Min Cost Climbing Stairs
LeetCode 747 | Difficulty: Easyβ
EasyProblem Descriptionβ
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.
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
Topics: Array, Dynamic Programming
Approachβ
Dynamic Programmingβ
Break the problem into overlapping subproblems. Define a state (what information do you need?), a recurrence (how does state[i] depend on smaller states?), and a base case. Consider both top-down (memoization) and bottom-up (tabulation) approaches.
When to use
Optimal substructure + overlapping subproblems (counting ways, min/max cost, feasibility).
Solutionsβ
Solution 1: C# (Best: 146 ms)β
| Metric | Value |
|---|---|
| Runtime | 146 ms |
| Memory | 36.8 MB |
| Date | 2022-01-24 |
Solution
public class Solution {
public int MinCostClimbingStairs(int[] cost) {
int n = cost.Length;
int[] dp = new int[n];
dp[0] = cost[0];
dp[1] = cost[1];
for (int i = 2; i < n; i++)
{
dp[i] = cost[i]+Math.Min(dp[i-1], dp[i-2]);
}
return Math.Min(dp[n-1], dp[n-2]);
}
}
Complexity Analysisβ
| Approach | Time | Space |
|---|---|---|
| Dynamic Programming |