<p>There are several consecutive houses along a street, each of which has some money inside. There is also a robber, who wants to steal money from the homes, but he <strong>refuses to steal from adjacent homes</strong>.</p>

<p>The <strong>capability</strong> of the robber is the maximum amount of money he steals from one house of all the houses he robbed.</p>

<p>You are given an integer array <code>nums</code> representing how much money is stashed in each house. More formally, the <code>i<sup>th</sup></code> house from the left has <code>nums[i]</code> dollars.</p>

<p>You are also given an integer <code>k</code>, representing the <strong>minimum</strong> number of houses the robber will steal from. It is always possible to steal at least <code>k</code> houses.</p>

<p>Return <em>the <strong>minimum</strong> capability of the robber out of all the possible ways to steal at least </em><code>k</code><em> houses</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,3,5,9], k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
There are three ways to rob at least 2 houses:
- Rob the houses at indices 0 and 2. Capability is max(nums[0], nums[2]) = 5.
- Rob the houses at indices 0 and 3. Capability is max(nums[0], nums[3]) = 9.
- Rob the houses at indices 1 and 3. Capability is max(nums[1], nums[3]) = 9.
Therefore, we return min(5, 9, 9) = 5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,7,9,3,1], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 7 ways to rob the houses. The way which leads to minimum capability is to rob the house at index 0 and 4. Return max(nums[0], nums[4]) = 2.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= (nums.length + 1)/2</code></li>
</ul>

class Solution:
    def minCapability(self, nums: List[int], k: int) -> int:
        n = len(nums)
        # 猜capacity
        # 如果这个capacity太小，那么说明大部分的房子都偷不了，最后偷的房子不超过k
        def check(target):
            # 通过dp来计算在capacity = target的情况下小偷能抢劫的房子的个数
            # dp[i] 表示前i个房子小偷能抢的房子的个数
            # dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
            dp = [0] * (n + 2)
            for i, num in enumerate(nums):
                # 当前房子只有在num <= target 情况下才能抢劫
                if num <= target:
                    # 这个情况可以抢劫，可以抢劫也要分可抢可不抢
                    dp[i + 2] = max(dp[i + 1], dp[i] + 1)
                else:
                    dp[i + 2] = dp[i + 1]
            return dp[n + 1] >= k
            
        
        left = 0
        right = max(nums)
        while left <= right:
            mid = (left + right) // 2
            if check(mid):
                right = mid - 1
            else:
                left = mid + 1
        return right + 1
                
                
            
        return 0

        

        
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