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 = [1,1,1,3,3,4,3,2,4,2]

Output: true

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

Output: false

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To solve this problem in linear time complexity (O(n)), you can use the following algorithm:

`duplicates`

.`x`

in the array: a. If`x`

is in`duplicates`

, return`true`

. b. Otherwise, add`x`

to`duplicates`

.`false`

.Here is some example code in Python that implements this algorithm:

`def containsDuplicate(nums):`

duplicates = set()

for x in nums:

if x in duplicates:

return True

duplicates.add(x)

return False

This algorithm has a time complexity of O(n), as it requires iterating through the array once and performing a constant number of operations for each element.

I hope this helps! Let me know if you have any questions.