How do you solve Convert an Array Into a 2D Array With Conditions on LeetCode?

Convert an Array Into a 2D Array With Conditions (LeetCode #2610) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency map allows us to efficiently distribute elements into rows based on their occurrences.

What is the brute force solution for Convert an Array Into a 2D Array With Conditions?

The brute force approach for Convert an Array Into a 2D Array With Conditions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element of the input array and attempting to place it in existing rows of a 2D array. If the element cannot be placed in any existing row due to duplicates, a new row is created. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Convert an Array Into a 2D Array With Conditions?

Convert an Array Into a 2D Array With Conditions uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Convert an Array Into a 2D Array With Conditions?

Always clarify the requirements and constraints before jumping into coding. Think about how to minimize the number of rows while ensuring distinct integers. Practice explaining your thought process as you code, as this can help you catch mistakes early.

What are common mistakes when solving Convert an Array Into a 2D Array With Conditions?

Not considering the need for distinct integers in each row, leading to incorrect placements. Failing to handle edge cases where all elements are the same.

#2610

Convert an Array Into a 2D Array With Conditions

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a frequency map to count occurrences of each number. We then distribute these numbers into rows based on their frequency, ensuring that each row contains distinct integers and minimizing the number of rows needed.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map to count how many times each number appears in the input array.
2Step 2: Determine the maximum frequency from the frequency map, which will dictate the number of rows needed.
3Step 3: Create an empty 2D array with the number of rows equal to the maximum frequency.
4Step 4: Distribute the numbers into the rows based on their frequency, ensuring each row contains distinct integers.

solution.py10 lines

1from collections import Counter
2
3def convertArray(nums):
4    freq = Counter(nums)
5    max_freq = max(freq.values())
6    result = [[] for _ in range(max_freq)]
7    for num, count in freq.items():
8        for i in range(count):
9            result[i].append(num)
10    return result

ℹ

Complexity note: The time complexity is O(n) because we traverse the input array to build the frequency map and then distribute the numbers into rows. The space complexity is O(n) due to the storage of the frequency map and the result array.

1Using a frequency map allows us to efficiently distribute elements into rows based on their occurrences.
2The number of rows needed is determined by the maximum frequency of any element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.