How do you solve Sum of Unique Elements on LeetCode?

Sum of Unique Elements (LeetCode #1748) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient counting of elements.

What is the brute force solution for Sum of Unique Elements?

The brute force approach for Sum of Unique Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element against every other element to determine if it is unique. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Unique Elements?

Sum of Unique Elements uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Unique Elements?

Always consider the data structure that best fits the problem. Explain your thought process clearly while coding. Test your solution with edge cases.

What are common mistakes when solving Sum of Unique Elements?

Not using a hash map and relying on nested loops. Forgetting to check the frequency before summing.

#1748

Sum of Unique Elements

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using a hash map allows us to count occurrences of each element efficiently, enabling us to find the unique elements in a single pass.

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Algorithm

2 steps

1Step 1: Create a hash map to count the frequency of each element.
2Step 2: Iterate through the hash map and sum the keys that have a value of 1.

solution.py4 lines

1def sumOfUnique(nums):
2    from collections import Counter
3    count = Counter(nums)
4    return sum(num for num, freq in count.items() if freq == 1)

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Complexity note: The time complexity is O(n) because we traverse the array once to build the hash map and then iterate over the map, which is efficient. The space complexity is O(n) due to the storage of the hash map.

1Using a hash map allows for efficient counting of elements.
2Unique elements can be identified by their frequency being exactly one.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.