＃981 Time Based Key-Value Store

Description

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 “”.

Examples

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 * 105 calls will be made to set and get.

Ideas

Is to use a binary search idea , Because in constrain It says “All the timestamps timestamp of set are strictly increasing.”, So I set up an extra one directly list To store timestamp As the basis for the following binary search （set Is chaotic ）

In addition, there is a precondition for binary search ： To find no more than target One of the biggest timestamp, Therefore, the judgment of termination conditions is different from the ordinary binary search （ another , A non recursive solution is attached , By specifying a point as answer Assignment solves this problem ）

Code

class TimeMap {
    
    public Map<String, Map<Integer, String>> dict;
    public Map<String, List<Integer>> stampMap; 
    
    public TimeMap() {
    
        dict = new HashMap<>();
        stampMap = new HashMap<>();
    }
    
    public int findNearestTimestamp(List<Integer> timestamps, int start, int end, int target) {
    
        int middle = (start + end) / 2;
        if (end <= start){
    
            if (timestamps.get(end) > target)
                return timestamps.get(end - 1);
            return timestamps.get(end);
        }
        
        if (timestamps.get(middle) == target)
            return target;
        
        if (timestamps.get(middle) > target) 
            return findNearestTimestamp(timestamps, start, middle - 1, target);
        else
            return findNearestTimestamp(timestamps, middle + 1, end, target);
    }
    
    public void set(String key, String value, int timestamp) {
    
        if (dict.containsKey(key)) {
    
            Map<Integer, String> inside = dict.get(key);
            inside.put(timestamp, value);
            dict.put(key, inside);
            
            List<Integer> stampInside = stampMap.get(key);
            stampInside.add(timestamp);
            stampMap.put(key, stampInside);
        }
        else {
    
            Map<Integer, String> newInside = new HashMap<>();
            newInside.put(timestamp, value);
            dict.put(key, newInside);
            
            List<Integer> stampInside = new ArrayList<>();
            stampInside.add(timestamp);
            stampMap.put(key, stampInside);
        }
    }
    
    public String get(String key, int timestamp) {
    
        if (!dict.containsKey(key))
            return "";
        List<Integer> timestamps = stampMap.get(key);
        if (timestamp < timestamps.get(0))
            return "";
        int ts = findNearestTimestamp(timestamps, 0, timestamps.size() - 1, timestamp);
        return dict.get(key).getOrDefault(ts, "");
    }
}

Non recursive binary search

public int findNearestTimestamp(List<Integer> timestamps, int target) {
    
        int start = 0;
        int end = timestamps.size() - 1;
        
        int answer = -1;
        
        while(start <= end) {
    
            int middle = (start + end) / 2;
            if (timestamps.get(middle) <= target) {
    
                answer = timestamps.get(middle);
                start = middle + 1;
            }
            
            else 
                end = middle - 1;
        }
        
        return answer;
    }