2610 Convert an Array Into a 2D Array With Conditions - Medium

Problem:

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

Problem Analysis:

High-Level Strategy:

• The goal of this solution seems to be organizing a given list of integers into a matrix based on their occurrences. The matrix is constructed such that each row represents a unique integer, and the elements of the row are the occurrences of that integer in the input list.

Complexity:

• Let n be the length of the input list nums.
• The solution iterates through each element in nums once, and for each element, it performs constant time operations (dictionary lookups, list appends). Therefore, the time complexity is O(n).
• The space complexity is also O(n) since the solution uses a dictionary (count) to store the count of occurrences and a list (res) to store the resulting matrix.

Solutions:

class Solution:
    def findMatrix(self, nums: List[int]) -> List[List[int]]:
        count = defaultdict(int)
        res = []
        for n in nums:
            # get current row from count 
            row = count[n]
            # if they matches, we need to add row
            if len(res) == row:
                res.append([])
            
            res[row].append(n)
            count[n] += 1

        return res

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