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2434. Design A Number Container System

Difficulty: Medium

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2434. Design a Number Container System

Medium


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 <= 109
  • At most 105 calls will be made in total to change and find.

Solution

class NumberContainers {
    private HashMap<Integer, TreeSet<Integer>> map;
    private HashMap<Integer, Integer> info;
    public NumberContainers() {
        map = new HashMap<>();
        info = new HashMap<>();
    }
    public void change(int index, int number) {
        if (info.containsKey(index)) {
            int prev_number = info.get(index);
            TreeSet<Integer> prev = map.get(prev_number);
            prev.remove(index);
            if (!map.containsKey(number)) map.put(number, new TreeSet<>());
            map.get(number).add(index);
            info.put(index, number);
        }
        else {
            if (!map.containsKey(number)) map.put(number, new TreeSet<>());
            map.get(number).add(index);
            info.put(index, number);
        }
    }
    public int find(int number) {
        if (map.containsKey(number)) {
            TreeSet<Integer> temp = new TreeSet<>();
            temp = map.get(number);
            if (temp.size() == 0) return -1;
            return temp.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);
 */

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here