746. Min Cost Climbing Stairs

Description

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Constraints

• cost will have a length in the range [2, 1000].

• Every cost[i] will be an integer in the range [0, 999].

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: cost = [10, 15, 20]

Output: 15

Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]

Output: 6

Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

Solutions

Solution 1

/**
 * Time complexity : O(N) where N is the length of cost.
 * Space complexity : O(1), the space used by f1, f2.
 */

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int f1 = 0, f2 = 0;
        for (int i = cost.length - 1; i >= 0; --i) {
            int f0 = cost[i] + Math.min(f1, f2);
            f2 = f1;
            f1 = f0;
        }
        return Math.min(f1, f2);
    }
}

Solution 2

/**
 * Time complexity : O(N) where N is the length of cost.
 * Space complexity : O(1)
 */

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        if(cost == null || cost.length == 0) {
            return 0;
        }
        int n = cost.length;
        if(n == 1) {
            return cost[0];
        }
        for(int i = 2; i < cost.length; i++) {
            cost[i] += Math.min(cost[i-2], cost[i-1]);
        }
        return Math.min(cost[n-2], cost[n-1]);
    }
}

Follow up