LeetCode Entry

3693. Climbing Stairs II

10.11.2026 medium 2026 kotlin rust

Min cost c[j]+(j-i)^2 to reach the end

3693. Climbing Stairs II medium substack youtube

https://dmitrysamoylenko.com/leetcode/

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Problem TLDR

Min cost c[j]+(j-i)^2 to reach the end

Intuition

Attention to the 1-indexed part. Otherwise it is a basic choice problem of the last 3 choices.

Approach

• we can modify c in place
• fold is the shortest

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(n)\)

Code

    fun climbStairs(n: Int, c: IntArray) =
    c.fold(List(3){0}){(a,b,d),x->listOf(x+minOf(a+1,b+4,d+9),a,b)}[0]

    pub fn climb_stairs(n: i32, c: Vec<i32>) -> i32 {
        c.iter().fold([0;3],|[a,b,d],x|[x+(a+1).min(b+4).min(d+9),a,b])[0]
    }