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 atindexwith thenumber. If there is already a number at thatindex, replace it.int find(int number)Returns the smallest index for the givennumber, or-1if there is no index that is filled bynumberin 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
105calls will be made in total tochangeandfind.
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