In the world of software engineering interviews, familiarity with coding challenges is a key to success. Among these challenges, LeetCode stands out as a prime resource for honing your problem-solving skills.

Today, I'm thrilled to guide you through a common yet intriguing problem from LeetCode: "Contains Duplicate" (LeetCode 217). This problem is a classic example that tests your ability to work with arrays and understand the nuances of data structures in Python.

Understanding the Problem

The "Contains Duplicate" problem is straightforward: you are given an integer array nums, and your task is to determine if any value appears at least twice in the array. If so, return true; otherwise, return false.

This challenge checks your ability to identify duplicates in an array—a fundamental skill in coding interviews and everyday programming.

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.

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

Strategy for Solution

To tackle this problem, understanding the implications of each potential solution is crucial. Here, I'll describe two primary approaches—utilizing sorting and hash tables (specifically, sets and counters)—and analyze their time and space complexities.

Sorting Approach

Description: By sorting the array, we ensure that any duplicate elements are positioned next to each other. Then, we simply iterate through the sorted array, checking if any adjacent elements are equal.

Big O Notation Analysis:

• Time Complexity: O(n log n) due to the sorting operation.

• Space Complexity: O(1), assuming the sort is in-place.

Hash Table Approach

Using a Set:

• Description: We iterate through the array, adding elements to a set. If we ever encounter an element that's already in the set, we know there's a duplicate.

• Big O Analysis: Time Complexity: O(n), Space Complexity: O(n).

Using a Counter:

• Description: Similar to the set, but we use a Counter to count occurrences of each element. If any count is greater than 1, we return true.

• Big O Analysis: Time Complexity: O(n), Space Complexity: O(n).

Python Solutions

Solution with Sorting

def containsDuplicate(self, nums: List[int]) -> bool:
    nums.sort()  # Sort the list to ensure duplicates are adjacent
    for i in range(len(nums)-1):  # Loop through the list
        if nums[i] == nums[i+1]:  # Check if adjacent elements are equal
            return True  # Duplicate found
    return False  # No duplicates found

Intuitive Approach Using a Set

def containsDuplicate(self, nums: List[int]) -> bool:
    values = set()  # Initialize an empty set
    for num in nums:  # Iterate over each number
        if num in values:  # If the number is already in the set, it's a duplicate
            return True
        values.add(num)  # Add the number to the set
    return False

Smart Use of a Set

def containsDuplicate(self, nums: List[int]) -> bool:
    return len(set(nums)) != len(nums)  # Compare set length to list length

Solution Using a Counter

from collections import Counter
def containsDuplicate(self, nums: List[int]) -> bool:
    freq = Counter(nums)  # Count occurrences of each number
    for num, freq in freq.items():  # Iterate through the Counter
        if freq > 1:  # If any number appears more than once
            return True
    return False

Conclusion

Solving the "Contains Duplicate" problem on LeetCode offers a fantastic opportunity to practice with arrays and data structures. By exploring multiple solutions, we not only sharpen our coding skills but also deepen our understanding of time and space complexities.

Remember, mastering these challenges is not just about finding a solution but about understanding all possible solutions and their trade-offs.

Happy coding, and may your journey through coding interviews be successful and enlightening!