Sum of Unique Elements
<p>You are given an integer array <code>nums</code>. The unique elements of an array are the elements that appear <strong>exactly once</strong> in the array.</p>

<p>Return <em>the <strong>sum</strong> of all the unique elements of </em><code>nums</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The unique elements are [1,3], and the sum is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,1,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no unique elements, and the sum is 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 15
<strong>Explanation:</strong> The unique elements are [1,2,3,4,5], and the sum is 15.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>
Array; Hash Table; Counting
impl Solution {
    pub fn sum_of_unique(nums: Vec<i32>) -> i32 {
        let mut cnt = [0; 101];
        for x in nums {
            cnt[x as usize] += 1;
        }
        let mut ans = 0;
        for x in 1..101 {
            if cnt[x] == 1 {
                ans += x;
            }
        }
        ans as i32
    }
}
