# leetcode\_5601

There are n (id, value) pairs, where id is an integer between 1 and n and value is a string. No two pairs have the same id.

Design a stream that takes the n pairs in an arbitrary order, and returns the values over several calls in increasing order of their ids.

Implement the OrderedStream class:

OrderedStream(int n) Constructs the stream to take n values and sets a current ptr to 1. String\[] insert(int id, String value) Stores the new (id, value) pair in the stream. After storing the pair: If the stream has stored a pair with id = ptr, then find the longest contiguous incrementing sequence of ids starting with id = ptr and return a list of the values associated with those ids in order. Then, update ptr to the last id + 1. Otherwise, return an empty list.

Example:

Input \["OrderedStream", "insert", "insert", "insert", "insert", "insert"] \[\[5], \[3, "ccccc"], \[1, "aaaaa"], \[2, "bbbbb"], \[5, "eeeee"], \[4, "ddddd"]] Output \[null, \[], \["aaaaa"], \["bbbbb", "ccccc"], \[], \["ddddd", "eeeee"]]

Explanation OrderedStream os= new OrderedStream(5); os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns \[]. os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns \["aaaaa"]. os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns \["bbbbb", "ccccc"]. os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns \[]. os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns \["ddddd", "eeeee"].

Constraints:

1 <= n <= 1000 1 <= id <= n value.length == 5 value consists only of lowercase letters. Each call to insert will have a unique id. Exactly n calls will be made to insert.

## Solutions

1. **treemap**

```cpp
class OrderedStream {
public:
    map<int, string> m;
    int ptr = 1;
    OrderedStream(int n) {

    }

    vector<string> insert(int id, string value) {
        m[id] = value;
        auto it = m.find(ptr);
        if (it == m.end()) return {};
        vector<string> res;
        while (it != m.end()) {
            res.push_back(it->second);
            auto nt = next(it);
            ptr++;
            if (nt == m.end() || nt->first != it->first + 1)
                break;
            it = nt;
        }
        return res;
    }
};

/**
 * Your OrderedStream object will be instantiated and called as such:
 * OrderedStream* obj = new OrderedStream(n);
 * vector<string> param_1 = obj->insert(id,value);
 */
```