**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.

Table of Contents

## [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"] [[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]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 <= 10`

^{9}- At most
`10`

calls will be made^{5}**in total**to`change`

and`find`

.