3447. Assign Elements to Groups with Constraints

MediumArrayHash Table

Problem Description

You are given an integer array groups, where each element groups[i] represents the size of the i-th group. Additionally, you have an integer array elements. Your task is to assign one element from the elements array to each group based on the following rules:

An element at index j can be assigned to a group i if groups[i] is divisible by elements[j].
If multiple elements can be assigned to a group, assign the element with the smallest index j.
If no elements satisfy the condition for a group, assign -1 to that group.

The result should be an integer array assigned, where assigned[i] is the index of the element chosen for group i, or -1 if no suitable element exists. Note that an element may be assigned to more than one group.

Quick Interview Experience

Help others by sharing your interview experience

Have you seen this problem before?

Intuition

To solve the problem, the approach involves first determining a suitable mapping strategy based on the constraints provided.

Understanding Divisibility Requirement:
- For each group identified by its size, we're interested in finding elements whose values divide the size of the group. This introduces a linkage between the values in elements and the sizes in groups.
Efficiency through Preprocessing:
- Since we seek to assign indices to group sizes with the smallest index priority, it is more efficient to preprocess potential candidates (elements) that can service multiple group sizes due to their divisibility nature.
Using a Dynamic Marker Array:
- Introduce an array d initialized with -1 to store the indices of elements. This array tracks the smallest index for which an element divides a given integer. By marking possible divisors, we ensure that once a suitable divisor is found, any greater multiples (including group sizes) can immediately use this pre-determined smallest index.
Iterative Mapping:
- For each element in the elements array, if it can be a divisor (x) for any size up to the maximum value present in groups, mark its index in array d. This creates a fast lookup mechanism.
- Finally, iterate over each size in groups, and retrieve the preassigned d index to build the final assigned array.

The idea is to leverage the breadth of divisibility to perform minimal calculation during the actual group-to-element mapping phase, optimizing both time complexity and logical clarity.

Solution Approach

The solution to the problem is implemented by utilizing a structured approach that combines enumeration with a strategic use of data structures. Here's a detailed walkthrough of the solution:

Identify the Maximum Group Size:
- We begin by determining the maximum value in the groups array, denoted as mx. This maximum value serves as the upper bound for possible indices in our marker array d.
Initialize a Marker Array:
- We set up an array d of size mx + 1, initialized with -1. This array will store indices of elements that can divide specific numbers. The initial -1 implies no element has been mapped to that number yet.
Processing Elements for Divisibility:
- Iterate over the elements array. For each element x at index j, we perform the following:
  - If x is greater than mx or d[x] is already set (not -1), we continue to the next element.
  - Otherwise, for every multiple y starting from x up to mx (range(x, mx + 1, x)), we set d[y] to j if d[y] is still -1. This action indicates that the element x can be used as a divisor for all these multiples and marks the lowest available index j.
Construct the Assigned Array:
- Iterate over the groups array. For each group size x, retrieve the index from d[x] and construct the assigned array. If no element from elements divides x (indicated by d[x] being -1), then -1 is assigned for that group.

The implementation efficiently maps elements to groups via a single traversal of each elements and groups array, leveraging preprocessed divisibility information. This approach optimizes the assignment process, satisfying the need for the smallest index assignment where applicable.

Ready to land your dream job?

Unlock your dream job with a 3-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's work through a small example to illustrate the solution approach.

Suppose we have the following input arrays:

groups = [6, 8, 14]
elements = [2, 3, 5]

The task is to assign indices of elements to each group based on the divisibility rules described.

Step 1: Determine the Maximum Group Size

First, find the maximum element in the groups array, which is mx = 14.

Step 2: Initialize the Marker Array

Create an array d of length mx + 1 (i.e., 15), initialized with -1:

d = [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1]

This array will help track which index of the elements can divide a particular number.

Step 3: Populate the Divisibility Markers

Iterate over the elements array to fill the d array:

For elements[0] = 2:
- Since 2 <= 14 and d[2] is -1, update all multiples of 2 within the range up to 14 with index 0:
  - d[2] = 0, d[4] = 0, d[6] = 0, d[8] = 0, d[10] = 0, d[12] = 0, d[14] = 0
For elements[1] = 3:
- Since 3 <= 14 and d[3] is -1, update all multiples of 3 up to 14 with index 1:
  - d[3] = 1, d[9] = 1, d[12] is already 0
For elements[2] = 5:
- 5 <= 14 and d[5] is -1, update all multiples of 5 up to 14 with index 2:
  - d[5] = 2, d[10] is already 0

After processing, the d array looks like: d = [-1, -1, 0, 1, 0, 2, 0, -1, 0, 1, 0, -1, 0, -1, 0]

Step 4: Assign Elements to Groups

Finally, for each element in groups, retrieve their indices from d:

groups[0] = 6: d[6] = 0, so assign index 0 (corresponding to elements[0] = 2)
groups[1] = 8: d[8] = 0, so assign index 0
groups[2] = 14: d[14] = 0, assign index 0

Thus, the resulting assigned array is [0, 0, 0].

Through this approach, each group is efficiently assigned an element index based on divisibility, while ensuring the smallest possible index is chosen where several elements can divide a group size.

