Leetcode 981. Time Based Key-Value Store

  • 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 "".
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 "ba2r" along with timestamp = 4.
timeMap.get("foo", 4); // return "bar2"
timeMap.get("foo", 5); // return "bar2"

Solution

  1. Don’t use unordered_multimap.
  • we should not keep equivalent keys
  • multimap does’t support [] operation. So inconvenient!
  • key needs to be associative
  • timestamp needs to be ordered for a time-bound search
class TimeMap {  
unordered_map<string, map<int, string>> hash;
// key : [ timestamp: value]

public:
/** Initialize your data structure here. */
TimeMap() {

}

void set(string key, string value, int timestamp) {
hash[key][timestamp] = value;
}

string get(string key, int timestamp) {
if (hash.find(key) != hash.end()) {
map<int, string>& m = hash.find(key)->second;
map<int, string>::iterator itr
= m.upper_bound(timestamp);
if (itr == m.begin()) {
return "";
} else {
return prev(itr)->second;
}
}
return "";
}
};

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