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Allocate Mailboxes

Hard

Asked by:

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Topics:

ArraysDynamic Programming

Given the array houses where houses[i] is the location of the i^th house along a street and an integer k, allocate k mailboxes in the street.

Return the minimum total distance between each house and its nearest mailbox.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: houses = [1,4,8,10,20], k = 3
Output: 5
Explanation: Allocate mailboxes in position 3, 9 and 20.
Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5

Example 2:

Input: houses = [2,3,5,12,18], k = 2
Output: 9
Explanation: Allocate mailboxes in position 3 and 14.
Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| + |5-3| + |12-14| + |18-14| = 9.

Constraints:

1 <= k <= houses.length <= 100
1 <= houses[i] <= 10⁴
All the integers of houses are unique.

Allocate Mailboxes

Hard

Asked by:

16 views

Topics:

ArraysDynamic Programming

Given the array houses where houses[i] is the location of the i^th house along a street and an integer k, allocate k mailboxes in the street.

Return the minimum total distance between each house and its nearest mailbox.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: houses = [1,4,8,10,20], k = 3
Output: 5
Explanation: Allocate mailboxes in position 3, 9 and 20.
Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5

Example 2:

Input: houses = [2,3,5,12,18], k = 2
Output: 9
Explanation: Allocate mailboxes in position 3 and 14.
Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| + |5-3| + |12-14| + |18-14| = 9.

Constraints:

1 <= k <= houses.length <= 100
1 <= houses[i] <= 10⁴
All the integers of houses are unique.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the mailbox locations (house positions)? Can they be negative, zero, or very large?
How many mailboxes (k) will we need to allocate, and what is the relationship between k and the number of houses? Can k ever be 0, 1, or greater than the number of houses?
If there are multiple equally optimal placements of mailboxes (resulting in the same minimum total distance), is any one of those placements acceptable, or is there a specific criteria for choosing amongst them?
What data type should I use to represent the total distance? Should I anticipate potential integer overflow issues?
What should I return if the input array of house locations is null or empty, or if k is invalid (e.g., k <= 0)?

Brute Force Solution

Approach

The brute force approach to the mailbox problem means trying every possible way to divide houses into groups, where each group gets a mailbox. We calculate the total distance for each arrangement and pick the arrangement with the smallest total distance.

Here's how the algorithm would work step-by-step:

Consider allocating mailboxes by placing the first mailbox at the first house.
Then, consider placing the first mailbox covering the first two houses, and then the first three houses, and so on.
For each of these potential locations for the first mailbox, consider all possible locations for the second mailbox.
Continue this pattern of trying all possible positions for each mailbox until all mailboxes are allocated to groups of houses.
For each arrangement, calculate the total distance each house is from its nearest mailbox.
Compare the total distance for each arrangement, and select the allocation that minimizes the overall distance.

Code Implementation

def allocate_mailboxes_brute_force(houses, number_of_mailboxes):
    number_of_houses = len(houses)
    minimum_distance = float('inf')

    def calculate_total_distance(mailbox_positions):
        total_distance = 0
        for house_position in houses:
            minimum_house_distance = float('inf')
            for mailbox_position in mailbox_positions:
                minimum_house_distance = min(minimum_house_distance, abs(house_position - houses[mailbox_position]))
            total_distance += minimum_house_distance
        return total_distance

    def find_minimum_distance(current_mailbox_index, current_mailbox_positions):
        nonlocal minimum_distance

        #If all mailboxes are placed, calculate the total distance
        if current_mailbox_index == number_of_mailboxes:
            total_distance = calculate_total_distance(current_mailbox_positions)
            minimum_distance = min(minimum_distance, total_distance)
            return

        #Try placing mailbox at each possible house
        start_position = 0 if not current_mailbox_positions else current_mailbox_positions[-1] # Each mailbox must cover at least one house
        for mailbox_position in range(start_position, number_of_houses):

            #Create a new arrangement with the new mailbox at this house.
            new_mailbox_positions = current_mailbox_positions + [mailbox_position]
            if len(new_mailbox_positions) <= number_of_mailboxes:
                find_minimum_distance(current_mailbox_index + 1, new_mailbox_positions)

    find_minimum_distance(0, [])
    return minimum_distance

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^k) – The provided brute force approach explores all possible ways to allocate k mailboxes among n houses. In the worst-case scenario, we are essentially iterating through all combinations of house groupings to assign mailboxes. The number of such combinations grows exponentially with the number of mailboxes (k) relative to the number of houses (n). Therefore the complexity is O(n^k) as we must iterate over the houses for each mailbox, where k is the number of mailboxes.

Space Complexity

O(N^K) – The brute force approach recursively explores all possible mailbox placements. For each possible position of the first mailbox, it considers all positions for the second, and so on. Since there can be up to N possible positions for each of the K mailboxes, the call stack can grow to a depth proportional to K, and at each level of the call stack, we are potentially storing intermediate results representing a potential arrangement of mailboxes across the houses. These arrangements effectively require storing K mailbox positions where each can be 1 to N which means worst case branching is N for each K levels. Therefore, space complexity is O(N^K).

