 [654223] but every number can be at any place
    //           so it doesnt matter
    // so it is (n-1)!
    // how to  calc factorial of 10^5?
    //
    //

• first should be the smallest
• other numbers doesn’t matter

Approach

• use longs

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(1)\)

Code

// 9ms
    fun countPermutations(c: IntArray) = (1..<c.size)
    .fold(1L) { r, t -> if (c[t] > c[0]) (r*t) % 1000000007 else 0 }

// 0ms
    pub fn count_permutations(c: Vec<i32>) -> i32 {
        (1..c.len()).fold(1, |r,t|
        if c[t] > c[0] {(r*t)%1000000007} else { 0 })as _
    }