Minimize Maximum Component Cost

Medium

Salesforce

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Topics:

GraphsBinary SearchGreedy Algorithms

You are given an undirected connected graph with n nodes labeled from 0 to n - 1 and a 2D integer array edges where edges[i] = [u_i, v_i, w_i] denotes an undirected edge between node u_i and node v_i with weight w_i, and an integer k.

You are allowed to remove any number of edges from the graph such that the resulting graph has at most k connected components.

The cost of a component is defined as the maximum edge weight in that component. If a component has no edges, its cost is 0.

Return the minimum possible value of the maximum cost among all components after such removals.

Example 1:

Input: n = 5, edges = [[0,1,4],[1,2,3],[1,3,2],[3,4,6]], k = 2

Output: 4

Explanation:

Remove the edge between nodes 3 and 4 (weight 6).
The resulting components have costs of 0 and 4, so the overall maximum cost is 4.

Example 2:

Input: n = 4, edges = [[0,1,5],[1,2,5],[2,3,5]], k = 1

Output: 5

Explanation:

No edge can be removed, since allowing only one component (k = 1) requires the graph to stay fully connected.
That single component’s cost equals its largest edge weight, which is 5.

Constraints:

1 <= n <= 5 * 10⁴
0 <= edges.length <= 10⁵
edges[i].length == 3
0 <= u_i, v_i < n
1 <= w_i <= 10⁶
1 <= k <= n
The input graph is connected.

Try coding it in LeetCode

Minimize Maximum Component Cost

Medium

Salesforce

1 view

Topics:

GraphsBinary SearchGreedy Algorithms

You are given an undirected connected graph with n nodes labeled from 0 to n - 1 and a 2D integer array edges where edges[i] = [u_i, v_i, w_i] denotes an undirected edge between node u_i and node v_i with weight w_i, and an integer k.

You are allowed to remove any number of edges from the graph such that the resulting graph has at most k connected components.

The cost of a component is defined as the maximum edge weight in that component. If a component has no edges, its cost is 0.

Return the minimum possible value of the maximum cost among all components after such removals.

Example 1:

Input: n = 5, edges = [[0,1,4],[1,2,3],[1,3,2],[3,4,6]], k = 2

Output: 4

Explanation:

Remove the edge between nodes 3 and 4 (weight 6).
The resulting components have costs of 0 and 4, so the overall maximum cost is 4.

Example 2:

Input: n = 4, edges = [[0,1,5],[1,2,5],[2,3,5]], k = 1

Output: 5

Explanation:

No edge can be removed, since allowing only one component (k = 1) requires the graph to stay fully connected.
That single component’s cost equals its largest edge weight, which is 5.

Constraints:

1 <= n <= 5 * 10⁴
0 <= edges.length <= 10⁵
edges[i].length == 3
0 <= u_i, v_i < n
1 <= w_i <= 10⁶
1 <= k <= n
The input graph is connected.

Try coding it in LeetCode

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 are the maximum sizes of the `nums` and `factors` arrays, and what is the range of values within those arrays?
Can `factors` be empty, and if so, what should the return value be?
If it's impossible to split `nums` such that all components have a cost less than or equal to a given maximum cost, should I return an error, a specific value like -1, or is this guaranteed not to happen?
Are there any constraints on the number of components I can split `nums` into? Can I make as many components as there are elements in `nums`?
Are all numbers in `nums` and `factors` positive integers? Should I handle negative numbers or zeros?

Brute Force Solution

Approach

The brute force approach to minimizing maximum component cost involves trying every possible combination of dividing the numbers into separate groups. We want to explore all possible ways to split the numbers, and then find the split that gives us the lowest possible maximum cost among the groups. Think of it like testing every possible seating arrangement to find the best one.

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

Consider all possible ways to split the numbers into different groups.
For each way you split the numbers, calculate the cost of each group by, for example, multiplying all the numbers in that group together.
Find the group with the highest cost for that particular way of splitting the numbers.
Keep track of the highest cost among all groups for each split.
Compare the highest costs from all the different splits.
The split with the smallest of these highest costs is the answer we are looking for.

Code Implementation

def minimize_maximum_component_cost_brute_force(numbers):
    best_cost = float('inf')
    
    def calculate_cost(grouping):
        max_cost = 0
        for group in grouping:
            group_cost = sum(group)
            max_cost = max(max_cost, group_cost)
        return max_cost

    def find_all_groupings(remaining_numbers, current_grouping):
        nonlocal best_cost
        
        if not remaining_numbers:
            cost = calculate_cost(current_grouping)
            best_cost = min(best_cost, cost)
            return

        for i in range(1, 1 << len(remaining_numbers)):
            # Create the first group based on the current bitmask
            first_group = []
            next_remaining_numbers = []
            for j in range(len(remaining_numbers)):
                if (i >> j) & 1:
                    first_group.append(remaining_numbers[j])
                else:
                    next_remaining_numbers.append(remaining_numbers[j])
            
            # Recursively find groupings for the remaining numbers
            find_all_groupings(next_remaining_numbers, current_grouping + [first_group])

    # Kick off the recursive process
    find_all_groupings(numbers, [])

    return best_cost

Big(O) Analysis

Time Complexity

O(B(n)) – The brute force approach involves considering all possible ways to split n numbers into groups. The number of ways to partition a set of n elements is given by the Bell number, denoted as B(n). For each partition, we need to calculate the cost of each group, which can take O(n) time in the worst case if all numbers are in one group, where n is the input size. Since we iterate through all B(n) possible partitions, and for each partition, we perform calculations that take O(n) time, the overall time complexity is O(n * B(n)). Because B(n) grows faster than any polynomial, we can approximate it as O(B(n)).

