#1748

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Sum of Unique Elements

Given an array, find the sum of all its unique elements by identifying those that appear only once.

Problem Statement

You are given an integer array nums. Your task is to identify the unique elements in the array, which are those that appear exactly once. Then, calculate the sum of these unique elements.

For example, with nums = [1,2,3,2], the unique elements are 1 and 3, and their sum is 4. If there are no unique elements, the result should be 0.

Examples

Example 1

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

Output: 4

The unique elements are [1,3], and the sum is 4.

Example 2

Input: nums = [1,1,1,1,1]

Output: 0

There are no unique elements, and the sum is 0.

Example 3

Input: nums = [1,2,3,4,5]

Output: 15

The unique elements are [1,2,3,4,5], and the sum is 15.

Constraints

1 <= nums.length <= 100
1 <= nums[i] <= 100

Solution Approach

Hash Table for Counting Frequencies

Create a hash table (dictionary) to count the frequency of each number in the array. This allows for quick lookup to check how many times an element occurs.

Iterate and Sum Unique Elements

After counting the frequencies, iterate through the hash table, adding elements with a frequency of 1 to the sum. This gives the sum of the unique elements in the array.

Optimize with Early Termination

To optimize, you can stop the iteration early if all unique elements are found or if an element is found more than once, reducing unnecessary checks.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of elements in the array, due to the single pass required to count frequencies and another pass to sum unique elements. The space complexity is O(n) for storing the frequencies in the hash table.

What Interviewers Usually Probe

Candidate should clearly explain the hash table usage and the need for frequency counting.
Look for knowledge of iterating over a hash map to filter out elements with frequency greater than 1.
Assess if the candidate considers early termination or optimizing the solution.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle edge cases like arrays with no unique elements or arrays with all elements the same.
Misunderstanding the concept of unique elements, leading to incorrect frequency counting.
Inefficient solutions that repeatedly scan the array, instead of using a hash table for counting.

Follow-up variants

What if negative numbers are allowed in the array?
How would the solution change if the array was sorted?
Can the problem be solved with constant space?

FAQ

How does array scanning plus hash lookup work in the Sum of Unique Elements problem?

In this problem, array scanning is used to iterate through the elements, and hash lookup (via a dictionary) efficiently counts the frequency of each element.

What is the time complexity of this approach?

The time complexity is O(n), where n is the number of elements in the array, since we only scan the array twice: once for counting and once for summing unique elements.

Can the algorithm be optimized further?

You can optimize the solution by stopping the iteration early if you find that all unique elements have been processed, saving time in some cases.

What happens if there are no unique elements in the array?

If there are no unique elements, the sum returned should be 0, as per the problem's requirements.

How does GhostInterview help with the Sum of Unique Elements problem?

GhostInterview assists by providing efficient ways to count frequencies and sum unique elements, ensuring quick implementation and test case handling.

terminal

Solution

Solution 1

#### Python3

1

2

3

4

class Solution:
    def sumOfUnique(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        return sum(x for x, v in cnt.items() if v == 1)

Solution 2

#### Java

1

2

3

4

class Solution:
    def sumOfUnique(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        return sum(x for x, v in cnt.items() if v == 1)

Continue Practicing

#1726 tuple with same product #1775 equal sum arrays with minimum number of operations #1782 count pairs of nodes