Convert an Array Into a 2D Array With Conditions

Medium

Google

1 view

Topics:

Arrays

You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions:

The 2D array should contain only the elements of the array nums.
Each row in the 2D array contains distinct integers.
The number of rows in the 2D array should be minimal.

Return the resulting array. If there are multiple answers, return any of them.

Note that the 2D array can have a different number of elements on each row.

Example 1:

Input: nums = [1,3,4,1,2,3,1]
Output: [[1,3,4,2],[1,3],[1]]
Explanation: We can create a 2D array that contains the following rows:
- 1,3,4,2
- 1,3
- 1
All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
It can be shown that we cannot have less than 3 rows in a valid array.

Example 2:

Input: nums = [1,2,3,4]
Output: [[4,3,2,1]]
Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= nums.length

How would you approach this problem? Can you provide a well-optimized solution?

Convert an Array Into a 2D Array With Conditions

Medium

Google

1 view

Topics:

Arrays

You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions:

The 2D array should contain only the elements of the array nums.
Each row in the 2D array contains distinct integers.
The number of rows in the 2D array should be minimal.

Return the resulting array. If there are multiple answers, return any of them.

Note that the 2D array can have a different number of elements on each row.

Example 1:

Input: nums = [1,3,4,1,2,3,1]
Output: [[1,3,4,2],[1,3],[1]]
Explanation: We can create a 2D array that contains the following rows:
- 1,3,4,2
- 1,3
- 1
All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
It can be shown that we cannot have less than 3 rows in a valid array.

Example 2:

Input: nums = [1,2,3,4]
Output: [[4,3,2,1]]
Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= nums.length

How would you approach this problem? Can you provide a well-optimized solution?

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:

Can the input array contain duplicate numbers, and if so, how should they be distributed across the rows of the 2D array?
What is the expected output if the input array cannot be converted into a 2D array that satisfies the conditions? Should I return an empty 2D array, or throw an exception?
Are there any constraints on the size or dimensions of the resulting 2D array (e.g., maximum number of rows or columns)?
Is the order of numbers within each row of the 2D array significant, or can they be in any order?
Can the input array be empty or null, and if so, what should the function return in those cases?

Brute Force Solution

Approach

The brute force method to create the 2D array involves exploring every possible way to group numbers. We repeatedly examine the original group of numbers and attempt different combinations to form the rows of the 2D array until a valid arrangement is found.

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

Start with the original group of numbers.
Create an empty 2D array to hold the rows.
Try picking a number from the original group and placing it as the first element of the first row of the 2D array.
Check if the number already exists in the first row.
If the number doesn't already exist, move onto creating the second element in the first row by picking another number from the original group.
Repeat this process for the first row, checking if each number picked already exists in that row.
Once the first row has been formed, create the second row, third row, etc. using the same method.
After creating a complete 2D array check if every number in each row is unique.
If all the numbers in each row are unique and if the total number of rows is the minimum possible, then it's a valid answer.
Repeat this entire process from the beginning, trying all the possible arrangements for the 2D array.
If no valid arrangement is found after exhausting all possibilities, return an empty 2D array.

Code Implementation

def convert_to_2d_array_brute_force(numbers):
    import itertools

    def is_valid_2d_array(array_2d, original_numbers):
        all_numbers = []
        for row in array_2d:
            all_numbers.extend(row)
        all_numbers.sort()
        original_numbers.sort()
        if all_numbers != original_numbers:
            return False
        for row in array_2d:
            if len(row) != len(set(row)):
                return False
        return True

    def find_minimum_rows(numbers):
        if not numbers:
            return 0
        counts = {}
        for number in numbers:
            counts[number] = counts.get(number, 0) + 1
        return max(counts.values())

    minimum_rows_needed = find_minimum_rows(numbers.copy())
    
    for permutation in itertools.permutations(numbers):
        for number_of_rows in range(minimum_rows_needed, len(numbers) + 1):
            
