# Solving Leetcode Interviews in Seconds with AI: Unique Number of Occurrences


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1207" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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. Example 2:  Input: arr = [1,2] Output: false  Example 3:  Input: arr = [-3,0,1,-3,1,1,1,-3,10,0] Output: true    Constraints:  1 <= arr.length <= 1000 -1000 <= arr[i] <= 1000  

	# Explanation
	Here's the solution:

*   **Count Occurrences:** Use a dictionary (hash map) to efficiently count the frequency of each number in the input array.
*   **Check Uniqueness of Counts:** Use a set to store the counts. If we encounter a count that's already in the set, it means the counts are not unique, and we return `False`. Otherwise, we add the count to the set and continue.
*   **Return Result:** If we successfully iterate through all the counts without finding duplicates, we return `True`.

*   Runtime Complexity: O(n), Storage Complexity: O(n)

	
	# Code
	```python
	def uniqueOccurrences(arr: list[int]) -> bool:
    """
    Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.
    """
    counts = {}
    for num in arr:
        counts[num] = counts.get(num, 0) + 1

    occurrences = set()
    for count in counts.values():
        if count in occurrences:
            return False
        occurrences.add(count)

    return True
	```