Design a Number Container System
<p>Design a number container system that can do the following:</p>

<ul>
	<li><strong>Insert </strong>or <strong>Replace</strong> a number at the given index in the system.</li>
	<li><strong>Return </strong>the smallest index for the given number in the system.</li>
</ul>

<p>Implement the <code>NumberContainers</code> class:</p>

<ul>
	<li><code>NumberContainers()</code> Initializes the number container system.</li>
	<li><code>void change(int index, int number)</code> Fills the container at <code>index</code> with the <code>number</code>. If there is already a number at that <code>index</code>, replace it.</li>
	<li><code>int find(int number)</code> Returns the smallest index for the given <code>number</code>, or <code>-1</code> if there is no index that is filled by <code>number</code> in the system.</li>
</ul>

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

<pre>
<strong>Input</strong>
[&quot;NumberContainers&quot;, &quot;find&quot;, &quot;change&quot;, &quot;change&quot;, &quot;change&quot;, &quot;change&quot;, &quot;find&quot;, &quot;change&quot;, &quot;find&quot;]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
<strong>Output</strong>
[null, -1, null, null, null, null, 1, null, 2]

<strong>Explanation</strong>
NumberContainers nc = new NumberContainers();
nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
nc.change(2, 10); // Your container at index 2 will be filled with number 10.
nc.change(1, 10); // Your container at index 1 will be filled with number 10.
nc.change(3, 10); // Your container at index 3 will be filled with number 10.
nc.change(5, 10); // Your container at index 5 will be filled with number 10.
nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. 
nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.
</pre>

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

<ul>
	<li><code>1 &lt;= index, number &lt;= 10<sup>9</sup></code></li>
	<li>At most <code>10<sup>5</sup></code> calls will be made <strong>in total</strong> to <code>change</code> and <code>find</code>.</li>
</ul>
Design; Hash Table; Ordered Set; Heap (Priority Queue)
use std::collections::{BTreeSet, HashMap};

struct NumberContainers {
    d: HashMap<i32, i32>,
    g: HashMap<i32, BTreeSet<i32>>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl NumberContainers {
    fn new() -> Self {
        Self {
            d: HashMap::new(),
            g: HashMap::new(),
        }
    }

    fn change(&mut self, index: i32, number: i32) {
        if let Some(&old_number) = self.d.get(&index) {
            if let Some(set) = self.g.get_mut(&old_number) {
                set.remove(&index);
            }
        }
        self.d.insert(index, number);
        self.g
            .entry(number)
            .or_insert_with(BTreeSet::new)
            .insert(index);
    }

    fn find(&self, number: i32) -> i32 {
        match self.g.get(&number) {
            Some(set) if !set.is_empty() => *set.iter().next().unwrap(),
            _ => -1,
        }
    }
}
