Min Cost Climbing Stairs

Description

You are given an integer array cost where cost[i] is the cost of ith 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.

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.

Constraints:

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

Code

Basically it is Climbing Stairs with costs.

Time Complexity: $O(n)$, Space Complexity: $O(n)$

寫出 DP 關係式： DP[i] = min(DP[i-1] + cost[i-1], DP[i-2] + cost[i-2]);

接著定義出 base case：DP[0] = 0; DP[1] = 0;

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

第二種 DP 關係式：DP[i] = cost[i] + min(DP[i-1], DP[i-2]);

定義出 base case：DP[0] = cost[0]; DP[1] = cost[1];

Problem Solved! 差別在於使用的 Spcae 為 $n$ 格（上面的解法是 $n+1$）。

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

Optimization

Space complexity 可以壓到 $O(1)$，因為觀察到在 DP 計算的過程中，只會需要用到三個值。

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

Source

• Min Cost Climbing Stairs - LeetCode