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Formatted question description: https://leetcode.ca/all/746.html

746. Min Cost Climbing Stairs (Easy)

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.

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].

Note:

  1. cost will have a length in the range [2, 1000].
  2. Every cost[i] will be an integer in the range [0, 999].

Companies:
Amazon

Related Topics:
Array, Dynamic Programming

Similar Questions:

Solution 1. DP

// OJ: https://leetcode.com/problems/min-cost-climbing-stairs/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int prev = 0, cur = 0;
        for (int i = 2; i <= cost.size(); ++i) {
            int newCur = min(cur + cost[i - 1], prev + cost[i - 2]);
            prev = cur;
            cur = newCur;
        }
        return cur;
    }
};

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

############

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int a = 0, b = 0;
        for (int i = 1; i < cost.length; ++i) {
            int c = Math.min(a + cost[i - 1], b + cost[i]);
            a = b;
            b = c;
        }
        return b;
    }
}

  • // OJ: https://leetcode.com/problems/min-cost-climbing-stairs/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minCostClimbingStairs(vector<int>& A) {
        int first = 0, second = 0;
        for (int i = 2; i <= A.size(); ++i) {
            int cur = min(first + A[i - 2], second + A[i - 1]);
            first = second;
            second = cur;
        }
        return second;
    }
};

  • class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        a = b = 0
        for i in range(1, len(cost)):
            a, b = b, min(a + cost[i - 1], b + cost[i])
        return b

############

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        costed = [0, 0]
        for i in xrange(2, len(cost)):
            costed.append(min(costed[i - 1] + cost[i - 1], costed[i - 2] + cost[i - 2]))
        return min(costed[-1] + cost[-1], costed[-2] + cost[-2])

  • func minCostClimbingStairs(cost []int) int {
	a, b := 0, 0
	for i := 1; i < len(cost); i++ {
		a, b = b, min(a+cost[i-1], b+cost[i])
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

  • function minCostClimbingStairs(cost: number[]): number {
    let a = 0,
        b = 0;
    for (let i = 1; i < cost.length; ++i) {
        [a, b] = [b, Math.min(a + cost[i - 1], b + cost[i])];
    }
    return b;
}

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