2560. House Robber IV - Detailed Explanation
Problem Statement
Description:
You are given an integer array nums representing the amount of money in each house and an integer k. In this variant of the house robber problem, you want to select exactly k houses to rob. However, there is a twist: you are only allowed to rob houses that have at least a certain threshold value.
Additionally, you cannot rob two adjacent houses. Your goal is to determine the maximum possible threshold such that there exists a way to choose k houses (without picking two consecutive houses) where each chosen house has money greater than or equal to that threshold.
In other words, define a threshold value x. The decision problem is: "Can you pick at least k non-adjacent houses from nums such that every chosen house has a value ≥ x?" The final answer is the maximum x for which this is possible.
Examples:
-
Example 1:
Input:
nums = [2, 3, 5, 9, 4] k = 2Output:
4Explanation:
- Consider threshold
x = 4.
Houses with values ≥ 4 are at indices 2, 3, and 4 (values 5, 9, and 4).
A valid selection is to choose the house at index 2 (value 5) and then skip index 3 (since adjacent) and choose index 4 (value 4). This gives 2 houses meeting the requirement. - Trying a higher threshold such as
x = 5is not feasible because even though houses at indices 2 and 3 have values 5 and 9, they are adjacent, and you cannot pick both. Thus, the maximum threshold for which you can pick 2 non-adjacent houses is 4.
- Consider threshold
-
Example 2:
Input:
nums = [5, 4, 3, 2, 1] k = 2Output:
3Explanation:
- For threshold
x = 3:
You can choose the house at index 0 (5) and then skip index 1; at index 2 the value 3 meets the requirement. Thus, 2 houses are chosen. - A threshold of 4 would only let you pick index 0 (5) while indices 1 (4) and others would be too close or not high enough when considering non-adjacency. Hence, 3 is the maximum threshold.
- For threshold
Constraints:
- ( 1 \leq \text{nums.length} \leq 10^5 )
- ( 1 \leq k \leq \text{nums.length} )
- ( 1 \leq \text{nums}[i] \leq 10^9 )
Hints
-
Hint 1:
Think about how the feasibility of a threshold is monotonic. In other words, if it’s possible to choosekhouses with a thresholdx, then any threshold lower thanxis also possible. This suggests that you can use binary search on the threshold value. -
Hint 2:
For a given thresholdx, use a greedy strategy: scan through the array and whenever you find a house with a value ≥x, choose it and skip the next house (to maintain the non-adjacency requirement). Count how many houses you can pick; if you reach at leastk, thenxis feasible.
Approaches
Approach 1: Brute-Force (Not Recommended)
One might try to test every possible combination of houses (subject to the non-adjacency constraint) and check the minimum value among the chosen houses. Then, maximize that minimum value over all valid selections.
- Drawback:
This approach is combinatorial (exponential in complexity) and infeasible for large input sizes.
Approach 2: Binary Search on Answer (Optimal)
Idea:
- Binary Search Range:
The potential threshold values lie between the minimum and maximum values innums. - Decision Function:
For a candidate thresholdx, simulate a greedy selection:- Initialize a counter and index
i = 0. - Traverse
nums: ifnums[i] >= x, count this house and skip the next house (i.e., incrementiby 2); otherwise, move to the next house (i++). - If the count reaches or exceeds
k, then it is feasible to achieve thresholdx.
- Initialize a counter and index
- Adjust the Search:
- If the candidate threshold is feasible, try a higher value.
- Otherwise, try a lower threshold.
- Result:
The highest threshold for which the decision function returns feasible is the answer.
Why It Works:
- Monotonicity:
If you can achieve thresholdx, you can achieve any threshold below it. - Efficiency:
Binary search reduces the number of candidate thresholds logarithmically while the greedy check runs in linear time.
Code Implementation
Python Code
Java Code
Complexity Analysis
-
Time Complexity:
-
The binary search runs in ( O(\log(\text{max(nums)} - \text{min(nums)})) ) iterations.
-
Each iteration runs a linear scan of the array ( O(n) ).
Thus, the overall time complexity is ( O(n \log M) ) where ( M ) is the range of values in
nums. -
-
Space Complexity:
The solution uses a constant amount of extra space ( O(1) ) (apart from the input).
Step-by-Step Walkthrough & Visual Example
Consider the array nums = [2, 3, 5, 9, 4] with ( k = 2 ).
-
Determine the Search Range:
- Minimum value: 2
- Maximum value: 9
- Binary search is performed between 2 and 9.
-
Binary Search Iteration:
- Iteration 1:
- mid = ( \lfloor (2+9)/2 \rfloor = 5 )
- Check feasibility:
- Traverse: indices 0 (2 < 5), index 1 (3 < 5), index 2 (5 ≥ 5) → count = 1, skip index 3; then index 4 (4 < 5).
- Count = 1, which is less than ( k = 2 ).
- Thus, threshold 5 is not feasible; set hi = mid - 1 = 4.
- Iteration 2:
- Now search between 2 and 4.
- mid = ( \lfloor (2+4)/2 \rfloor = 3 )
- Check feasibility:
- Traverse: index 0 (2 < 3), index 1 (3 ≥ 3) → count = 1, skip index 2; index 3 (9 ≥ 3) → count = 2, skip index 4.
- Count = 2, which meets ( k ).
- Threshold 3 is feasible; record answer = 3, and try a higher threshold by setting lo = mid + 1 = 4.
- Iteration 3:
- Now search between 4 and 4.
- mid = 4
- Check feasibility:
- Traverse: index 0 (2 < 4), index 1 (3 < 4), index 2 (5 ≥ 4) → count = 1, skip index 3; index 4 (4 ≥ 4) → count = 2.
- Count = 2, feasible.
- Record answer = 4 and set lo = mid + 1 = 5.
- Now lo (5) > hi (4), so the binary search ends and the final answer is 4.
- Iteration 1:
-
Result:
The maximum threshold is 4.
Common Mistakes
- Overlooking Non-adjacency:
Failing to skip the next index after choosing a house can lead to an invalid selection. - Binary Search Boundaries:
Not correctly setting or updating the low and high bounds in the binary search may lead to an incorrect answer. - Edge Handling:
Ensure the code correctly handles cases where ( k ) equals the number of houses or when no valid selection exists (though per constraints, a solution is guaranteed).
Edge Cases & Alternative Variations
-
Edge Case 1:
Whennumscontains only one house and ( k = 1 ). The answer is simply the value of that house. -
Edge Case 2:
When all houses have the same value; the maximum threshold is that common value. -
Alternative Variation:
Instead of exactly ( k ) houses, some variations might ask for at least ( k ) houses. The feasibility check would be similar, with only a slight adjustment in the condition.
Related Problems for Further Practice
- House Robber I / II / III:
These classic problems explore different constraints on which houses can be robbed. - Binary Search on Answer:
Practice problems that require using binary search over a range of possible answers, such as "Capacity To Ship Packages Within D Days" or "Koko Eating Bananas." - Greedy Algorithms:
Problems where a greedy selection strategy (combined with binary search) is needed.
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