2560. House Robber IV
Problem Description
In this problem, we are considering a street lined with consecutive houses, each containing a certain amount of money. A robber aims to steal money from these houses, but with one condition: they are not willing to rob two adjacent houses. This constraint is dictated by the need to minimize the risk of getting caught. Additionally, there is a requirement that the robber must steal from at least k
houses. The robber's "capability" is defined as the maximum amount of money stolen from a single house during the heist.
We are given an array of integers nums
, where nums[i]
represents the amount of money in the i
th house. We also have an integer k
, which represents the minimum number of houses from which the robber must steal.
Our task is to calculate the minimum "capability" of the robber to achieve the heist of at least k
houses. In other words, among all the possible strategies to rob at least k
houses while avoiding adjacent homes, we want to find the one that involves the robber stealing the least amount of money from any single house they choose to rob. This will be the smallest maximum amount of money taken from one house in any valid robbing strategy.
Intuition
The intuition behind the solution is to use binary search along with a custom feasibility function to find the desired minimum capability. The first insight is that if the robber can complete the heist with a certain capability, they can also complete it with any higher capability. This makes the problem a good candidate for binary search, where the answer is the smallest capability that allows the robber to rob at least k
non-adjacent houses.
Here's the step-by-step approach:
-
Define a feasibility function
f(x)
that takes a capability valuex
and returnsTrue
if it's possible for the robber to complete the heist with capabilityx
, orFalse
otherwise. This function simulates the robber's actions, stealing from non-adjacent houses with values less than or equal to 'x' until they have reached the target ofk
houses. -
Apply binary search to find the minimum
x
for whichf(x)
isTrue
. The search space is from 0 to the maximum amount in any house (inclusive), as the robber could potentially need to rob the house with the most money ifk
equals the total number of houses. -
The feasibility function keeps two variables: a counter
cnt
for the number of successfully robbed houses, andj
to track the index of the last house the robber stole from. -
For each house, if its value is greater than
x
or it's adjacent to the last robbed house (indexj
), the robber skips it. -
Whenever a house is eligible (it satisfies the conditions and is not next to the last robbed house), increment
cnt
and updatej
to the current house's index. -
If
cnt
reaches at leastk
,f(x)
returnsTrue
, indicating that robbing at leastk
houses is feasible with the current capabilityx
. -
Use Python's
bisect_left
function with a range of capabilities and the feasibility function as the key. This will find the leftmost index wheref(x)
isTrue
, which corresponds to the minimum capability required.
By using binary search, we efficiently narrow down the minimum capability required rather than checking every possible capability value.
Learn more about Binary Search patterns.
Solution Approach
The solution is implemented in Python and utilizes a binary search algorithm. Binary search is a highly efficient algorithm used to find an element in a sorted sequence by repeatedly dividing the search interval in half.
Hereās a detailed explanation of the solution implementation:
-
Binary Search Rationale: Since we are looking for the minimum capability, we set up a binary search over a range of possible capabilities, starting from 0 to
max(nums)
. Binary search works on a sorted sequence, and while capability values aren't inherently sorted, their feasibility is monotonically increasing: if it's possible to rob at leastk
houses with capabilityx
, it's possible with anyx' > x
as well. -
Feasibility Function
f(x)
: This function takes a capability valuex
as input and determines whether it is possible to rob at leastk
houses without robbing adjacent ones and not exceeding capabilityx
. The function iterates over all the houses, maintaining a countcnt
of houses robbed and an indexj
of the last house robbed to enforce the non-adjacent constraint. -
Conditional House Selection: While iterating through the houses, two conditions must be checked for each house to decide whether it can be robbed:
- Its value is less than or equal to the capability
x
. - It is not adjacent to the last house robbed, which is checked by comparing the current index
i
withj + 1
.
- Its value is less than or equal to the capability
-
Incrementing and Checking Count: Whenever a house meets the above conditions, we increment
cnt
and updatej
to the current index. Ifcnt
becomes greater or equal tok
,f(x)
returnsTrue
, signaling that this capability level is feasible. -
Binary Search Execution: Python's
bisect_left
function is used to find the smallest index in the range[0, max(nums) + 1)
wheref(x)
returnsTrue
. Here,bisect_left(range(max(nums) + 1), True, key=f)
is searching for the position to insertTrue
in a sorted sequence of boolean values obtained by applyingf(x)
on eachx
within the range so that the list remains sorted. Sincef(x)
returnsTrue
for allx
greater than or equal to the minimum capability,bisect_left
effectively finds the smallest suchx
. -
Time Complexity: The time complexity of this solution is
O(n log m)
, wheren
is the number of houses andm
is the maximum amount of money in any house. The functionf(x)
takesO(n)
time per check, and the binary search requiresO(log m)
checks.
The use of binary search greatly optimizes the search process compared to a brute-force approach, which would have been significantly slower and less efficient.
The provided Python solution leverages the algorithmic strength of binary search in combination with a custom-designed feasibility checker to solve this problem of optimizing the robber's capability within the given constraints.
