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

Problem Link

https://leetcode.com/problems/convert-an-array-into-a-2d-array-with-conditions/

Problem Statement

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:

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

Approach 1: Frequency Count

We can distribute the elements by frequency count. Taking $nums = [1,3,4,1,2,3,1]$ as an example, we can do the followings:

• there are three 1s, we can put them into first three rows.
• there are one 2, we can put it into the first row.
• there are two 3s, we can put them into first two rows.
• there are one 4, we can put it into the first row.

As we can see, we need to distribute $K$ elements into first $K$ rows, which means we will need $max_freq$ rows in total. We can iterate $nums$ and check if the frequency count is greater than / equal to the target row or not. If so, we need to add a new row. Then we can push the element to the target row and increase the frequncy count.

C++

Written by @wingkwong

class Solution {
public:
    vector<vector<int>> findMatrix(vector<int>& nums) {
        vector<vector<int>> ans;
        // or use vector<int> m(nums.size()) since we only need max_freq rows
        unordered_map<int, int> m;
        for (auto x : nums) {
            // we need a new row to store `x`
            if (m[x] >= ans.size()) ans.push_back({});
            // add `x` to the target row
            ans[m[x]].push_back(x);
            // increase its frequncy count
            m[x]++;
        }
        return ans;
    }
};

Elixir

Written by @jit

@spec find_matrix(nums :: [integer]) :: [[integer]]
# Distribute elements by frequency count:
def find_matrix(nums) do
  cnts = nums |> Enum.frequencies() |> Enum.to_list()
  max_cnt = cnts |> Enum.map(&elem(&1, 1)) |> Enum.max()

  for cnt <- 1..max_cnt do
    cnts
    |> Enum.filter(fn {_elt, c} -> c >= cnt end)
    |> Enum.map(fn {elt, _c} -> elt end)
  end
end