Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Example 1:


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

Example 2:


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

Example 3:


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

Source: leetcode | Difficulty: EASY
Tags: Sorting

Data structures

Starter code: class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        

Solutions:
class Solution:

	def containsDuplicate(self, nums):
		num_set = set(nums)
		if len(nums) == len(num_set):
			return False
		return True


class Solution:

	def containsDuplicate(self, nums):
		dict1 = {}
		for num in nums:
			dict1[num] = dict1.get(num, 0) + 1
		for key in dict1:
			if dict1[key] > 1:
				return True
		return False


class Solution:

	def containsDuplicate(self, nums):
		nums.sort()
		if len(nums) - 1 == 1:
			if nums[0] == nums[1]:
				return True
		for i in range(1, len(nums)):
			print(i)
			if nums[i] == nums[i - 1]:
				return True
		return False


class Solution:

	def containsDuplicate(self, nums):
		myDict = collections.Counter(nums)
		print(myDict)
		for value in list(myDict.values()):
			if value > 1:
				return True
		else:
			return False


class Solution:

	def containsDuplicate(self, nums):
		diction = {}
		flag = 0
		for number in nums:
			if number not in diction:
				diction[number] = 1
			else:
				flag = 1
				break
		if flag == 1:
			return True
		else:
			return False


class Solution:

	def containsDuplicate(self, nums):
		if len(nums) < 2:
			return False
		return True if len(set(nums)) < len(nums) else False


class Solution:

	def containsDuplicate(self, nums):
		if nums == None or len(nums) <= 1:
			return False
		map = {}
		for i in nums:
			if i in map:
				return True
			else:
				map[i] = 0
		return False


class Solution:

	def containsDuplicate(self, nums):
		myDict = {}
		for num in nums:
			if myDict.get(num):
				return True
			else:
				myDict[num] = 1
		else:
			return False


class Solution:

	def containsDuplicate(self, nums):
		n_map = {}
		for n in nums:
			if n in n_map:
				n_map[n] += 1
			else:
				n_map[n] = 1
		for (n, count) in n_map.items():
			if count != 1:
				return True
		return False


class Solution:

	def containsDuplicate(self, nums):
		hashmap = {}
		for i in nums:
			if i in hashmap.keys():
				return True
			else:
				hashmap[i] = hash(i)
		return False


class Solution:

	def containsDuplicate(self, nums):
		if len(nums) > 1:
			count = 0
			giru = []
			nums.sort()
			for i in range(len(nums)):
				if i + 1 < len(nums):
					if nums[i] == nums[i + 1]:
						count = count + 1
			if count > 0:
				return True
			else:
				return False
		else:
			return False


class Solution:

	def containsDuplicate(self, nums):
		d = {}
		for num in nums:
			if num in d:
				return True
			else:
				d[num] = 1
		return False


class Solution:

	def containsDuplicate(self, nums):
		xx = dict()
		for i in range(len(nums)):
			if nums[i] in xx:
				return True
			else:
				xx[nums[i]] = 1
		return False


class Solution:

	def containsDuplicate(self, nums):
		seen = dict()
		for num in nums:
			if str(num) in seen:
				return True
			else:
				seen[str(num)] = 1
		return False