Problem Description

You have two integer arrays: groups and elements.

Each groups[i] represents the size of the i-th group. Your task is to assign exactly one element from the elements array to each group following these rules:

1. Assignment Rule: An element at index j can be assigned to group i only if groups[i] is divisible by elements[j]. In other words, groups[i] % elements[j] == 0.

2. Priority Rule: If multiple elements can be assigned to a group, choose the element with the smallest index j.

3. No Match Rule: If no element from the elements array can divide a group's size, assign -1 to that group.

4. Reusability: The same element can be assigned to multiple groups.

Your goal is to return an array assigned where assigned[i] contains the index of the element chosen for group i, or -1 if no suitable element exists.

For example, if groups = [6, 4] and elements = [3, 2, 5]:

• For group 0 with size 6: Both element at index 0 (value 3) and element at index 1 (value 2) can divide 6. We choose index 0 (smallest index).
• For group 1 with size 4: Only element at index 1 (value 2) can divide 4.
• Result: [0, 1]

Intuition

The naive approach would be to check every group against every element to find valid assignments. However, we can optimize this by thinking about the problem from a different angle.

Instead of checking each group individually, we can precompute which element should be assigned to every possible group size up to the maximum group size. This way, when we need to find the assignment for a specific group, we can simply look it up.

The key insight is that if an element elements[j] has value x, it can divide all multiples of x. So for element x, the divisible group sizes would be x, 2x, 3x, 4x, ... up to the maximum group size.

By iterating through elements in order (from smallest index to largest), we can ensure that when multiple elements can divide a group size, the one with the smallest index gets recorded first. We only update the assignment for a group size if it hasn't been assigned yet (still -1).

For example, if elements = [2, 3] and maximum group size is 6:

• Element at index 0 (value 2) can handle group sizes: 2, 4, 6
• Element at index 1 (value 3) can handle group sizes: 3, 6

Since we process index 0 first, group size 6 gets assigned to index 0 even though index 1 could also handle it.

This precomputation approach reduces redundant calculations, especially when there are multiple groups with the same size. We compute the assignment for each possible size once, then simply look up the result for each group.

Solution Approach

We implement the precomputation strategy using the following steps:

Step 1: Find Maximum Group Size First, we find the maximum value in the groups array, denoted as mx. This helps us determine the range of group sizes we need to consider.

mx = max(groups)

Step 2: Initialize Assignment Array We create an array d of size mx + 1 to store the element index assigned to each possible group size. Initially, all values are set to -1, indicating no assignment yet.

d = [-1] * (mx + 1)

Step 3: Process Elements and Assign to Group Sizes We iterate through the elements array with both index j and value x:

for j, x in enumerate(elements):

For each element, we first check two conditions:

• If x > mx: The element is larger than any group size, so it can't divide any group
• If d[x] != -1: This element value has already been assigned by an earlier element with a smaller index

If either condition is true, we skip this element:

if x > mx or d[x] != -1:
    continue

Otherwise, we iterate through all multiples of x up to mx and assign index j to those positions in array d that haven't been assigned yet:

for y in range(x, mx + 1, x):
    if d[y] == -1:
        d[y] = j

The range (x, mx + 1, x) generates the sequence x, 2x, 3x, ... up to mx.

Step 4: Build Final Result Finally, we construct the answer by looking up each group size in our precomputed array d:

return [d[x] for x in groups]

For each group size in groups, we retrieve its assigned element index from d[x]. If no element was assigned to that size, d[x] remains -1.

Time Complexity: O(n + m × (mx/min_element)) where n is the length of groups, m is the length of elements, and mx is the maximum group size.

Space Complexity: O(mx) for the assignment array d.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given:

• groups = [6, 4, 8, 3]
• elements = [3, 2, 5]

Step 1: Find Maximum Group Size

mx = max([6, 4, 8, 3]) = 8

Step 2: Initialize Assignment Array Create array d of size 9 (indices 0-8), all initialized to -1:

d = [-1, -1, -1, -1, -1, -1, -1, -1, -1]
     0    1    2    3    4    5    6    7    8

Step 3: Process Each Element

Processing element at index 0 (value 3):

• Check: 3 ≤ 8 ✓ and d[3] == -1 ✓
• Mark multiples of 3: positions 3, 6

d = [-1, -1, -1, 0, -1, -1, 0, -1, -1]
     0    1    2   3    4    5   6    7    8

Processing element at index 1 (value 2):

