Contains Duplicate LeetCode Problem 217 [Python Solution]

In this guide, we’re going to tackle the Contains Duplicate LeetCode problem.

This is a fantastic problem for both beginners and experienced programmers alike, as it offers multiple solutions, each with its own set of trade-offs.

Question

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Problem Overview

The Contains Duplicate problem is a common coding challenge. In this problem, you are given an array of integers, and your task is to determine if there are any duplicate values in the array.

If there is at least one duplicate, you should return true; otherwise, return false.

Example 1:

Input: nums = [1,2,3,1]
Output: true

Example 2:

Input: nums = [1,2,3,4]
Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true

Understanding the Constraints

Now, let’s understand the constraints provided:

• The length of the input array nums will be between 1 and 105. This tells us that we can have arrays of varying sizes, but not extremely large ones.
• The values in the array nums can range from -109 to 109. This means the integers can be both positive and negative and cover a wide range.

Contains Duplicate LeetCode Problem Solution

How do we solve this problem?

1. Bruteforce Approach

Apparently, the easiest and most straightforward approach to solving this problem is to use a brute force method.

We can iterate through each element in the array and, for each element, check if it appears again in the rest of the array.

If we find any element that appears more than once, we return true, indicating the presence of duplicates.

Otherwise, if we iterate through the entire array without finding any duplicates, we return false.

The time complexity of this brute force approach is O(n^2), where n is the length of the input array.

This is because, in the worst case, we might need to compare each element with every other element in the array.

However, this approach doesn’t require any additional memory, as we are only using the original array.

2. Efficient Approach with Hash Set

A more efficient approach is to use a data structure called a hash set to keep track of the unique values we have encountered so far.

Here’s how this approach works:

1. Initialize an empty hash set.
2. Iterate through the elements in the input array nums.
3. For each element num, check if it already exists in the hash set.
   • If it does, return true because we have found a duplicate.
   • If it doesn’t, add num to the hash set.
4. If we complete the loop without finding any duplicates, return false.

This approach has a time complexity of O(n) because we iterate through the array once, and the average time complexity for hash set operations is O(1).

However, it does use extra memory to store the hash set, which can have a space complexity of O(n) in the worst case if all elements in the array are distinct.

Python Code Solution

Here’s a Python implementation of the efficient approach using a hash set:

def containsDuplicate(nums):
    # Initialize an empty hash set
    num_set = set()
    
    # Iterate through the elements in the array
    for num in nums:
        # Check if the current element is already in the set
        if num in num_set:
            return True  # Found a duplicate
        # Add the current element to the set
        num_set.add(num)
    
    # If we reach this point, no duplicates were found
    return False

Time and Space Complexity

• Time Complexity: O(n) – We iterate through the array once.
• Space Complexity: O(n) – In the worst case, the hash set can contain all unique elements from the input array.

n here is the size of the input array.

Reasoning Behind Our Approach

The efficient approach using a hash set optimizes the time complexity by avoiding unnecessary comparisons.

As we iterate through the input array, we use the hash set to keep track of unique elements we have encountered so far.

If we ever encounter an element that already exists in the hash set, we know we’ve found a duplicate, and we can return true immediately.

This approach guarantees that we find duplicates in linear time, making it more efficient than the brute force approach.

Conclusion

Solving the Contains Duplicate problem efficiently involves using a hash set to track unique elements, resulting in a time complexity of O(n) while using some extra memory.

This approach strikes a balance between time and space complexity and is suitable for handling the given constraints.

I hope you found this explanation helpful in understanding how to approach and solve this problem.

Solve on LeetCode now.

If you have any questions or would like further clarification, please feel free to ask.

Happy coding!