Leetcode Daily Challenge Solutions

This is my attemp to make the coding experience easier for you guys so that you can easily learn what to do in today's leetcode challenge.

Always here to assist you guys.

Today's 02-01-24 Problem Link

Intuition

Unitary way to think.

Approach

I counted the frequencies of every number in the given array
- used HashMap for that
The maximum frequency of all element will be the number of answer rows
- as in every row there should be unique elements
Created and initialised answerlist with empty sub-lists
Now, according to their frequencies adding elements to their rows
- elements with frequency '1' will be present in only first row
- elements with frequency '2' will be present in first and second row
- ... and so on

Have a look at the code , still have any confusion then please let me know in the comments Keep Solving.:)

Complexity

Time complexity : $$O(cm)$$

Space complexity : $$O(c)$$

$$c$$ : number of unique elements in array $$m$$ : maximum frequency of any element

Code

class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        HashMap<Integer, Integer> m = new HashMap<>();
        int mf = 0; // to store maximum frequency of any integer : it will determine the number of rows
    
        // adding numbers in hashmap along with its frequencies
        for( int i = 0; i < nums.length; i++){
            m.putIfAbsent(nums[i], 0);
            m.put( nums[i], m.get(nums[i]) + 1 );
            mf = Math.max(mf, m.get(nums[i]));
        }
        List<List<Integer>> l = new ArrayList<>(); // to store the answer list
        // the element with maximum frequency will be present in every sub-list of answerlist
        // making 'mf' rows in answer list
        for( int i = 0; i < mf; i++){              
            List<Integer> t = new ArrayList<>();
            l.add(t);
        }
        
        // according to their frequencies adding elements to their rows
        // elements with frequency '1' will be present in only first row
        // elements with frequency '2' will be present in first and second row
        // ... and so on

        for( int c : m.keySet() ){
            for( int f = 0; f < m.get(c); f++){
                l.get(f).add(c);
            }
        }
        return l;
    }
}