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Author: NamVr

2690-house-robber-iv/README.md

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+<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>

2690-house-robber-iv/solution.py

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+class Solution:
+    def minCapability(self, nums, k):
+        # Store the maximum nums value in maxReward.
+        min_reward, max_reward = 1, max(nums)
+        total_houses = len(nums)
+
+        # Use binary search to find the minimum reward possible.
+        while min_reward < max_reward:
+            mid_reward = (min_reward + max_reward) // 2
+            possible_thefts = 0
+
+            index = 0
+            while index < total_houses:
+                if nums[index] <= mid_reward:
+                    possible_thefts += 1
+                    index += 2  # Skip the next house to maintain the non-adjacent condition
+                else:
+                    index += 1
+
+            if possible_thefts >= k:
+                max_reward = mid_reward
+            else:
+                min_reward = mid_reward + 1
+
+        return min_reward