easyArraysPattern: Hashing
Count Unique Elements Solution
Problem Statement
You are given an array of integers `nums` and need to implement a function that returns the count of unique elements in the array.
Examples
Example 1:
Input:[1, 2, 2, 3, 4, 4, 5]
Output:5
Explanation: Unique elements are 1, 2, 3, 4, 5. So the count is 5.
Example 2:
Input:[1, 1, 1, 1, 1]
Output:1
Explanation: Only one unique element, which is 1.
Constraints
- 1 <= nums.length <= 1000
- -1000 <= nums[i] <= 1000
- The array may contain duplicate elements
- The array may be empty
- The function should return the count of unique elements
Time: O(n) Space: O(n)
A more optimal approach is to utilize a data structure like a hash set, which allows for constant time complexity lookups and insertions. By iterating through the array and adding each element to the set, we can easily count the unique elements in linear time complexity. This approach provides a significant performance improvement over the brute force method.
Run, Test & Submit Code
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Solve on Interactive WorkspaceTested Solutions
import sys
input = sys.stdin.readline
nums = list(map(int, input().split()))
unique_count = len(set(nums))
print(unique_count)