Grind 75 - 24 - Contains Duplicate

• Leetcode Number: 217
• Difficulty Level: Easy
• 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.Input: nums = [1,2,3,4]
• Output: false

Questions to be Asked the Interviewer:

• Can the input array contain negative integers?
• What is the expected range of values in the array?
• How should I handle an empty array?

Edge Cases for this Problem:

• An empty array or an array with only one element, where the answer will always be false.

Explain the Available Approaches to Solve this Problem:

1. Using Sorting: Sort the array and then iterate through it to check if any adjacent elements are the same.
2. Using a Set: Utilize a set to check for duplicates as we iterate through the array.

My Approach

• I will use the Set approach as it provides an efficient solution.
• Iterate through the array, and for each element, check if it's already in the set.
• If yes, return true, else add the element to the set.
• If the loop completes without finding duplicates, return false.

Algorithm and Data Structure:

• Algorithm: Linear Search
• Data Structure: Hash Set

def containsDuplicate(nums):
? ? seen = set()
? ? for num in nums:
? ? ? ? if num in seen:
? ? ? ? ? ? return True
? ? ? ? seen.add(num)
? ? return False        

Explain the Solution to a Non-Programmer :

• Imagine you have a basket of numbers, and you want to find out if any number appears more than once.
• You start with an empty plate and pick numbers from the basket one by one.
• Each time you pick a number, you check if it's already on the plate.If it is, you've found a number that appears more than once, so you can stop and say "yes, there is a duplicate."
• If it's not on the plate, you put it there and continue with the next number.
• If you go through all the numbers in the basket without finding any that appear more than once, you can say "no, there are no duplicates."

Code in Detail :

• Initialize a Set: We create an empty set called seen to store the numbers we have encountered.
• Iterate Through the Numbers: We go through each number in the given list nums.Check for Duplicates: For each number, we check if it is already in the set seen.If Duplicate Found: If the number is already in the set, we return True, indicating that a duplicate is found.
• If Not a Duplicate: If the number is not in the set, we add it to the set seen to keep track of it.
• No Duplicates Found: If the loop completes without finding any duplicates, we return False, indicating that there are no duplicates.

Pattern of This Problem:

• This problem follows the pattern of using a Hash Set to track unique elements.
• By using a set to store the elements encountered so far, we can efficiently determine whether a duplicate exists.

Big O Notation:

• Time Complexity: The time complexity of the code is O(n)
• O(n), where - n is the number of elements in the list. We iterate through the list once, performing constant-time operations.
• Space Complexity: The space complexity is also O(n) -O(n), as we might store all unique elements in the set in the worst case (when there are no duplicates).

Points to Remember to Solve this Problem:

• Using a set to track unique elements is a powerful pattern that comes in handy in many problems.
• Always consider the trade-off between space and time. In this case, the use of additional space enables a significant improvement in time complexity.
• It's essential to understand what the problem requires in terms of duplicates (e.g., finding any duplicate or all duplicates) as this may affect the solution design.

Code Line by Line with Some Sample Data:

Suppose nums = [1, 2, 3, 4, 2].

• seen = set(): Initialize an empty set to store numbers seen so far.
• Loop through the numbers in nums:1: Add to seen, continue.
• 2: Add to seen, continue.
• 3: Add to seen, continue.
• 4: Add to seen, continue.
• 2: Already in seen, return True (duplicate found).

Key Syntax Used in This Solution :

• Set Operations: Using a set (seen = set()) to store unique elements and checking for existence with in (e.g., if num in seen:).
• Looping Through List: Iterating through the list using a for-loop (for num in nums:).
• Returning Values: Using return True or return False to return the result.

The next problem is

Maximum Subarray