Neetcode 1: Duplicate Integer

Given an integer array nums, return true if any value appears more than once in the array, otherwise return false.

Example 1:

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

Output: true

Example 2:

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

Output: false

Python Duplicate Integer

Understand: We need to return true if we find more than 1 of any variable. so [1,1,2,3] returns true and [1,2,3,4] returns false

Match: We see this as a simple hash table problem.

~~Plan: Create a hash table that has integers as key and values as incremented amount~~ Plan: Create a set() because we can automatically remove duplicate numbers in this use case. If set() is less than length of original list, return false. Else return true

Return false at the end of our loop... We couldnt find duplicates.

Implement:

class Solution:
    def hasDuplicate(self, nums: List[int]) -> bool:
        hashset = set() # Python can define a set like so
        for n in nums: # Create a loop for num in nums
            if n in hashset: # Check existence in hashset at O(1)
                return True # Return True if already exists
            hashset.add(n) # Continue and add element to the set
        return False # Finally return False if our loop exits

Result: We successfully return True on duplicates

Evaluation: We return at O(n) speed where the search for the integer within the hashset is of course at O(1) search.

C++ Duplicate Integer

Understand:

• Again, we have an array of nums and need to return true if there's a duplicate.
• C++ has iterator and cannot be easily outputted as a string

class Solution {
public:
    bool hasDuplicate(vector<int>& nums) {
	    // C++ declares unordered set like so
        unordered_set<int> s; 
        // Loop based on size of the nums
        for (int i = 0; i < nums.size(); i++) { 
	        // s.find(nums[i]) aka the iterator
	        // DOES NOT return s.end(), 
            if (s.find(nums[i]) != s.end()) { 
                return true; // then the number already exists in the unordered set so let's return true
            }
            s.insert(nums[i]);
        }
        return false;
    }
};

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