Solution 1:
Time complexity : O(nlogn)
Space complexity : O(1)
After sorting an array, iterate through it using Two Pointers.
If two pointers (index and index + 1) that are next to each other have the same value, there is a duplicate element, and the function returns true.
If there are no duplicate elements, both pointers continue to move simultaneously.
If the iteration is complete and there are no duplicate elements, return false.
class Solution {
func containsDuplicate(_ nums: [Int]) -> Bool {
var newNums = nums
newNums.sort()
for index in 0..<newNums.count - 1 {
if newNums[index] == newNums[index + 1] {
return true
}
}
return false
}
}
Solution 2:
Use Set:
Iterate over the array and put the numbers in the Set.
If the number is already in the Set, return true.
If the iteration is complete and there are no duplicate elements, return false.
class Solution {
func containsDuplicate(_ nums: [Int]) -> Bool {
var distinctSet = Set<Int>()
for num in nums {
if distinctSet.contains(num) {
return true
} else {
distinctSet.insert(num)
}
}
return false
}
}
Solution 3:
Time complexity : O(n)
Space complexity : O(n)
This problem can be solved with one line of code.
class Solution {
func containsDuplicate(_ nums: [Int]) -> Bool {
nums.count != Set(nums).count
}
}
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