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Let's go through a small example to illustrate the solution approach. Suppose we are given the following array of integers representing the amounts of money in the houses and a minimum number of houses k
the robber needs to steal from:
nums = [2, 3, 1, 1, 5, 1, 2] k = 3
Our task is to find the minimum "capability" for the robber so that they can rob at least k
non-adjacent houses.
Step 1: Binary Search Rationale
We initiate a binary search for capabilities ranging from 0 to the maximum value in nums
, which is 5 in this case. Hence, our search space is 0 to 5
.
Step 2: Feasibility Function f(x)
The feasibility function checks if the robber can rob k
non-adjacent houses with a capability x
. Let us see how this function works with an example capability of x = 3
.
Step 3: Conditional House Selection
When iterating through nums using the feasibility function with x = 3
, the robber can rob houses with value less than or equal to 3 and must skip adjacent houses.
- House 0: Value is 2 (ā¤ 3), robber can steal from this house.
- House 1: Value is 3 (ā¤ 3), but it's adjacent to House 0, so the robber skips this house.
- House 2: Value is 1 (ā¤ 3), and not adjacent to the last robbed house (House 0), so the robber can steal.
- House 3: Value is 1 (ā¤ 3), but adjacent to House 2, so it's skipped.
- House 4: Value is 5 (> 3), so this house can't be robbed with capability of 3.
- House 5: Value is 1 (ā¤ 3), and not adjacent to the last robbed house (House 2), so the robber can steal.
After this process, the robber has stolen from 3 houses (0, 2, and 5), which means f(3) = True
.
Step 4: Incrementing and Checking Count
We check if at least k
houses can be robbed without exceeding the capability x
. Since f(3)
returned True
, we can proceed with the binary search to see if there's a lower capability that still allows for k
houses to be robbed.
Step 5: Binary Search Execution
Using the bisect_left
function, we find that the smallest capability where f(x)
returns True
is actually 2, because the robber could steal from houses 0, 2, and 5 with amounts of 2, 1, and 1, respectively, satisfying the requirement to steal from at least k = 3
houses without robbing adjacent houses.
Step 6: Time Complexity
The time complexity of this operation for our list nums
is O(n log m)
where n
is 7 and m
is 5, resulting from the binary search operation and feasibility checks for each capability value.
By applying this algorithm, we have efficiently determined the minimum "capability" for the robber, allowing them to complete their heist according to the given constraints. In our example, the minimum capability is 2. This is the least amount of money that the robber would need to steal from any single house in their strategy to rob at least k = 3
non-adjacent houses.
Solution Implementation
1from typing import List
2from bisect import bisect_left
3
4class Solution:
5 # This function aims to find the minimum capability required to form k pairs
6 def minCapability(self, nums: List[int], k: int) -> int:
7
8 # The inner function 'is_feasible' checks if a given capability 'capability'
9 # is sufficient to form at least k pairs.
10 def is_feasible(capability):
11 count, last_used_index = 0, -2 # Initialize counters
12 for current_index, value in enumerate(nums):
13 # Skip if the value is greater than the capability or if the element
14 # is right after the previously used element (to ensure non-adjacency).
15 if value > capability or current_index == last_used_index + 1:
16 continue
17 # If neither condition is met, increment the count
18 # and update 'last_used_index'.
19 count += 1
20 last_used_index = current_index
21 # Check if the capability is sufficient to form k pairs.
22 return count >= k
23
24 # Perform a binary search over the range [0, max(nums) + 1),
25 # using the 'is_feasible' function as the key.
26 # The 'bisect_left' will find the leftmost value in the range
27 # where 'is_feasible' returns True (i.e., the minimum capability).
28 return bisect_left(range(max(nums) + 1), True, key=is_feasible)
29
1class Solution {
2
3 // Determine the minimum capability to partition the array in such a way that
4 // the sum of each sub-array is less than or equal to k
5 public int minCapability(int[] nums, int k) {
6 // Start with the least possible capability
7 int left = 0;
8 // Set an upper limit for the search space, assuming the max value according to problem constraints
9 int right = (int) 1e9;
10
11 // Perform a binary search to find the minimum capability
12 while (left < right) {
13 // Get the midpoint of the current search space
14 int mid = (left + right) >> 1;
15
16 // Check if the current capability can achieve the required partition
17 if (calculatePartitionCount(nums, mid) >= k) {
18 // If it qualifies, search the lower half to find a smaller capability
19 right = mid;
20 } else {
21 // Otherwise, search the upper half
22 left = mid + 1;
23 }
24 }
25
26 // left is now the minimum capability that can achieve the required partition
27 return left;
28 }
29
30 // Helper method to calculate the number of partitions formed by capability x
31 private int calculatePartitionCount(int[] nums, int x) {
32 int count = 0; // Initialize the partition count
33 int lastPartitionIndex = -2; // Initialize the index of the last partition start
34
35 // Iterate over the array
36 for (int i = 0; i < nums.length; ++i) {
37 // Skip if the current number exceeds the capability or is the next immediate number after the last partition
38 if (nums[i] > x || i == lastPartitionIndex + 1) {
39 continue;
40 }
41 // Increment the partition count and update lastPartitionIndex
42 ++count;
43 lastPartitionIndex = i;
44 }
45
46 // Return the total number of partitions that can be made with capability x
47 return count;
48 }
49}
50
1#include <vector>
2#include <algorithm>
3
4class Solution {
5public:
6 // Function to find the minimum capability needed to meet the condition.
