# Solving Leetcode Interviews in Seconds with AI: Degree of an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "697" 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 a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements. Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.   Example 1:  Input: nums = [1,2,2,3,1] Output: 2 Explanation:  The input array has a degree of 2 because both elements 1 and 2 appear twice. Of the subarrays that have the same degree: [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2] The shortest length is 2. So return 2.  Example 2:  Input: nums = [1,2,2,3,1,4,2] Output: 6 Explanation:  The degree is 3 because the element 2 is repeated 3 times. So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.    Constraints:  nums.length will be between 1 and 50,000. nums[i] will be an integer between 0 and 49,999.  

	# Explanation
	Here's the solution to the problem:

*   **Find the Degree:** Iterate through the array to determine the degree (maximum frequency of any element). Use a hash map (dictionary) to store the frequency of each element.
*   **Identify Elements with Maximum Frequency:** Identify all elements that have the maximum frequency (the degree).
*   **Find Shortest Subarray:** Iterate through the array again. For each element with the maximum frequency, find the first and last occurrences in the array. The difference between these occurrences + 1 will give the length of the shortest subarray containing that element with the maximum frequency.  Maintain a running minimum of these lengths.

*   **Runtime Complexity:** O(n), where n is the length of the input array.
*   **Storage Complexity:** O(n), where n is the length of the input array (in the worst case, all elements are unique).

	
	# Code
	```python
	def find_shortest_subarray(nums):
    """
    Finds the smallest possible length of a contiguous subarray of nums that has the same degree as nums.
    """
    counts = {}
    first_occurrence = {}
    last_occurrence = {}

    degree = 0
    for i, num in enumerate(nums):
        if num not in counts:
            counts[num] = 0
            first_occurrence[num] = i
        counts[num] += 1
        last_occurrence[num] = i
        degree = max(degree, counts[num])

    min_length = float('inf')
    for num, count in counts.items():
        if count == degree:
            length = last_occurrence[num] - first_occurrence[num] + 1
            min_length = min(min_length, length)

    return min_length
	```