Space Complexity

O(2^N) – The algorithm considers all possible ways to split the N numbers into groups. In the worst case, it needs to store all possible subsets of the numbers, leading to 2^N possible splits. Although not explicitly stated, managing or representing each of these splits requires memory. Because each split must store an arbitrary number of the original N values in arbitrary group arrangements. The total number of split combinations dominates all other auxiliary variable space, leading to a space complexity of O(2^N).

Optimal Solution

Approach

The goal is to find the smallest possible cost of the largest connected group after removing some connections. We can solve this by trying out different maximum cost limits, checking if we can achieve that limit, and narrowing down our search.

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

Instead of directly finding the best answer, we'll guess a possible maximum cost for a group.
Given our guess, we'll go through all the possible connections. If a connection makes a group larger than our guess, we'll temporarily remove it.
After removing connections, we'll check if any remaining connected group is larger than our guess. If so, our guess was too low, and we need to try a higher one.
If all groups are small enough, our guess was potentially too high, and we can try a smaller guess.
By repeatedly guessing and adjusting our guess based on whether it's too high or too low, we can quickly find the smallest maximum cost possible using binary search.

Code Implementation

def minimize_maximum_component_cost(numbers, number_of_components):
    def find_factors(numbers):
        factors = set()
        for number in numbers:
            for i in range(1, int(number**0.5) + 1):
                if number % i == 0:
                    factors.add(i)
                    factors.add(number // i)
        return sorted(list(factors))

    def can_split(maximum_cost):
        group_count = 0
        current_group = []
        for number in numbers:
            current_group.append(number)
            
            # Check if the greatest common factor of this group is <= maximum_cost.
            if len(current_group) == 1:
                group_greatest_common_factor = current_group[0]
            else:
                group_greatest_common_factor = current_group[0]
                for i in range(1, len(current_group)):
                    group_greatest_common_factor = gcd(group_greatest_common_factor, current_group[i])

            if group_greatest_common_factor <= maximum_cost:
                group_count += 1
                current_group = []

        # If there are remaining numbers, a new group should be created.
        if current_group:
          group_count+=1

        return group_count <= number_of_components

    def gcd(a, b):
        if b == 0:
            return a
        return gcd(b, a % b)

    factors = find_factors(numbers)
    minimum_maximum_cost = float('inf')

    # Iterate through the sorted factors to test each as a maximum cost.
    for maximum_cost in factors:
        
        # Check if we can split the numbers into the specified number of components.
        if can_split(maximum_cost):
            minimum_maximum_cost = maximum_cost
            break

    return minimum_maximum_cost

Big(O) Analysis

Time Complexity

O(E * log(SUM) * α(N)) – The binary search on the possible maximum cost ranges from 0 to SUM (where SUM is the sum of all nums array elements). This contributes a log(SUM) factor. Inside the binary search, we iterate through each edge (E) to potentially remove it if it connects components exceeding the current maximum cost guess. After potential edge removals, we check connected components using a Union-Find data structure, which has a near-constant time complexity of α(N) per union or find operation, where N is the number of nodes, and α is the inverse Ackermann function. Since we potentially perform a union operation for each edge, the overall complexity of the Union-Find operations within the binary search is E * α(N). Therefore, the total time complexity is O(E * log(SUM) * α(N)).

Space Complexity

O(N) – The algorithm uses a Union-Find data structure to track connected components after removing edges. The Union-Find data structure requires an array of size N, where N is the number of elements (or nodes) to represent the parent of each node. Additionally, it might use another array of size N to store the size or rank of each component. Therefore, the auxiliary space is proportional to the number of elements, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 if nums is null or empty, as there are no components and thus no cost.
factors is null or empty	If factors is null or empty, each component's cost is 0, therefore, the minimum possible maximum cost is 0.
nums contains only one element	The array consists of a single component; check if the element is divisible by any factor, the cost would be 1 or 0 appropriately.
All elements in nums are divisible by at least one factor	The entire array can be one component with a cost equal to the number of elements in nums.
No element in nums is divisible by any factor	Each element will be in its own component of cost 0, thus the minimum possible maximum cost is 0.
Large values in nums or factors that could lead to integer overflow during divisibility checks	Use appropriate data types (long) to prevent integer overflow during the divisibility checks.
nums contains duplicate elements, some divisible by factors, some not	The algorithm should correctly count the number of divisible elements within a component even with duplicates.
Binary Search fails to converge if search space is not properly defined	Define low and high properly to prevent infinite loops in the binary search (low=0, high=n).