Contains Duplicate II Leetcode Problem 219 [Python Solution]

Welcome back to another exciting coding adventure!

In this blog post, we're going to tackle the Contains Duplicate II problem, a LeetCode question that falls under the category of Sliding Window.

By the end of this post, you'll have a clear understanding of the problem, its constraints, and an efficient Python solution to it.

Problem Overview

The problem statement goes like this: Given an integer array nums and an integer k, we need to determine if there are two distinct indices i and j in the array such that nums[i] is equal to nums[j], and the absolute difference between their indices (|i - j|) is less than or equal to k.

In simple terms, we're looking for duplicate values within a certain "window" of size k.

The key is understanding what constitutes a valid window and the concept of duplicates.

Let's dive into this with an example.

Example 1:

Input:

nums = [1, 2, 3, 1]
k = 3


Output:

True


In this example, we have the array [1, 2, 3, 1].

We can see that there are duplicate values (1) at different indices (0 and 3).

The question is whether the size of the window is less than or equal to k.

• For the window between the elements 1 at indices 0 and 3, the size is 4.

Is |0 - 3| <= 3?

Yes, it is.
– For the window between the elements 1 at indices 3 and 3, the size is 1.

Is |3 - 3| <= 3?

Yes, it is.

So, this is a valid window.

The key takeaway is that when i and j point to the same element, the size of the window is 1, and the absolute difference between their indices is 0, making it valid.

Now, let's look at the constraints that apply to this problem.

Understanding Constraints

The problem comes with the following constraints:

• 1 <= nums.length <= 105: The length of the array nums is between 1 and 105.
• -109 <= nums[i] <= 109: The values within the array can range from -109 to 109.
• 0 <= k <= 105: The value of k can range from 0 to 105. These constraints help us understand the potential scale of the problem and the need for an efficient solution.

Efficient Python Code Solution

To solve this problem efficiently, we can use a sliding window approach along with a hash set (or hash map) to keep track of the values within the window.

The sliding window technique allows us to avoid checking every possible window and instead iteratively slide the window over the array.

Let's implement this solution in Python:

def containsNearbyDuplicate(self, nums, k):
    # Initialize a set to keep track of values within the window
    window = set()
    # Initialize the left pointer of the window
    L = 0

    for R in range(len(nums)):
        # If the window size exceeds k, remove the leftmost value
        if R - L > k:
            window.remove(nums[L])
            L += 1
        # If the current value is already in the window, we found a duplicate
        if nums[R] in window:
            return True
        # Add the current value to the window
        window.add(nums[R])

    # If no duplicates were found, return False
    return False

Let's break down the code step by step:

1. We start by initializing an empty set called window to keep track of the values within our sliding window.

We also initialize the left pointer L to 0.

1. We iterate through the nums array using a for loop.

The variable R represents the right pointer of our sliding window.

1. If the size of the window exceeds k, it means we need to remove the leftmost value to maintain a valid window.

We remove the leftmost value from the window set and increment the left pointer L.

1. Next, we check if the current value nums[R] is already in our window.

If it is, we've found a duplicate within the window, and we return True.

1. If the current value is not a duplicate, we add it to our window set.

2. After looping through the entire array, if no duplicates were found, we return False.

This efficient solution has a time complexity of O(n), where n is the length of the input array nums, and a memory complexity of O(K), where K represents the maximum value of k (in this problem, K can be at most equal to the length of the array nums).

Time and Space Complexity

Time Complexity: O(n)

The time complexity of this solution is O(n) because we iterate through the entire nums array exactly once.

The operations performed within the loop, such as adding and removing elements from the window set, have a constant time complexity, and we never add the same element to the set more than once.

Space Complexity: O(K)

The space complexity of this solution is O(K), where K represents the maximum value of k.

In the worst case, our window set may contain up to k unique elements, which are all stored in memory.

Reasoning Behind Our Approach

The reasoning behind this approach lies in the efficient use of a sliding window and a hash set.

By maintaining a window of size k as we iterate through the array, we can quickly identify duplicates.

The hash set allows us to check for duplicates in constant time, making the solution both time and memory-efficient.

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Conclusion

In this blog post, we've explored the Contains Duplicate II problem, understood its constraints, and implemented an efficient Python solution using the sliding window technique and a hash set.

This solution provides a time complexity of O(n) and a space complexity of O(K), where K represents the maximum value of k.

To further enhance your coding skills, practice this solution and experiment with different test cases.

Remember, the ability to efficiently handle sliding windows and hash sets is a valuable skill in competitive programming and real-world software development.

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Happy coding! 🚀

LeetCode Problem Link