Optimal Solution

Approach

The problem asks us to minimize the total distance people have to walk to the nearest mailbox. The optimal approach involves cleverly identifying the best locations for mailboxes by considering the median house location within each group. By minimizing the distance for smaller groups, we build up to a solution that minimizes the overall distance.

Here's how the algorithm would work step-by-step:

If we only have one mailbox, we want to put it in the location that minimizes the total walking distance for everyone. The best spot for one mailbox is at the median house location.
If we have more mailboxes than houses, just put a mailbox at each house - the walking distance will be zero.
Now, imagine we need to find the best spots for multiple mailboxes. We'll use the single mailbox case as a building block.
Think about dividing the houses into groups, with each group getting its own mailbox. The key is finding the best way to split up these groups and where to put the mailbox for each group.
We can find the optimal solution by building it up piece by piece. First, figure out the best spot for one mailbox serving the first few houses; then, figure out the best spot for two mailboxes serving a slightly larger group, and so on.
To efficiently calculate the distance for each group, use previous calculations. We can calculate it more efficiently by subtracting out the part you no longer want and adding in the part you do.
By working our way from small groups to larger groups, we can arrive at the best locations for all the mailboxes that minimize the total distance for everyone.

Code Implementation

def allocate_mailboxes(houses, number_of_mailboxes):    houses.sort()    number_of_houses = len(houses)
    # If more mailboxes than houses, each house gets one    if number_of_mailboxes >= number_of_houses:
        return 0
    # Cost to put one mailbox between house i and j inclusive    cost = [[0] * number_of_houses for _ in range(number_of_houses)]
    for i in range(number_of_houses):
        for j in range(i, number_of_houses):
            median = houses[(i + j) // 2]
            for k in range(i, j + 1):
                cost[i][j] += abs(houses[k] - median)
    dp_table = [[float('inf')] * number_of_houses for _ in range(number_of_mailboxes + 1)]
    # If only one mailbox, find the cost for each house
    for i in range(number_of_houses):
        dp_table[1][i] = cost[0][i]
    # Build the solution iteratively for each mailbox count.
    for mailbox_count in range(2, number_of_mailboxes + 1):
        for house_index in range(mailbox_count - 1, number_of_houses):

            # Find the optimal split between previous mailboxes and the current one
            for split_point in range(mailbox_count - 2, house_index):

                dp_table[mailbox_count][house_index] = min(dp_table[mailbox_count][house_index], \
                    dp_table[mailbox_count - 1][split_point] + cost[split_point + 1][house_index])
    return dp_table[number_of_mailboxes][number_of_houses - 1]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^2 * k) – The solution employs dynamic programming to determine the optimal placement of k mailboxes among n houses. The outermost loop iterates up to k times to consider the different possible number of mailboxes. The nested loops iterate through all possible subarrays of houses (i to j), where i and j can range from 0 to n, leading to O(n^2) iterations to calculate the cost of placing a mailbox at the median house for each subarray. The innermost operation calculates the distance of all houses within the range to their median location, but this distance calculation is precomputed and stored for efficiency. The overall complexity is thus dominated by the triple-nested loop structure, resulting in O(n^2 * k) time complexity.

Space Complexity

O(N*K) – The primary space usage comes from the dynamic programming table used to store the minimum distances. This table has dimensions related to the number of houses (N) and the number of mailboxes (K), resulting in a table of size N*K. Auxiliary space is also used in pre-calculating the distances between houses when determining the optimal mailbox location for a single mailbox serving some houses which can be done in O(N^2) at worst but can be optimized to O(N). The N*K term dominates. Therefore, the space complexity is O(N*K).

Edge Cases

Case	How to Handle
Null or empty houses array	Return 0 or throw an IllegalArgumentException, depending on the problem constraints.
Number of mailboxes (k) is zero or negative	Return 0 or throw an IllegalArgumentException as it's an invalid input.
Number of mailboxes (k) is greater than the number of houses	The minimal total distance will be zero, because each house has its own mailbox.
Houses array contains duplicate positions	The median calculation for the same position must consider multiple houses at the same location.
Houses array contains very large or very small coordinate values (integer overflow potential)	Use long data type to prevent integer overflow during distance calculations.
Houses are already sorted optimally for 1 mailbox (check if further partitioning will result in improved efficiency when k > 1)	The algorithm should be able to correctly identify and handle a pre-sorted optimal state.
When k = number of houses, distance is 0	The code should return 0 directly without unnecessary calculations.
Houses array contains only one element	If k >= 1, the minimal total distance is 0, otherwise it is an invalid input.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the mailbox locations (house positions)? Can they be negative, zero, or very large?
How many mailboxes (k) will we need to allocate, and what is the relationship between k and the number of houses? Can k ever be 0, 1, or greater than the number of houses?
If there are multiple equally optimal placements of mailboxes (resulting in the same minimum total distance), is any one of those placements acceptable, or is there a specific criteria for choosing amongst them?
What data type should I use to represent the total distance? Should I anticipate potential integer overflow issues?
What should I return if the input array of house locations is null or empty, or if k is invalid (e.g., k <= 0)?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider allocating mailboxes by placing the first mailbox at the first house.
Then, consider placing the first mailbox covering the first two houses, and then the first three houses, and so on.
For each of these potential locations for the first mailbox, consider all possible locations for the second mailbox.
Continue this pattern of trying all possible positions for each mailbox until all mailboxes are allocated to groups of houses.
For each arrangement, calculate the total distance each house is from its nearest mailbox.
Compare the total distance for each arrangement, and select the allocation that minimizes the overall distance.

