#3447

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Assign Elements to Groups with Constraints

Assign elements to groups by size constraints with a sieve-like approach to find suitable element matches efficiently.

Problem Statement

You are given two integer arrays: 'groups' and 'elements'. The 'groups' array represents the size of each group, and 'elements' holds available element sizes. You need to assign an element to each group based on specific rules.

Return an array where each index represents the group, and the value is the index of the chosen element from the 'elements' array, or -1 if no suitable element exists that fits the group's size.

Examples

Example 1

Input: groups = [8,4,3,2,4], elements = [4,2]

Output: [0,0,-1,1,0]

Example 2

Input: groups = [2,3,5,7], elements = [5,3,3]

Output: [-1,1,0,-1]

Example 3

Input: groups = [10,21,30,41], elements = [2,1]

Output: [0,1,0,1]

elements[0] = 2 is assigned to the groups with even values, and elements[1] = 1 is assigned to the groups with odd values.

Constraints

1 <= groups.length <= 105
1 <= elements.length <= 105
1 <= groups[i] <= 105
1 <= elements[i] <= 105

Solution Approach

Array Scanning

Iterate over the 'groups' array, checking each group size to see if there is an element that fits. Use efficient array scanning to track group positions and element availability.

Hash Table Lookup

Implement a hash table to quickly look up the available elements. This speeds up the process of matching elements to the groups, reducing the need for repeated scans.

Sieve-like Approach

Consider applying a sieve-like method to filter through groups and elements, marking off the ones already assigned, and avoiding unnecessary repeated checks.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach, where the sieve-like method can potentially reduce the number of redundant checks. Space complexity depends on the size of the hash table or any other auxiliary structures used to keep track of available elements and assigned groups.

What Interviewers Usually Probe

Test if the candidate can optimize array scanning with hashing.
Look for understanding of sieve-like techniques applied to grouping problems.
Assess if the candidate avoids redundant work by using appropriate data structures.

Common Pitfalls or Variants

Common pitfalls

Failing to use efficient data structures like hash tables leads to inefficient solutions.
Overlooking cases where no suitable element is available, leading to incorrect -1 assignments.
Not handling the edge cases where groups or elements are exhausted early.

Follow-up variants

Change the problem to require exact group size matching instead of any available element.
Introduce additional constraints on elements, like limited reuse or an order of assignment.
Allow multiple elements per group, adding complexity to the assignment logic.

FAQ

What is the most efficient way to solve the "Assign Elements to Groups with Constraints" problem?

The most efficient way involves using array scanning combined with hash table lookups, possibly applying a sieve-like method to reduce redundant checks.

How do I handle cases where no element fits a group?

Return -1 for that group, indicating that no element could be assigned based on the group size constraint.

What is the time complexity of the solution?

The time complexity depends on the approach chosen, but it is typically linear with respect to the number of groups and elements, O(n + m), where n is the number of groups and m is the number of elements.

Can I optimize the solution further?

Yes, using a hash table for element lookups can significantly speed up the process, and applying a sieve-like filtering technique can reduce unnecessary operations.

What happens if elements are smaller than the group size?

If no element fits, the group should be assigned -1, meaning no suitable element is available for that group.

terminal

Solution

Solution 1: Enumeration

First, we find the maximum value in the array $\textit{groups}$, denoted as $\textit{mx}$. We use an array $\textit{d}$ to record the index corresponding to each element. Initially, $\textit{d}[x] = -1$ indicates that the element $x$ has not been assigned yet.

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class Solution:
    def assignElements(self, groups: List[int], elements: List[int]) -> List[int]:
        mx = max(groups)
        d = [-1] * (mx + 1)
        for j, x in enumerate(elements):
            if x > mx or d[x] != -1:
                continue
            for y in range(x, mx + 1, x):
                if d[y] == -1:
                    d[y] = j
        return [d[x] for x in groups]

Continue Practicing

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