# [Solution] Design a Number Container System solution leetcode

Design a Number Container System solution leetcode – Design a number container system that can do the following:

• Insert or Replace a number at the given index in the system.
• Return the smallest index for the given number in the system.

## [Solution] Design a Number Container System solution leetcode

Implement the `NumberContainers` class:

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

Example 1:

```Input
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], , [2, 10], [1, 10], [3, 10], [5, 10], , [1, 20], ]
Output
[null, -1, null, null, null, null, 1, null, 2]```

## [Solution] Design a Number Container System solution leetcode

```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.
```

Constraints:

• `1 <= index, number <= 109`
• At most `105` calls will be made in total to `change` and `find`.