            # We will construct the array and test if it is valid
            array_2d = [[] for _ in range(number_of_rows)]
            index = 0
            valid_permutation = True
            for number in permutation:
                array_2d[index % number_of_rows].append(number)
                index += 1
            
            # Ensure the array is valid before returning it
            if is_valid_2d_array(array_2d, numbers.copy()):
                return array_2d

    return []

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach involves exploring all possible arrangements of numbers to form the 2D array. In the worst-case scenario, this requires generating all permutations of the input array of size n. Generating all permutations of n elements takes O(n!) time. For each permutation, we also need to validate if the array meets the constraints described which takes additional O(n) steps, however this is dominated by the initial O(n!) complexity. Therefore, the overall time complexity is O(n!).

Space Complexity

O(N) – The brute force approach described uses a 2D array to store the result, which in the worst case, could potentially store all N input numbers. The algorithm also keeps track of numbers already placed in each row. To avoid duplicates in each row, a set or a similar data structure might be used to keep track of numbers already present in a given row, contributing O(N) in the worst-case scenario where each number is unique. Considering all of these data structures, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The core idea is to count how many times each number appears. We then use this information to create the 2D array, placing each number in a new row if we've already seen it in a previous row. This ensures we meet the specific conditions efficiently.

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

First, we need to know how many times each number shows up in the original list.
Then, we create the structure that will hold our result: a list of lists (the 2D array).
Next, we go through the original list number by number.
For each number, we check if we already have that many rows in our 2D array. If not, we add new rows until we do.
Then, we put the number in the next available row. It's like placing each number in the lowest possible row where it doesn't already exist in that row.
We repeat this process for every number in the original list.
Finally, we return the 2D array we've created.

Code Implementation

def convert_to_2d_array(original_list):
    number_counts = {}
    for number in original_list:
        number_counts[number] = number_counts.get(number, 0) + 1

    result_2d_array = []

    for number in original_list:
        # Need to add rows if not enough exist for current number count.
        while len(result_2d_array) < number_counts[number]:

            result_2d_array.append([])

        # Place the number in the next available row.
        result_2d_array[number_counts[number] - 1].append(number)

        # Decrement count to put same number in next row.
        number_counts[number] -= 1

    return result_2d_array

Big(O) Analysis

Time Complexity

O(n) – We iterate through the input array nums of size n once to count the frequency of each number. Then, we iterate through nums again. In the second loop, for each element, we are potentially adding it to existing rows in the 2D array. Adding to a list takes constant time. Therefore, the dominant operation is iterating through the input array nums twice, which is 2n. Hence, the time complexity is O(n).

Space Complexity

O(N) – The algorithm uses a frequency count, which requires storing the count of each unique number in the input array. In the worst case, all numbers in the input array nums are unique, leading to a frequency count potentially storing N key-value pairs, where N is the length of the input array nums. Additionally, the resulting 2D array can have at most N elements stored within its rows. Therefore, the auxiliary space used grows linearly with the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty 2D array to indicate no solution is possible.
Array with only one unique element	The single element will be placed in the first row; subsequent rows will be empty.
Array with all identical elements	Each row will contain only that identical element; the number of rows will equal the frequency of the element.
Large input array size impacting performance	Use a frequency counting approach (hash map) to achieve O(n) time complexity, improving scalability.
Array with negative numbers or zeros	The frequency counting hash map will correctly handle non-positive integers.
No valid 2D array is possible (e.g., [1,1,2,2,3,3,4] cannot form rows with distinct numbers)	The algorithm will create rows until no more distinct numbers can be added; subsequent rows are empty.
Input array contains duplicate counts exceeding array length	The algorithm ensures a correct arrangement by creating rows based on the maximum frequency of any number.
Integer overflow when counting frequencies in extremely large arrays	Use appropriate data types (e.g., long) for frequency counts to prevent overflow.