Sum of Unique Elements

Sum of Unique Elements - Problem

You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.

Return the sum of all the unique elements of nums.

Input & Output

Example 1 — Basic Case

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

› Output: 4

💡 Note: Elements 1 and 3 appear exactly once, so sum = 1 + 3 = 4. Element 2 appears twice so it's not unique.

Example 2 — All Unique

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

› Output: 15

💡 Note: All elements appear exactly once, so sum = 1 + 2 + 3 + 4 + 5 = 15

Example 3 — No Unique Elements

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

› Output: 0

💡 Note: Every element appears exactly twice, so no element is unique. Sum = 0

Constraints

1 ≤ nums.length ≤ 100
-100 ≤ nums[i] ≤ 100

Visualization

Tap to expand

Asked in

a Amazon 35 G Google 28

The key insight is to identify elements that appear exactly once and sum them up. The best approach uses a hash map to count frequencies in O(n) time, or the clever one-pass method that adds on first occurrence and subtracts on duplicates. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each element, scan the entire array to count its occurrences. If it appears exactly once, add it to the sum.

Frequency Count (Hash Map)

⏱️ Time: O(n) Space: O(n)

First pass through the array to count frequency of each element using a hash map. Second pass to sum elements that appear exactly once.

One-Pass Optimized

⏱️ Time: O(n) Space: O(n)

Use a hash map to track seen elements and their contribution to sum. When we see an element for the first time, add it to sum. When we see it again, subtract it from sum.

Brute Force — Algorithm Steps

For each element in the array
Count how many times it appears by checking against all other elements
If count is exactly 1, add to sum

Visualization

Tap to expand

Step-by-Step Walkthrough

Check Element

For each element, count occurrences

Count Frequency

Compare against all elements

Sum Unique

Add elements with count = 1

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int totalSum = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int count = 0;
        // Count occurrences of nums[i]
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == nums[i]) {
                count++;
            }
        }
        
        // If appears exactly once, add to sum
        if (count == 1) {
            totalSum += nums[i];
        }
    }
    
    return totalSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array
    int nums[1000];
    int size = 0;
    char* ptr = line + 1; // Skip opening bracket
    
    while (*ptr && *ptr != ']') {
        if (*ptr == ',' || *ptr == ' ') {
            ptr++;
            continue;
        }
        nums[size++] = atoi(ptr);
        while (*ptr && *ptr != ',' && *ptr != ']') ptr++;
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - for each of n elements, we check against all n elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for sum and counters

✓ Linear Space

89.2K Views

Medium Frequency

~10 min Avg. Time

1.5K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler