LeetCode 3447 - Assign Elements to Groups with Constraints

Difficulty: 🟡 Medium
Topics: Array, Hash Table

Solution

Problem Understanding

The problem requires assigning elements from one array, elements, to groups defined by another array, groups, under a divisibility constraint. Specifically, for each group groups[i], we must find the smallest index j in elements such that groups[i] % elements[j] == 0. If no such element exists, we assign -1 for that group. Multiple groups can share the same element, but each group gets only one element.

The input arrays groups and elements can be up to 10^5 in length, and individual values can also reach 10^5. This tells us that any naive solution that checks every pair (group, element) in a double loop would be too slow because it would involve up to 10^10 operations. Edge cases include groups that have no divisible element, elements that are all larger than some groups, or groups that are 1, which is divisible by all elements.

Approaches

Brute Force

The brute-force approach iterates over each group, then iterates over all elements to find the first divisible element. This works because it directly implements the problem statement, but it is too slow. For n = 10^5 groups and m = 10^5 elements, the time complexity would be O(n * m) which is 10^10 operations, exceeding practical limits.

Optimal Approach

The key insight is that instead of checking all elements for each group, we can precompute the smallest index of elements for all potential divisors. Using a mapping from element values to their smallest index, we can efficiently check, for each group, which element divides it by iterating over divisors of the group only.

We can enumerate all divisors of a number g in O(sqrt(g)) time. For each divisor d, if d exists in the element-to-index mapping, it is a candidate. We pick the candidate with the smallest index. This reduces the complexity from O(n*m) to O(n * sqrt(max(groups))), which is acceptable given the constraints.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n * m)	O(1)	Check every element for every group
Optimal	O(n * sqrt(maxG) + m)	O(m)	Precompute element indices, iterate divisors only

Algorithm Walkthrough

Initialize a dictionary element_index mapping each element value to its smallest index. This allows constant-time lookups to see if an element exists and its index.
Create an empty list assigned to store results for each group.
For each group g in groups, iterate over all its divisors d using 1..sqrt(g):
If d is in element_index, it is a candidate.
If g // d is also in element_index, it is another candidate (avoid duplicate if d*d == g).
Among all candidates, pick the one with the smallest index. If no candidates exist, append -1.
Return the assigned array.

Why it works: This algorithm ensures that we only consider valid elements that divide the group, and by storing the smallest index for each element, we always select the earliest occurrence. Enumerating divisors ensures all possibilities are checked efficiently.

Python Solution

from typing import List
import math

class Solution:
    def assignElements(self, groups: List[int], elements: List[int]) -> List[int]:
        element_index = {}
        for i, val in enumerate(elements):
            if val not in element_index:
                element_index[val] = i

        assigned = []
        for g in groups:
            min_idx = float('inf')
            for d in range(1, int(math.isqrt(g)) + 1):
                if g % d == 0:
                    if d in element_index:
                        min_idx = min(min_idx, element_index[d])
                    other = g // d
                    if other != d and other in element_index:
                        min_idx = min(min_idx, element_index[other])
            assigned.append(min_idx if min_idx != float('inf') else -1)
        return assigned

Explanation: First, we build a mapping from element values to their first occurrence. Then, for each group, we enumerate divisors and check if they exist in elements. We maintain the smallest index found. If no divisor is present in elements, we append -1.

Go Solution

import (
    "math"
)

func assignElements(groups []int, elements []int) []int {
    elementIndex := make(map[int]int)
    for i, val := range elements {
        if _, exists := elementIndex[val]; !exists {
            elementIndex[val] = i
        }
    }

    assigned := make([]int, len(groups))
    for i, g := range groups {
        minIdx := int(^uint(0) >> 1) // Max int
        sqrtG := int(math.Sqrt(float64(g)))
        for d := 1; d <= sqrtG; d++ {
            if g%d == 0 {
                if idx, exists := elementIndex[d]; exists {
                    if idx < minIdx {
                        minIdx = idx
                    }
                }
                other := g / d
                if other != d {
                    if idx, exists := elementIndex[other]; exists {
                        if idx < minIdx {
                            minIdx = idx
                        }
                    }
                }
            }
        }
        if minIdx == int(^uint(0)>>1) {
            assigned[i] = -1
        } else {
            assigned[i] = minIdx
        }
    }
    return assigned
}

Go-specific notes: We use int(^uint(0) >> 1) to represent infinity. The map structure is similar to Python dict. Slices are used to store results, and care is taken to avoid duplicates when divisors are equal.

Worked Examples

Example 1: groups = [8,4,3,2,4], elements = [4,2]

Group	Divisors	Candidates (element index)	Assigned
8	1,2,4,8	4->0, 2->1	0
4	1,2,4	4->0, 2->1	0
3	1,3	none	-1
2	1,2	2->1	1
4	1,2,4	4->0, 2->1	0

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

Group	Divisors	Candidates	Assigned
2	1,2	none	-1
3	1,3	3->1	1
5	1,5	5->0	0
7	1,7	none	-1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * sqrt(maxG) + m)	`m` to build element index map, `n` groups each with O(sqrt(g)) divisors
Space	O(m)	Store smallest index for each unique element

The dominant factor is iterating divisors for each group, but sqrt(maxG) is at most 317 for maxG = 10^5, which is manageable.

Test Cases

# Provided examples
assert Solution().assignElements([8,4,3,2,4], [4,2]) == [0,0,-1,1,0]
assert Solution().assignElements([2,3,5,7], [5,3,3]) == [-1,1,0,-1]
assert Solution().assignElements([10,21,30,41], [2,1]) == [0,1,0,1]

# Edge cases
assert Solution().assignElements([1], [1]) == [0]  # single group and element
assert Solution().assignElements([100000], [1]) == [0]  # large group divisible by 1
assert Solution().assignElements([7,11,13], [2,3,5]) == [-1,-1,-1]  # no divisible elements
assert Solution().assignElements([12,18], [2,3,6]) == [0,0]  # multiple divisors, choose smallest index
assert Solution().assignElements([36], [2,3,4,6]) == [0]  # smallest index among divisors

Test	Why
Provided examples	Validate correctness with problem statement
Single element/group	Minimal input size
Large group with divisor 1	Tests upper bound
No divisible elements	Edge case returns -1
Multiple divisors	Ensure smallest index chosen
Multiple candidates	Confirms algorithm picks correct index