Introduction

In this blog post, we will explore how to solve the LeetCode problem "26" 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

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums. Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things: Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums. Return k. Custom Judge: The judge will test your solution with the following code: int[] nums = [...]; // Input array int[] expectedNums = [...]; // The expected answer with correct length int k = removeDuplicates(nums); // Calls your implementation assert k == expectedNums.length; for (int i = 0; i < k; i++) { assert nums[i] == expectedNums[i]; } If all assertions pass, then your solution will be accepted. Example 1: Input: nums = [1,1,2] Output: 2, nums = [1,2,] Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively. It does not matter what you leave beyond the returned k (hence they are underscores). Example 2: Input: nums = [0,0,1,1,1,2,2,3,3,4] Output: 5, nums = [0,1,2,3,4,,,,,] Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively. It does not matter what you leave beyond the returned k (hence they are underscores). Constraints: 1 <= nums.length <= 3 * 104 -100 <= nums[i] <= 100 nums is sorted in non-decreasing order.

Explanation

• Two Pointers: Utilize two pointers, i (slow runner) and j (fast runner), to iterate through the array.
  • In-place Modification: When nums[i] and nums[j] are different, move nums[j] to nums[i+1] and increment i.
  • Return Unique Count: The final value of i + 1 represents the number of unique elements (k).

• Time Complexity: O(n), where n is the length of the array. Space Complexity: O(1).

Code

    def removeDuplicates(nums: list[int]) -> int:
    """
    Removes duplicate elements from a sorted array in-place.

    Args:
        nums: A sorted list of integers.

    Returns:
        The number of unique elements in the array.
    """
    if not nums:
        return 0

    i = 0  # Slow runner
    for j in range(1, len(nums)):  # Fast runner
        if nums[i] != nums[j]:
            i += 1
            nums[i] = nums[j]

    return i + 1