Introduction

In this blog post, we will explore how to solve the LeetCode problem "594" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1. Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences. Example 1: Input: nums = [1,3,2,2,5,2,3,7] Output: 5 Explanation: The longest harmonious subsequence is [3,2,2,2,3]. Example 2: Input: nums = [1,2,3,4] Output: 2 Explanation: The longest harmonious subsequences are [1,2], [2,3], and [3,4], all of which have a length of 2. Example 3: Input: nums = [1,1,1,1] Output: 0 Explanation: No harmonic subsequence exists. Constraints: 1 <= nums.length <= 2 * 104 -109 <= nums[i] <= 109

Explanation

Here's the breakdown of the approach, complexities, and Python code:

• High-Level Approach:

  • Use a hash map (dictionary in Python) to store the frequency of each number in the input array.
  • Iterate through the hash map. For each number, check if its adjacent number (number + 1) also exists in the hash map.
  • If both the number and its adjacent number exist, calculate the sum of their frequencies. Keep track of the maximum sum encountered so far, which represents the length of the longest harmonious subsequence.

• Complexity:

  • Runtime Complexity: O(N), where N is the length of the input array.
  • Storage Complexity: O(N) in the worst case, to store the frequency map.

Code

    def findLHS(nums):
    """
    Finds the length of the longest harmonious subsequence in an array.

    Args:
        nums: A list of integers.

    Returns:
        The length of the longest harmonious subsequence.
    """

    freq_map = {}
    for num in nums:
        freq_map[num] = freq_map.get(num, 0) + 1

    max_len = 0
    for num in freq_map:
        if num + 1 in freq_map:
            max_len = max(max_len, freq_map[num] + freq_map[num + 1])

    return max_len