Code Implementation

def allocate_mailboxes_brute_force(houses, number_of_mailboxes):
    number_of_houses = len(houses)
    minimum_distance = float('inf')

    def calculate_total_distance(mailbox_positions):
        total_distance = 0
        for house_position in houses:
            minimum_house_distance = float('inf')
            for mailbox_position in mailbox_positions:
                minimum_house_distance = min(minimum_house_distance, abs(house_position - houses[mailbox_position]))
            total_distance += minimum_house_distance
        return total_distance

    def find_minimum_distance(current_mailbox_index, current_mailbox_positions):
        nonlocal minimum_distance

        #If all mailboxes are placed, calculate the total distance
        if current_mailbox_index == number_of_mailboxes:
            total_distance = calculate_total_distance(current_mailbox_positions)
            minimum_distance = min(minimum_distance, total_distance)
            return

        #Try placing mailbox at each possible house
        start_position = 0 if not current_mailbox_positions else current_mailbox_positions[-1] # Each mailbox must cover at least one house
        for mailbox_position in range(start_position, number_of_houses):

            #Create a new arrangement with the new mailbox at this house.
            new_mailbox_positions = current_mailbox_positions + [mailbox_position]
            if len(new_mailbox_positions) <= number_of_mailboxes:
                find_minimum_distance(current_mailbox_index + 1, new_mailbox_positions)

    find_minimum_distance(0, [])
    return minimum_distance

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

If we only have one mailbox, we want to put it in the location that minimizes the total walking distance for everyone. The best spot for one mailbox is at the median house location.
If we have more mailboxes than houses, just put a mailbox at each house - the walking distance will be zero.
Now, imagine we need to find the best spots for multiple mailboxes. We'll use the single mailbox case as a building block.
Think about dividing the houses into groups, with each group getting its own mailbox. The key is finding the best way to split up these groups and where to put the mailbox for each group.
We can find the optimal solution by building it up piece by piece. First, figure out the best spot for one mailbox serving the first few houses; then, figure out the best spot for two mailboxes serving a slightly larger group, and so on.
To efficiently calculate the distance for each group, use previous calculations. We can calculate it more efficiently by subtracting out the part you no longer want and adding in the part you do.
By working our way from small groups to larger groups, we can arrive at the best locations for all the mailboxes that minimize the total distance for everyone.

Code Implementation

def allocate_mailboxes(houses, number_of_mailboxes):    houses.sort()    number_of_houses = len(houses)
    # If more mailboxes than houses, each house gets one    if number_of_mailboxes >= number_of_houses:
        return 0
    # Cost to put one mailbox between house i and j inclusive    cost = [[0] * number_of_houses for _ in range(number_of_houses)]
    for i in range(number_of_houses):
        for j in range(i, number_of_houses):
            median = houses[(i + j) // 2]
            for k in range(i, j + 1):
                cost[i][j] += abs(houses[k] - median)
    dp_table = [[float('inf')] * number_of_houses for _ in range(number_of_mailboxes + 1)]
    # If only one mailbox, find the cost for each house
    for i in range(number_of_houses):
        dp_table[1][i] = cost[0][i]
    # Build the solution iteratively for each mailbox count.
    for mailbox_count in range(2, number_of_mailboxes + 1):
        for house_index in range(mailbox_count - 1, number_of_houses):

            # Find the optimal split between previous mailboxes and the current one
            for split_point in range(mailbox_count - 2, house_index):

                dp_table[mailbox_count][house_index] = min(dp_table[mailbox_count][house_index], \
                    dp_table[mailbox_count - 1][split_point] + cost[split_point + 1][house_index])
    return dp_table[number_of_mailboxes][number_of_houses - 1]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty houses array	Return 0 or throw an IllegalArgumentException, depending on the problem constraints.
Number of mailboxes (k) is zero or negative	Return 0 or throw an IllegalArgumentException as it's an invalid input.
Number of mailboxes (k) is greater than the number of houses	The minimal total distance will be zero, because each house has its own mailbox.
Houses array contains duplicate positions	The median calculation for the same position must consider multiple houses at the same location.
Houses array contains very large or very small coordinate values (integer overflow potential)	Use long data type to prevent integer overflow during distance calculations.
Houses are already sorted optimally for 1 mailbox (check if further partitioning will result in improved efficiency when k > 1)	The algorithm should be able to correctly identify and handle a pre-sorted optimal state.
When k = number of houses, distance is 0	The code should return 0 directly without unnecessary calculations.
Houses array contains only one element	If k >= 1, the minimal total distance is 0, otherwise it is an invalid input.