Minimum Operations to Collect Elements
<p>You are given an array <code>nums</code> of positive integers and an integer <code>k</code>.</p>

<p>In one operation, you can remove the last element of the array and add it to your collection.</p>

<p>Return <em>the <strong>minimum number of operations</strong> needed to collect elements</em> <code>1, 2, ..., k</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,1,5,4,2], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,5,4,2], k = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong> After 5 operations, we collect elements 2, 4, 5, 1, and 3, in this order. Our collection contains elements 1 through 5. Hence, the answer is 5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,5,3,1], k = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> After 4 operations, we collect elements 1, 3, 5, and 2, in this order. Our collection contains elements 1 through 3. Hence, the answer is 4.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
	<li>The input is generated such that you can collect elements <code>1, 2, ..., k</code>.</li>
</ul>
Bit Manipulation; Array; Hash Table
class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        int n = nums.size();
        vector<bool> isAdded(n);
        int count = 0;
        for (int i = n - 1;; --i) {
            if (nums[i] > k || isAdded[nums[i] - 1]) {
                continue;
            }
            isAdded[nums[i] - 1] = true;
            if (++count == k) {
                return n - i;
            }
        }
    }
};