• Check: 2 ≤ 8 ✓ and d[2] == -1 ✓
• Mark multiples of 2: positions 2, 4, 6, 8
• Note: position 6 already has 0, so we skip it

d = [-1, -1, 1, 0, 1, -1, 0, -1, 1]
     0    1   2   3   4    5   6    7   8

Processing element at index 2 (value 5):

• Check: 5 ≤ 8 ✓ and d[5] == -1 ✓
• Mark multiples of 5: position 5

d = [-1, -1, 1, 0, 1, 2, 0, -1, 1]
     0    1   2   3   4   5   6    7   8

Step 4: Build Final Result Look up each group size in array d:

• Group size 6: d[6] = 0 (element at index 0 divides 6)
• Group size 4: d[4] = 1 (element at index 1 divides 4)
• Group size 8: d[8] = 1 (element at index 1 divides 8)
• Group size 3: d[3] = 0 (element at index 0 divides 3)

Result: [0, 1, 1, 0]

This demonstrates how the precomputation ensures that each group size gets assigned the element with the smallest index that can divide it, and how the same element index can be reused for multiple groups.

Solution Implementation

Time and Space Complexity

Time Complexity: O(M × log M + n)

The time complexity consists of two main parts:

• Finding the maximum value in groups: O(n) where n is the length of groups
• Iterating through elements and for each element x, iterating through its multiples up to M: For each element x in elements, we iterate through values x, 2x, 3x, ... up to M. The number of multiples of x up to M is M/x. Summing over all possible values from 1 to M, we get M/1 + M/2 + M/3 + ... + M/M = M × (1 + 1/2 + 1/3 + ... + 1/M) = M × H(M), where H(M) is the harmonic series which is approximately O(log M). Since we process at most m elements (length of elements), and in the worst case we process divisors up to M, the total is O(M × log M)
• Building the result array: O(n)

Therefore, the total time complexity is O(M × log M + n).

Space Complexity: O(M)

The space complexity is determined by:

• The array d of size M + 1: O(M)
• The output array of size n: O(n) (not counted as auxiliary space since it's the required output)

Therefore, the auxiliary space complexity is O(M).

Where:

• n is the length of the groups array
• m is the length of the elements array
• M is the maximum value in the groups array

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

Common Pitfalls

Pitfall 1: Overwriting Previously Assigned Values

The Problem: A common mistake is to unconditionally assign element indices to all multiples without checking if a better (smaller index) assignment already exists. For example:

# WRONG: This overwrites existing assignments
for multiple in range(element_value, max_group_value + 1, element_value):
    assignments[multiple] = element_index  # Overwrites without checking!

Why It Happens: When processing elements, if we don't check whether a position has already been assigned, we might replace a smaller index with a larger one, violating the priority rule.

Example Scenario:

• groups = [6], elements = [2, 3]
• Element at index 0 (value 2) assigns index 0 to position 6
• Element at index 1 (value 3) would overwrite position 6 with index 1
• Wrong result: [1] instead of correct [0]

Solution: Always check if a position is unassigned before updating it:

if assignments[multiple] == -1:
    assignments[multiple] = element_index

Pitfall 2: Not Handling the Element Value Itself

The Problem: Forgetting to check and potentially assign the element value itself (not just its multiples) can lead to incorrect results when a group size exactly matches an element value.

Why It Happens: The condition if element_value > max_group_value or assignments[element_value] != -1 might skip processing when assignments[element_value] is already set, even though we still need to process the element's other multiples.

Example Scenario:

• groups = [4, 8], elements = [4, 2]
• Element at index 0 (value 4) should assign index 0 to positions 4 and 8
• If we skip entirely when assignments[4] != -1, position 8 might not get assigned

Solution: The current implementation handles this correctly by checking assignments[element_value] != -1 as an optimization - if the element value itself is already assigned by a smaller index, then ALL its multiples must also be assigned by that same smaller index (since any divisor of x is also a divisor of multiples of x).

Pitfall 3: Incorrect Range Boundaries

The Problem: Using incorrect loop boundaries when iterating through multiples:

# WRONG: Off-by-one error
for multiple in range(element_value, max_group_value, element_value):  # Missing +1

Why It Happens: Python's range() function excludes the end value, so we need max_group_value + 1 to include max_group_value itself.

Example Scenario:

• groups = [10], elements = [5]
• Without +1, the range would be range(5, 10, 5) producing only [5]
• Group size 10 wouldn't get assigned even though 5 divides 10

Solution: Always use range(element_value, max_group_value + 1, element_value) to ensure the maximum group value is included in the iteration.