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House Robber IV

Problem Description​

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 refuses to steal from adjacent homes.

The capability of the robber is the maximum amount of money he steals from one house of all the houses he robbed.

You are given an integer array nums representing how much money is stashed in each house. More formally, the ith house from the left has nums[i] dollars.

You are also given an integer k, representing the minimum number of houses the robber will steal from. It is always possible to steal at least k houses.

Return the minimum capability of the robber out of all the possible ways to steal at least k houses.

Examples​

Example 1:

Input: nums = [2,3,5,9], k = 2
Output: 5
Explanation:
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.

Example 2:

Input: nums = [2,7,9,3,1], k = 2
Output: 2
Explanation: 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.

Constraints​

  • 1 <= nums.length <= 10^5
  • 1 <= nums[i] <= 10^9
  • 1 <= k <= (nums.length + 1)/2

Approach​

The problem is asking for the minimum stealing ability of the thief. We can use binary search to enumerate the stealing ability of the thief. For the enumerated ability xx, we can use a greedy approach to determine whether the thief can steal at least kk houses. Specifically, we traverse the array from left to right. For the current house ii we are traversing, if nums[i]≀xnums[i] \leq x and the difference between the index of ii and the last stolen house is greater than 11, then the thief can steal house ii. Otherwise, the thief cannot steal house ii. We accumulate the number of stolen houses. If the number of stolen houses is greater than or equal to kk, it means that the thief can steal at least kk houses, and at this time, the stealing ability xx of the thief might be the minimum. Otherwise, the stealing ability xx of the thief is not the minimum.

The time complexity is O(nΓ—log⁑m)O(n \times \log m), and the space complexity is O(1)O(1). Where nn and mm are the length of the array numsnums and the maximum value in the array numsnums, respectively.

Python3​

class Solution:
def minCapability(self, nums: List[int], k: int) -> int:
def f(x):
cnt, j = 0, -2
for i, v in enumerate(nums):
if v > x or i == j + 1:
continue
cnt += 1
j = i
return cnt >= k

return bisect_left(range(max(nums) + 1), True, key=f)

Java​

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;
}
}

C++​

class Solution {
public:
int minCapability(vector<int>& nums, int k) {
auto f = [&](int x) {
int cnt = 0, j = -2;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] > x || i == j + 1) {
continue;
}
++cnt;
j = i;
}
return cnt >= k;
};
int left = 0, right = *max_element(nums.begin(), nums.end());
while (left < right) {
int mid = (left + right) >> 1;
if (f(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
};