<p>You are given an integer array <code>cost</code> where <code>cost[i]</code> is the cost of <code>i<sup>th</sup></code> step on a staircase. Once you pay the cost, you can either climb one or two steps.</p>

<p>You can either start from the step with index <code>0</code>, or the step with index <code>1</code>.</p>

<p>Return <em>the minimum cost to reach the top of the floor</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> cost = [10,<u>15</u>,20]
<strong>Output:</strong> 15
<strong>Explanation:</strong> You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> cost = [<u>1</u>,100,<u>1</u>,1,<u>1</u>,100,<u>1</u>,<u>1</u>,100,<u>1</u>]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= cost.length &lt;= 1000</code></li>
	<li><code>0 &lt;= cost[i] &lt;= 999</code></li>
</ul>

// Approach 3: Bottom-Up, Constant Space
class Solution {
public:
    
    int minCostClimbingStairs(vector<int>& cost) {
        int oneStep = 0;
        int twoStep = 0;
        for(int i = 2; i < cost.size()+1; i++) {
            int temp = oneStep;
            oneStep = min(oneStep + cost[i-1], twoStep + cost[i-2]);
            twoStep = temp;
        }
        return oneStep;
    }
};
// Given N as the length of cost,
// Time complexity: O(N)
// minimumCost gets called with each index from 0 to N. Because of our memoization, each call will only take O(1) time.
// Space complexity: O(N)
// The extra space used by this algorithm is the recursion call stack. For example, minimumCost(10000) will call minimumCost(9999), which calls minimumCost(9998) etc., all the way down until the base cases at minimumCost(0) and minimumCost(1). In addition, our hash map memo will be of size N at the end, since we populate it with every index from 0 to N.


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