LeetCode: Time Based Key-Value Store Posted on February 5, 2018July 26, 2020 by braindenny Time Based Key-Value Store Similar Problems: CheatSheet: Leetcode For Code Interview CheatSheet: Common Code Problems & Follow-ups Tag: #binarysearch Create a timebased key-value store class TimeMap, that supports two operations. set(string key, string value, int timestamp) Stores the key and value, along with the given timestamp. get(string key, int timestamp) Returns a value such that set(key, value, timestamp_prev) was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the one with the largest timestamp_prev. If there are no values, it returns the empty string (“”). Example 1: Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]] Output: [null,null,"bar","bar",null,"bar2","bar2"] Explanation: TimeMap kv; kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1 kv.get("foo", 1); // output "bar" kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar" kv.set("foo", "bar2", 4); kv.get("foo", 4); // output "bar2" kv.get("foo", 5); //output "bar2" Example 2: Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]] Output: [null,null,null,"","high","high","low","low"] Note: All key/value strings are lowercase. All key/value strings have length in the range [1, 100] The timestamps for all TimeMap.set operations are strictly increasing. 1 <= timestamp <= 10^7 TimeMap.set and TimeMap.get functions will be called a total of 120000 times (combined) per test case. Github: code.dennyzhang.com Credits To: leetcode.com Leave me comments, if you have better ways to solve. Solution: // https://code.dennyzhang.com/time-based-key-value-store // Basic Ideas: Binary search // // Complexity: Time O(log(n)), Space O(n) type MyNode struct { value string timestamp int } type TimeMap struct { m map[string][]MyNode } /** Initialize your data structure here. */ func Constructor() TimeMap { return TimeMap{make(map[string][]MyNode)} } func (this *TimeMap) Set(key string, value string, timestamp int) { // timestamps are strictly increasing this.m[key] = append(this.m[key], MyNode{value, timestamp}) } func (this *TimeMap) Get(key string, timestamp int) string { // assume key is valid l, _ := this.m[key] // corner case if l[0].timestamp > timestamp { return "" } // set high to len(l), instead of len(l)-1 low, high := 0, len(l) // binary search for low < high { mid := (high-low)/2 + low if l[mid].timestamp == timestamp { return l[mid].value } if l[mid].timestamp < timestamp { low = mid+1 } else { high = mid } } return l[low-1].value } /** * Your TimeMap object will be instantiated and called as such: * obj := Constructor(); * obj.Set(key,value,timestamp); * param_2 := obj.Get(key,timestamp); */ Post Views: 10