Solution Implementation

1from typing import List
2
3class Solution:
4    def assignElements(self, groups: List[int], elements: List[int]) -> List[int]:
5        # Find the maximum value in groups
6        max_group_value = max(groups)
7      
8        # Initialize a list to store indices, with -1 indicating unassigned
9        assigned_indices = [-1] * (max_group_value + 1)
10      
11        # Iterate through elements, associating indices where conditions meet
12        for index, element in enumerate(elements):
13            # Skip if the element is greater than max_group_value or already assigned
14            if element > max_group_value or assigned_indices[element] != -1:
15                continue
16            # Assign index of element to all its multiples within range
17            for multiple in range(element, max_group_value + 1, element):
18                if assigned_indices[multiple] == -1:
19                    assigned_indices[multiple] = index
20      
21        # Map groups to their respective assigned element indices
22        return [assigned_indices[value] for value in groups]
23

1import java.util.Arrays;
2
3class Solution {
4    public int[] assignElements(int[] groups, int[] elements) {
5        // Find the maximum element in the 'groups' array
6        int maxGroupValue = Arrays.stream(groups).max().getAsInt();
7      
8        // Initialize an array to track the smallest index of element assigned to each value
9        // Fill initially with -1 indicating no assignments
10        int[] smallestIndexAssignments = new int[maxGroupValue + 1];
11        Arrays.fill(smallestIndexAssignments, -1);
12      
13        // Iterate over each element in the 'elements' array
14        for (int j = 0; j < elements.length; ++j) {
15            int element = elements[j];
16          
17            // Skip processing if element is outside the range or already assigned
18            if (element > maxGroupValue || smallestIndexAssignments[element] != -1) {
19                continue;
20            }
21          
22            // Assign elements to positions that are multiples of the current 'element' 
23            for (int multiple = element; multiple <= maxGroupValue; multiple += element) {
24                if (smallestIndexAssignments[multiple] == -1) {
25                    smallestIndexAssignments[multiple] = j;
26                }
27            }
28        }
29      
30        int numberOfGroups = groups.length;
31        int[] results = new int[numberOfGroups];
32      
33        // Assign the smallest index to each group from the precomputed assignments
34        for (int i = 0; i < numberOfGroups; ++i) {
35            results[i] = smallestIndexAssignments[groups[i]];
36        }
37      
38        return results;
39    }
40}
41

1#include <vector>
2#include <algorithm> // For std::max
3
4class Solution {
5public:
6    std::vector<int> assignElements(std::vector<int>& groups, std::vector<int>& elements) {
7        // Find the maximum value in the 'groups' vector
8        int maxGroupValue = *std::max_element(groups.begin(), groups.end());
9      
10        // Create a vector 'assignments' initialized with -1, with size maxGroupValue + 1
11        std::vector<int> assignments(maxGroupValue + 1, -1);
12
13        // Iterate through each element in the 'elements' vector
14        for (int j = 0; j < elements.size(); ++j) {
15            int currentElement = elements[j];
16
17            // If the current element is greater than maxGroupValue or already assigned, continue to the next element
18            if (currentElement > maxGroupValue || assignments[currentElement] != -1) {
19                continue;
20            }
21
22            // Assign the current element to all multiples of its value up to maxGroupValue
23            for (int multiple = currentElement; multiple <= maxGroupValue; multiple += currentElement) {
24                if (assignments[multiple] == -1) {
25                    assignments[multiple] = j;
26                }
27            }
28        }
29
30        // Initialize the result vector to collect the assignment of each group
31        std::vector<int> result(groups.size());
32
33        // Assign each group its corresponding element index
34        for (int i = 0; i < groups.size(); ++i) {
35            result[i] = assignments[groups[i]];
36        }
37
38        // Return the result vector
39        return result;
40    }
41};
42

1// Function to assign elements to groups
2function assignElements(groups: number[], elements: number[]): number[] {
3    // Determine the maximum value in the groups array
4    const mx = Math.max(...groups);
5
6    // Create an array d initialized with -1, of size mx + 1
7    const d: number[] = Array(mx + 1).fill(-1);
8
9    // Iterate through each element in elements array
10    for (let j = 0; j < elements.length; ++j) {
11        const x = elements[j];
12
13        // Skip the loop if x greater than mx or already assigned in array d
14        if (x > mx || d[x] !== -1) {
15            continue;
16        }
17
18        // Assign element positions in array d for multiples of x
19        for (let y = x; y <= mx; y += x) {
20            if (d[y] === -1) {
21                d[y] = j;
22            }
23        }
24    }
25
26    // Map each group to the corresponding index from elements array stored in array d
27    return groups.map(x => d[x]);
28}
29

Time and Space Complexity

The time complexity is O(M * (n / m) + n) where n is the length of elements, m is the number of unique divisors up to M, and M is the maximum value in groups. The space complexity is O(M) as the array d of size M + 1 is used to store results.

Learn more about how to find time and space complexity quickly.

Discover Your Strengths and Weaknesses: Take Our 3-Minute Quiz to Tailor Your Study Plan:

Consider the classic dynamic programming of fibonacci numbers, what is the recurrence relation?

dp[i] = dp[i - 1] + dp[i - 2]

dp[i] = dp[i - 1] + 1

dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]

dp[i] = dp[i - 1]

I don't know