Introduction

In this blog post, we will explore how to solve the LeetCode problem "219" 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 and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k. Example 1: Input: nums = [1,2,3,1], k = 3 Output: true Example 2: Input: nums = [1,0,1,1], k = 1 Output: true Example 3: Input: nums = [1,2,3,1,2,3], k = 2 Output: false Constraints: 1 <= nums.length <= 105 -109 <= nums[i] <= 109 0 <= k <= 105

Explanation

Here's a breakdown of the approach, complexity, and the Python code solution:

• High-Level Approach:

  • Use a hash map (dictionary in Python) to store each number encountered along with its index.
  • Iterate through the array. For each number, check if it's already in the hash map.
  • If the number is in the hash map, check if the absolute difference between the current index and the stored index is less than or equal to k. If so, return True. Otherwise, update the stored index with the current index. If the number is not in the hash map store it with its index.

• Complexity:

  • Runtime: O(n), where n is the length of the input array.
  • Storage: O(min(n,k)), as in the worst case we'll need to store n elements.

Code

    def containsNearbyDuplicate(nums, k):
    """
    Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.

    Args:
        nums (List[int]): The input array of integers.
        k (int): The maximum allowed difference between the indices.

    Returns:
        bool: True if there are two such indices, False otherwise.
    """
    num_map = {}  # Dictionary to store number and its index

    for i, num in enumerate(nums):
        if num in num_map:
            if abs(i - num_map[num]) <= k:
                return True
            else:
                num_map[num] = i  # Update with the new index
        else:
            num_map[num] = i  # Store the number and its index

    return False