Time Based Key-Value Store

Time Based Key-Value Store: Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key's value at a certain timestamp.

Implement the TimeMap class:

TimeMap() Initializes the object of the data structure.
void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp.
String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns "".

Example 1:

Input
["TimeMap", "set", "get", "get", "set", "get", "get"]
[[], ["foo", "bar", 1], ["foo", 1], ["foo", 3],
["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
Output
[null, null, "bar", "bar", null, "bar2", "bar2"]

Explanation
TimeMap timeMap = new TimeMap();
timeMap.set("foo", "bar", 1);
// store the key "foo" and value "bar" along with timestamp = 1.
timeMap.get("foo", 1);         // return "bar"
timeMap.get("foo", 3);
// return "bar", since there is no value corresponding to foo at
timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar".
timeMap.set("foo", "bar2", 4);
// store the key "foo" and value "bar2" along with timestamp = 4.
timeMap.get("foo", 4);         // return "bar2"
timeMap.get("foo", 5);         // return "bar2"

Constraints:

1 <= key.length, value.length <= 100
key and value consist of lowercase English letters and digits.
1 <= timestamp <= 10^7
All the timestamps timestamp of set are strictly increasing.
At most 2 * 10^5 calls will be made to set and get.

Try this Problem on your own or check similar problems:

Stock Price Fluctuation

Solution:

Java
JavaScript
Python
C++

class TimeMap {
    private class Pair{
        String value;
        int timestamp;
        public Pair(String value, int timestamp){
            this.value = value;
            this.timestamp = timestamp;
        }
    }
    Map<String, List<Pair>> map;
    public TimeMap() {
        map = new HashMap<>();
    }

    public void set(String key, String value, int timestamp) {
        map.computeIfAbsent(key, k -> new ArrayList<Pair>()).add(new Pair(value, timestamp));
    }

    public String get(String key, int timestamp) {
        if(!map.containsKey(key)) return "";

        int idx = Collections.binarySearch(map.get(key), new Pair("", timestamp), (a, b) -> a.timestamp - b.timestamp);

        if(idx >= 0) return map.get(key).get(idx).value;
        else if(idx == -1) return "";
        return map.get(key).get(~idx-1).value;
    }
}

/**
 * Your TimeMap object will be instantiated and called as such:
 * TimeMap obj = new TimeMap();
 * obj.set(key,value,timestamp);
 * String param_2 = obj.get(key,timestamp);
 */

var TimeMap = function () {
  this.map = new Map();
};

/**
 * @param {string} key
 * @param {string} value
 * @param {number} timestamp
 * @return {void}
 */
TimeMap.prototype.set = function (key, value, timestamp) {
  const values = this.map.get(key) ?? [];

  values.push({ value, timestamp });
  this.map.set(key, values);
};

/**
 * @param {string} key
 * @param {number} timestamp
 * @return {string}
 */
TimeMap.prototype.get = function (key, timestamp) {
  const values = this.map.get(key);
  if (!values) return "";
  if (values[0].timestamp > timestamp) return "";
  let left = 0;
  let right = values.length - 1;

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    const current = values[mid];

    if (current.timestamp === timestamp) return current.value;
    current.timestamp > timestamp ? (right = mid - 1) : (left = mid + 1);
  }
  return values[right]?.value ?? "";
};

/**
 * Your TimeMap object will be instantiated and called as such:
 * var obj = new TimeMap()
 * obj.set(key,value,timestamp)
 * var param_2 = obj.get(key,timestamp)
 */

class TimeMap:

    def __init__(self):
        self.data = defaultdict(list)


    def set(self, key: str, value: str, timestamp: int) -> None:
        self.data[key].append((timestamp, value))


    def get(self, key: str, timestamp: int) -> str:
        if key not in self.data:
            return ""

        values = self.data[key]
        idx = bisect.bisect(values, timestamp, key=lambda x: x[0])

        if idx == 0:
            return ""

        return values[idx - 1][1]



# Your TimeMap object will be instantiated and called as such:
# obj = TimeMap()
# obj.set(key,value,timestamp)
# param_2 = obj.get(key,timestamp)

class TimeMap {
private:
    struct Pair {
        string value;
        int timestamp;

        Pair(const string& value, int timestamp) : value(value), timestamp(timestamp) {}
    };
    unordered_map<string, vector<Pair>> map;
public:
    TimeMap() {

    }

    void set(string key, string value, int timestamp) {
        map[key].emplace_back(value, timestamp);
    }

    string get(string key, int timestamp) {
        if (map.count(key) == 0) {
            return "";
        }

        const vector<Pair>& pairs = map[key];

        auto it = upper_bound(pairs.begin(), pairs.end(), timestamp, [](int ts, const Pair& p) {
            return ts < p.timestamp;
        });

        if (it == pairs.begin()) {
            return "";
        }

        return prev(it)->value;
    }
};

/**
 * Your TimeMap object will be instantiated and called as such:
 * TimeMap* obj = new TimeMap();
 * obj->set(key,value,timestamp);
 * string param_2 = obj->get(key,timestamp);
 */

Time/Space Complexity:

Time Complexity: for get O(x(n + logm)) and O(yn) for set
Space Complexity: O(kmn)

where x is number of calls to function get, y is the number of calls to set function, n is the average string length (value and key), m is the average number of inserted elements for keys in map and k is the number of keys. Note that we spend n to hash the string both on get and set.

Explanation:

We should first note that timestamps are strictly increasing, which means their records are sorted on insertion. Since we can have multiple values for one key, we need to create a HashMap that will map the key to the list of values. We also decided to store each insertion as pair of value and timestamp. set function is simple since it only checks if we already have the key in the map, and if not, it creates a new list for that key and adds the value to it. get is a bit tricker but we use the ordered nature of the list for each of the keys and perform a binary search finding the closest (largest smaller or equal to the requested one) timestamp and returns the value for it. To understand how binary search works, you can check out Binary Search. A real time example where this algorithm could be handy is in snapshot isolation when on read, we get the most recent snapshot of data in database that has timestamp closest (smaller or equal) to that of the read operation.

Example 1:​

Constraints:​

Solution:​

Time/Space Complexity:​

Explanation:​

Example 1:

Constraints:

Solution:

Time/Space Complexity:

Explanation: