How do you solve House Robber IV on LeetCode?

House Robber IV (LeetCode #2560) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Binary search can efficiently narrow down the search space for the minimum capability.

What is the brute force solution for House Robber IV?

The brute force approach for House Robber IV is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of houses the robber can rob, ensuring that at least k houses are selected while avoiding adjacent houses. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve House Robber IV?

House Robber IV uses the following concepts: Array, Binary Search, Dynamic Programming, Greedy. The recommended patterns to study are: Binary Search, Greedy, Dynamic Programming.

What are the interview tips for House Robber IV?

Always consider edge cases, such as the smallest or largest possible inputs. Explain your thought process clearly during the interview; it shows your understanding. Practice both brute force and optimal solutions to be prepared for different problem types.

What are common mistakes when solving House Robber IV?

Not checking for adjacency correctly when generating combinations. Failing to optimize the brute force solution when a more efficient approach is available.

#2560

House Robber IV

Medium

ArrayBinary SearchDynamic ProgrammingGreedyBinary SearchGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

The optimal solution utilizes binary search combined with a greedy approach to efficiently find the minimum capability. By searching for the minimum possible maximum value that allows robbing at least k houses, we reduce the problem size significantly.

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Algorithm

3 steps

1Step 1: Initialize low as the minimum value in nums and high as the maximum value.
2Step 2: Perform binary search while low is less than high: - Calculate mid as the average of low and high. - Check if it's possible to rob at least k houses with the maximum capability of mid. - If possible, update high to mid; otherwise, update low to mid + 1.
3Step 3: Return low as the minimum capability.

solution.py18 lines

1def canRob(nums, k, max_cap):
2    count = 0
3    prev_index = -2
4    for i in range(len(nums)):
5        if nums[i] <= max_cap and i - prev_index > 1:
6            count += 1
7            prev_index = i
8    return count >= k
9
10def minCapability(nums, k):
11    low, high = min(nums), max(nums)
12    while low < high:
13        mid = (low + high) // 2
14        if canRob(nums, k, mid):
15            high = mid
16        else:
17            low = mid + 1
18    return low

ℹ

Complexity note: The time complexity is O(n log m) where n is the number of houses and m is the range of money values. We perform binary search on the money values while checking each value against the houses in linear time.

1Binary search can efficiently narrow down the search space for the minimum capability.
2Greedy checks ensure that we can rob the required number of houses without violating adjacency rules.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.