Leetcode 2610: Convert an Array Into a 2D Array With Conditions

Input: You are given an array of integers nums. The array consists of integers, and some integers may appear more than once.

Example: nums = [3, 1, 2, 3, 1, 4, 2]

Constraints:

• 1 <= nums.length <= 200

• 1 <= nums[i] <= nums.length

Output: Return a 2D array where each row contains distinct integers and all elements of the input array are used. The number of rows should be minimal.

Example: Output: [[3, 1, 2, 4], [3, 1], [2]]

Constraints:

• The output should be a valid 2D array where each row contains distinct integers.

Goal: The goal is to arrange the elements of the input array into a minimal 2D array where each row contains distinct integers and all the elements of the input array are used.

Steps:

• Count the frequency of each integer in the input array.

• Create a 2D array where each row contains distinct integers from the input array.

• Ensure that each row has as few elements as possible while still containing distinct integers.

Goal: Handle edge cases, such as when the input array is very small or contains a large number of duplicates.

Steps:

• The array length will be between 1 and 200.

Assumptions:

• The input array contains integers that may repeat.

• Input: nums = [1, 3, 4, 1, 2, 3, 1]

• Explanation: We can create a 2D array like [[1, 3, 4, 2], [1, 3], [1]], where each row contains distinct integers, and all elements of the array are used.

• Input: nums = [1, 2, 3, 4]

• Explanation: Since all integers are distinct, we can simply place all of them in the first row, resulting in [[4, 3, 2, 1]].

Approach: To solve this problem, we need to group repeated numbers into rows and ensure that each row has distinct integers, while minimizing the number of rows.

Observations:

• We need to find the frequency of each integer in the input array and ensure that each integer appears in a different row for each repetition.

• A possible approach is to first calculate the frequency of each number in the array. Then, we can iterate through the array and add the number to different rows based on how many times it appears.

Steps:

• Step 1: Count the frequency of each number in the array.

• Step 2: Create a 2D array with the appropriate number of rows.

• Step 3: Place each number in a row based on its frequency, ensuring that each row contains distinct integers.

Empty Inputs:

• An empty array should return an empty 2D array.

Large Inputs:

• For larger input arrays, ensure the algorithm handles the size within the given constraints.

Special Values:

• Arrays with only distinct values will have a single row, containing all elements.

Constraints:

• Handle arrays with a mix of unique and repeated values efficiently.

vector<vector<int>> findMatrix(vector<int>& nums) {
    map<int, int> mp;
    int mx = 0;
    for(int x: nums) {
        mp[x]++;
        mx = max(mp[x], mx);
    }
    vector<vector<int>> grid(mx);
    for(auto it: mp) {
        for(int i = 0; i < it.second; i++) {
            grid[i].push_back(it.first);
        }
    }
    return grid;
}

1 : Function Definition

vector<vector<int>> findMatrix(vector<int>& nums) {

Defines the function that takes a reference to a vector of integers and returns a 2D vector of integers.

2 : Variable Declaration

    map<int, int> mp;

Declares a map to store the frequency of each integer from the input vector.

3 : Variable Initialization

    int mx = 0;

Initializes the variable 'mx' to keep track of the maximum frequency of any number in the input vector.

4 : Loop - Iteration

    for(int x: nums) {

Starts a loop to iterate through each integer 'x' in the input vector.

5 : Map Update

        mp[x]++;

Increments the frequency count of the integer 'x' in the map.

6 : Maximum Frequency Update

        mx = max(mp[x], mx);

Updates the maximum frequency 'mx' if the frequency of the current integer 'x' is greater.

7 : Grid Initialization

    vector<vector<int>> grid(mx);

Initializes a 2D vector 'grid' with 'mx' rows to hold the elements grouped by their frequencies.

8 : Loop - Mapping Elements

    for(auto it: mp) {

Starts a loop to iterate through each key-value pair in the map 'mp'.

9 : Inner Loop - Frequency Processing

        for(int i = 0; i < it.second; i++) {

Starts an inner loop to iterate based on the frequency of the current element.

10 : Grid Update

            grid[i].push_back(it.first);

Adds the current integer (it.first) to the corresponding row in the 'grid'.

11 : Return Statement

    return grid;

Returns the 2D grid containing the elements grouped by their frequencies.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is linear in the size of the input array as we iterate through the array to count the frequency and then create the 2D array.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is linear as we store the frequency of the elements and the 2D array.

Leetcode 2610: Convert an Array Into a 2D Array With Conditions

Explore →