3693 - Climbing Stairs II
problem
You are climbing a staircase with n + 1 steps, numbered from 0 to n.
You are also given a 1-indexed integer array costs of length n, where costs[i] is the cost of step i.
From step i, you can jump only to step i + 1, i + 2, or i + 3. The cost of jumping from step i to step j is defined as: costs[j] + (j - i)2
You start from step 0 with cost = 0.
Return the minimum total cost to reach step n.
Example 1:
Input: n = 4, costs = [1,2,3,4]
Output: 13
Explanation:
One optimal path is 0 → 1 → 2 → 4
| Jump | Cost Calculation | Cost |
|---|---|---|
| 0 → 1 | costs[1] + (1 - 0)2 = 1 + 1 | 2 |
| 1 → 2 | costs[2] + (2 - 1)2 = 2 + 1 | 3 |
| 2 → 4 | costs[4] + (4 - 2)2 = 4 + 4 | 8 |
Thus, the minimum total cost is 2 + 3 + 8 = 13
Example 2:
Input: n = 4, costs = [5,1,6,2]
Output: 11
Explanation:
One optimal path is 0 → 2 → 4
| Jump | Cost Calculation | Cost |
|---|---|---|
| 0 → 2 | costs[2] + (2 - 0)2 = 1 + 4 | 5 |
| 2 → 4 | costs[4] + (4 - 2)2 = 2 + 4 | 6 |
Thus, the minimum total cost is 5 + 6 = 11
Example 3:
Input: n = 3, costs = [9,8,3]
Output: 12
Explanation:
The optimal path is 0 → 3 with total cost = costs[3] + (3 - 0)2 = 3 + 9 = 12
Constraints:
1 <= n == costs.length <= 1051 <= costs[i] <= 104
submission
// dynamic programming
// let a, b, c be the previous 3 minimal costs
impl Solution {
pub fn climb_stairs(n: i32, costs: Vec<i32>) -> i32 {
costs.into_iter().fold(
(0, 0, 0), // its ok to assume they are all 0 initially
|(a, b, c), cost| (
b,
c,
// the real transition happens here
cost + (a + 9).min(b + 4).min(c + 1),
),
)
// use the last one
.2
}
}