746. Min Cost Climbing Stairs

1. Description

You are given an integer array cost where cost[i] is the cost of $i^{th}$ step on a staircase. Once you pay the cost, you can either climb one or two steps.
You can either start from the step with index 0, or the step with index 1.
Return the minimum cost to reach the top of the floor.

2. Example

Example 1

Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.

• Pay 15 and climb two steps to reach the top.

The total cost is 15.

Example 2

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.

• Pay 1 and climb two steps to reach index 2.
• Pay 1 and climb two steps to reach index 4.
• Pay 1 and climb two steps to reach index 6.
• Pay 1 and climb one step to reach index 7.
• Pay 1 and climb two steps to reach index 9.
• Pay 1 and climb one step to reach the top.

The total cost is 6.

3. Constraints

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

4. Solutions

Dynamic Programming

n = cost.size()
Time complexity: O(n)
Space complexity: O(1)

// dp[i] represents the minimum cost to reach step i (standing on step i)
// You can reach step i from step i-1 or step i-2
// Transition:
//   dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])
//
// We use rolling variables to optimize space:
//   a = dp[i - 2], b = dp[i - 1]

class Solution {
public:
    int minCostClimbingStairs(const vector<int> &cost) {
        int a = 0, b = 0, min_sum = min(cost[0], cost[1]);
        for (int i = 2; i <= cost.size(); ++i) {
            min_sum = min(a + cost[i - 2], b + cost[i - 1]);
            a = b;
            b = min_sum;
        }

        return min_sum;
    }
};