Program to find sum of unique elements in Python

Suppose we have an array nums with few duplicate elements and some unique elements. We have to find the sum of all the unique elements present in nums.

So, if the input is like nums = [5,2,1,5,3,1,3,8], then the output will be 10 because only unique elements are 8 and 2, so their sum is 10.

Approach

To solve this, we will follow these steps −

• count := a dictionary holding all unique elements and their frequency

• ans := 0

• for each index i and value v in nums, do

  • if count[v] is same as 1, then

    • ans := ans + v

• return ans

Using Counter from Collections

Let us see the following implementation to get better understanding −

from collections import Counter

def solve(nums):
    count = Counter(nums)
    ans = 0
    for index, value in enumerate(nums):
        if count[value] == 1:
            ans += value
    return ans

nums = [5, 2, 1, 5, 3, 1, 3, 8]
print(solve(nums))

The output of the above code is −

10

Using Dictionary Approach

We can also solve this without using Counter by manually counting frequencies −

def solve(nums):
    # Count frequency of each element
    count = {}
    for num in nums:
        count[num] = count.get(num, 0) + 1
    
    # Sum elements that appear only once
    ans = 0
    for num, freq in count.items():
        if freq == 1:
            ans += num
    
    return ans

nums = [5, 2, 1, 5, 3, 1, 3, 8]
print(solve(nums))

The output of the above code is −

10

Optimized One-Pass Solution

Here's a more efficient approach that processes the array in a single pass −

def solve(nums):
    from collections import defaultdict
    
    count = defaultdict(int)
    unique_sum = 0
    
    for num in nums:
        if count[num] == 0:
            # First occurrence - add to sum
            unique_sum += num
        elif count[num] == 1:
            # Second occurrence - remove from sum
            unique_sum -= num
        # For count > 1, do nothing
        
        count[num] += 1
    
    return unique_sum

nums = [5, 2, 1, 5, 3, 1, 3, 8]
print(solve(nums))

The output of the above code is −

10

Comparison

Conclusion

Use Counter for clean readable code when working with frequency counts. The one-pass solution is most efficient for large datasets as it processes elements in a single iteration.