LeetCode 1207 solution

problem

Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.

Example 1:

• Input: arr = [1,2,2,1,1,3]
• Output: true
• Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

python

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        # hash table
        hash_t = {}
        for i in arr:
            if i in hash_t:
                hash_t[i] += 1
            else:
                hash_t[i] = 1
        hash_t_ = {}
        for j in hash_t.values():
            if j in hash_t_:
                hash_t_[j] += 1
            else:
                hash_t_[j] = 1
        for k in hash_t_.values():
            if k > 1:
                return False
        return True

The first thought that come up with my mind is that "I should use hash table with this problem". Occurrences of each number needs to be checked. And also, the recorded hash tables needs to be checked one more time with hash table. With this two step, I could solve the problem.

The best thing about hash table is that it is Fast data retrieval structure. At very most time for algorithm problem, using hash table for data search and store is faster than just search with for loop. Because of the "key-value pairs and hash function", this advantage($O(1)$) is possible.

other solution

This is solution from MarkSPhilips31

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        freq = {}
        for x in arr:
            freq[x] = freq.get(x, 0) + 1

        return len(freq) == len(set(freq.values()))

The code is so short... and clean. Making hash table is same as me, but to check whether there is over 1 occurrence happen, the len(freq) == len(set(freq.values())) part is soooo dope. Eduacative