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Leetcode Problem 2560. House Robber IV

2560. House Robber IV

Leetcode Solutions

Binary Search with Greedy Checking

  1. Initialize left to 1 and right to the maximum value in nums (1e9).
  2. Perform binary search: a. Calculate mid as the average of left and right. b. Use a greedy approach to count the number of houses that can be robbed with capability mid without robbing adjacent houses. c. If the count is at least k, update right to mid. Otherwise, update left to mid + 1.
  3. Continue the binary search until left is equal to right, which will be the minimum capability required.
  4. Return left as the answer.
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Greedy Approach with Sorting and Union Find

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