resource Access with in 5 min
http://202.79.172.224/id/880895
给了 list of <usr_id, resource_id, time_stamp>
问每个 resource 被访问了多少
follow up 返回那个5分钟内被访问最多的resource id 和次数
--请问 follow-up， 是把每个resource的ts先sort, 然后for loop 每个ts, 利用binary search找到五分钟内的最大index，然后求得此时的个数, 最后依次update max吗？
--用 queue 然后每次append的时候 check head 根据情况pop
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有一些资源的使用记录 比如 [{"ts": xxx, "user": xxx, "resource": xxx}. {"ts": xxx, "user": xxx, "resource": xxx}. {"ts": xxx, "user": xxx, "resource": xxx}. ] 然后目标是找到每一个resource 一定时间长度内 最多被使用了多少次 比如 我们有 r1: 10 11 12 20 25 r2: 5 10 15 20 21 22 然后让找5个ts内的 那么 r1: 3 (10-11-12) r2: 3 (20-21-22)
第二题同样的data 要返回 一个图一样的结构，大概是 reource： 到可能别的resource的可能性,这个是根据user来算的，基本就是一个反向bfs这样，时间不够了没做出来
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输入是一串<时间(秒)，用户id，资源id>，返回时间窗口为5分钟的被访问次数最多的资源。题目不难，但是有两个坑：输入是没有按照时间排序的；五分钟的定义是：0-300 秒，NOT 0-299秒。我掉到第一个坑里，有个测试案例没通过，时间结束了才意识到。
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public class Solution {
public static void main(String[] args) {
String[][] logs1 = new String[][] {
{ "58523", "user_1", "resource_1" },
{ "62314", "user_2", "resource_2" },
{ "54001", "user_1", "resource_3" },
{ "200", "user_6", "resource_5" },
{ "215", "user_6", "resource_4" },
{ "54060", "user_2", "resource_3" },
{ "53760", "user_3", "resource_3" },
{ "58522", "user_22", "resource_1" },
{ "53651", "user_5", "resource_3" },
{ "2", "user_6", "resource_1" },
{ "100", "user_6", "resource_6" },
{ "400", "user_7", "resource_2" },
{ "100", "user_8", "resource_6" },
{ "54359", "user_1", "resource_3"},
};
String[][] logs2 = new String[][] {
{"300", "user_1", "resource_3"},
{"599", "user_1", "resource_3"},
{"900", "user_1", "resource_3"},
{"1199", "user_1", "resource_3"},
{"1200", "user_1", "resource_3"},
{"1201", "user_1", "resource_3"},
{"1202", "user_1", "resource_3"}
};
String[][] logs3 = new String[][] {
{"300", "user_10", "resource_5"}
};
System.out.println(mostRequestedResource(logs1));
System.out.println(mostRequestedResource(logs2));
System.out.println(mostRequestedResource(logs3));
}
private static List<String> mostRequestedResource(String[][] logs) {
Map<String,ArrayList<Integer>> accessMap = new HashMap<>();
for (String[] entry : logs) {
Integer accessTime = Integer.parseInt(entry[0]);
String resourceId = entry[2];
if (accessMap.containsKey(resourceId)) {
var exList = accessMap.get(resourceId);
exList.add(accessTime);
accessMap.put(resourceId, exList);
} else {
ArrayList<Integer> newList = new ArrayList<>();
newList.add(accessTime);
accessMap.put(resourceId, newList);
}
}
String maxResource = "";
Integer maxWindowTime = 0;
for (var entry : accessMap.entrySet()) {
Integer entryTime = windowCount(entry.getValue(), 300);
if (entryTime > maxWindowTime) {
maxResource = entry.getKey();
maxWindowTime = entryTime;
}
}
List<String> result = new ArrayList<>();
result.add(maxResource);
result.add(maxWindowTime.toString());
return result;
}
private static Integer windowCount(ArrayList<Integer> accessTimes, Integer window) {
Collections.sort(accessTimes);
int maxWindowTime = 0;
for (Integer testVal : accessTimes) {
List<Integer> timesInWindow = accessTimes.stream()
.filter(val -> val >= testVal)
.filter(val -> val <= (testVal + window)).toList();
int testWindowTime = timesInWindow.size();
maxWindowTime = Integer.max(maxWindowTime, testWindowTime);
}
return maxWindowTime;
}Version 2:
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