7 int minCapability(std::vector<int>& nums, int k) {
8 // Lambda function to check if a given capability meets the condition.
9 auto isCapable = [&](int capability) {
10 int count = 0; // Initialize the count of qualified elements.
11 int prevIndex = -2; // Initialize the previous index (-2 is out of possible index range).
12
13 // Iterate through the numbers in the vector.
14 for (int i = 0; i < nums.size(); ++i) {
15 // Skip the number if it's greater than the capability or
16 // if it's the direct neighbor of the previously considered element.
17 if (nums[i] > capability || i == prevIndex + 1) {
18 continue;
19 }
20 // Increment the count of elements and update the previous index.
21 ++count;
22 prevIndex = i;
23 }
24 // Return true if the count is greater than or equal to k.
25 return count >= k;
26 };
27
28 // Binary search setup:
29 // Set the left boundary of the search to 0.
30 int left = 0;
31 // Find the maximum number in the vector and use it as the right boundary.
32 int right = *std::max_element(nums.begin(), nums.end());
33
34 // Perform binary search.
35 while (left < right) {
36 // Calculate the middle value between left and right.
37 int mid = (left + right) >> 1;
38 // If the current capability meets the condition, adjust the right boundary.
39 if (isCapable(mid)) {
40 right = mid;
41 } else {
42 // Otherwise, adjust the left boundary.
43 left = mid + 1;
44 }
45 }
46 // After exiting the loop, the smallest capability is found.
47 return left;
48 }
49};
50
1/**
2 * Determine the minimum capability required to select at least k elements,
3 * such that every two selected ones have at least one other element between them.
4 *
5 * @param capabilities - Array of capabilities.
6 * @param k - Number of elements to select.
7 * @return - The minimum capability required.
8 */
9function minCapability(capabilities: number[], k: number): number {
10 /**
11 * Helper function to determine if it's possible to select k elements
12 * given the constraint of at least one element between selected ones.
13 *
14 * @param maxCapability - The maximum capability to allow for selection.
15 * @return - True if at least k elements can be selected, false otherwise.
16 */
17 const canSelectKElements = (maxCapability: number): boolean => {
18 let count = 0; // Count of elements selected
19 let lastSelectedIndex = -2; // Index of last selected element
20
21 // Loop through all elements to determine if we can select them
22 for (let i = 0; i < capabilities.length; ++i) {
23 // Check if the current element can be selected (has desired capability and has gap)
24 if (capabilities[i] <= maxCapability && i - lastSelectedIndex > 1) {
25 ++count; // Increase the selected count
26 lastSelectedIndex = i; // Update the index of last selected element
27 }
28 }
29
30 // If we have selected at least k elements, return true.
31 return count >= k;
32 };
33
34 let left = 1;
35 // Find the element with the maximum capability.
36 let right = Math.max(...capabilities);
37
38 // Binary search to find the minimum capability required.
39 while (left < right) {
40 // Take the middle of the current range as the candidate capability.
41 const mid = (left + right) >> 1; // Equivalent to Math.floor((left + right) / 2)
42
43 // Use the helper function to check if mid can be a solution.
44 if (canSelectKElements(mid)) {
45 // We can select k elements with the mid capability, try lower capabilities.
46 right = mid;
47 } else {
48 // We cannot select k elements with the mid capability, try higher capabilities.
49 left = mid + 1;
50 }
51 }
52
53 // Return the lowest capability found by the binary search.
54 return left;
55}
56
Time and Space Complexity
Time Complexity
The given Python function minCapability
uses binary search through the bisect_left
function to find the minimum capability required. It applies a helper function f
as a key which processes the nums
list on each iteration.
-
The
bisect_left
function performs a binary search over a range of sizemax(nums) + 1
, which involvesO(log(Max))
iterations, whereMax
is the maximum element innums
. -
Within each binary search iteration, the helper function
f
performs a linear scan over thenums
list of sizen
, to check how many elements can be skipped without exceeding capabilityx
. The time complexity for this will beO(n)
.
Combining these two, the overall time complexity will be O(n * log(Max))
, where n
is the number of elements in nums
, and Max
is the maximum element in nums
.
Space Complexity
The space complexity of the minCapability
function is O(1)
assuming that the list nums
is given and does not count towards the space complexity (as it's an input). There are no additional data structures used that grow with the input size. The variables cnt
, j
, i
, and v
use a constant amount of space.
Learn more about how to find time and space complexity quickly using problem constraints.
Which technique can we use to find the middle of a linked list?
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