House Robber IV
<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>
Greedy; Array; Binary Search; Dynamic Programming
class Solution {
    public int minCapability(int[] nums, int k) {
        int left = 0, right = (int) 1e9;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (f(nums, mid) >= k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private int f(int[] nums, int x) {
        int cnt = 0, j = -2;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] > x || i == j + 1) {
                continue;
            }
            ++cnt;
            j = i;
        }
        return cnt;
    }
}