2349. Design a Number Container System

Problem

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.

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]

Explanation
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 10^5 calls will be made in total to change and find.

Solution (Java)

class NumberContainers {
    private Map<Integer, TreeSet<Integer>> indices = new HashMap<>();
    private Map<Integer, Integer> vals = new HashMap<>();

    public NumberContainers() {}

    public void change(int index, int number) {
        if (vals.containsKey(index)) {
            int old = vals.get(index);
            indices.get(old).remove(index);
            if (indices.get(old).isEmpty()) {
                indices.remove(old);
            }
        }
        vals.put(index, number);
        indices.computeIfAbsent(number, s -> new TreeSet<>()).add(index);
    }

    public int find(int number) {
        if (indices.containsKey(number)) {
            return indices.get(number).first();
        }
        return -1;
    }
}

/**
 * Your NumberContainers object will be instantiated and called as such:
 * NumberContainers obj = new NumberContainers();
 * obj.change(index,number);
 * int param_2 = obj.find(number);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).