217-ContainsDuplicate

Contains Duplicate

Easy

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

Constraints:

• $1$ <= nums.length <= $10^5$
• $-10^9$ <= nums[i] <= $10^9$

Problem can be found in here!

Solution

Basic Solution: Brute Force

def containsDuplicate(nums: List[int]) -> bool:
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            if nums[i] == nums[j]:
                return True

    return False

Time Complexity: $O(n^2)$, Space Complexity: O(1)

Improved Solution: Sort

def containsDuplicate(nums: List[int]) -> bool:
    nums.sort()
    left, right = 0, 1
    while right < len(nums):
        if nums[left] == nums[right]:
            return True
        else:
            left, right = right, right + 1

    return False

Time Complexity: $O(nlogn)$, Space Complexity: $O(1)$

Optimized Solution: Set or Hash Table

def containsDuplicate(nums: List[int]) -> bool:
    memo = set()
    for number in nums:
        if number in memo:
            return True
        else:
            memo.add(number)

    return False

Time Complexity: $O(n)$, Space Complexity: $O(n)$