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+// https://leetcode.com/problems/sum-of-unique-elements/
+
+//Test  cases
+
+// Example 1:
+// Input: nums = [1,2,3,2]
+// Output: 4
+// Explanation: The unique elements are [1,3], and the sum is 4.
+
+// Example 2:
+// Input: nums = [1,1,1,1,1]
+// Output: 0
+// Explanation: There are no unique elements, and the sum is 0.
+
+// Example 3:
+// Input: nums = [1,2,3,4,5]
+// Output: 15
+// Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.
+
+// Approach
+/*
+We will create an array of 101 elements, all sets to 0 and proceed to store the frequency there.
+
+Despite looping through 100 elements at each run to update result in the second loop.
+*/
+
+// Code
+class Solution
+{
+public:
+    int sumOfUnique(vector<int> &nums)
+    {
+        // support variables
+        int result = 0, arr[101] = {};
+        // parsing nums
+        for (int n : nums)
+        {
+            arr[n]++;
+        }
+        // updating result
+        for (int i = 1; i < 101; i++)
+        {
+            result += arr[i] == 1 ? i : 0;
+        }
+        return result;
+    }
+};