This solution is designed to check for duplicate values in an array of integers. The core idea revolves around utilizing a Set
data structure, specifically a HashSet
, to track the uniqueness of elements in the array. Here's a breakdown of the solution:
The primary intuition behind using a Set
is that this data structure is inherently designed to store unique elements. When an element is added to a Set
, it checks if the element already exists. If it does, the Set
remains unchanged. This property of a Set
is leveraged in this solution to identify duplicates in the array.
-
Initialize a
HashSet
: AHashSet
nameduniqueNumbers
is created to store the elements of the array. The choice ofHashSet
is due to its efficientO(1)
average time complexity foradd
and contains operations. -
Iterate Through the Array: The method iterates over each element in the nums array.
-
Check for Duplicates: For every element (
num
) in the array:- The method checks if
num
is already in theHashSet
usinguniqueNumbers.contains(num)
. - If
num
is found in theHashSet
, it meansnum
is a duplicate, and the method immediately returnstrue
.
- The method checks if
-
add
Element toHashSet
: Ifnum
is not found in theHashSet
, it is added usinguniqueNumbers.add(
num)
. This step ensures that theHashSet
always contains the unique elements encountered so far in the array. -
Return False if No Duplicates Found: If the method completes iterating through the array without finding any duplicates, it returns
false
.
-
Time Complexity:
O(N)
, where N is the number of elements in the array. Each element is checked exactly once to see if it is in theHashSet
, and this check isO(1)
on average due to the nature of hash-based data structures. -
Space Complexity:
O(1)
in the worst case. This happens when all elements in the array are distinct, causing theHashSet
to grow to the size of the array.
The solution is efficient and straightforward, using the unique properties of a HashSet
to detect duplicates with a single pass through the array. The O(N)
time complexity makes it suitable for large arrays, and the approach avoids the need for nested loops or additional sorting, which could increase the time complexity.