Kth User to Visit Website- Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers requests and an integer k.
Each element requests[i] contains [start, end] representing an inclusive interval for the possible times when user i visited the website.
Each user visits the site exactly once.
Return the users who may have been the kth (0-indexed) user to visit the site, sorted in ascending order. You can assume that the site can serve an unlimited number of users at any time.
Constraints
1 ≤ n ≤ 100,000 where n is the length of requests
0 ≤ k < n
Example 1
Input
requests = [
[3, 4],
[1, 3],
[4, 4]
]
k = 1
Output
[0, 1, 2]
Explanation
If users visited the site at timestamps [3, 3, 4], then user 0 and 1 may have been the kth user. If they
visited the site at [4, 3, 4], then user 0 and 2 may have been the kth user.
Example 2
Input
requests = [
[0, 0],
[0, 0],
[1, 1],
[2, 2],
[2, 2]
]
k = 3
Output
[3, 4]
Explanation
The only possible timestamps are [0, 0, 1, 2, 2]. Only the last two users could've been the 3rd user to
visit the site.
Example 3
Input
requests = [
[60, 70],
[0, 10],
[11, 30],
[22, 40]
]
k = 1
Output
[2, 3]
Explanation
If users visited the site at these timestamps [60, 5, 21, 22], then user 2 would've been the kth user. If they visited the site at [60, 5, 23, 22], then 3 would've been the kth user.
Solution :
Solution in C++ :
bool cmo(vector<int> &a, vector<int> &b) {
return a[1] < b[1];
}
vector<int> solve(vector<vector<int>> &re, int k) {
for (int i = 0; i < re.size(); i++) {
re[i].push_back(i);
}
sort(re.begin(), re.end());
int tmin = re[k][0];
sort(re.begin(), re.end(), cmo);
int tmax = re[k][1];
// cout<<tmin<<" "<<tmax<<endl;
vector<int> res;
for (auto &t : re) {
if (t[0] <= tmax && t[1] >= tmin) {
res.push_back(t[2]);
}
}
sort(begin(res), end(res));
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[][] requests, int k) {
long l = 0, h = Integer.MAX_VALUE;
while (l <= h) {
long m = (l + h) >>> 1;
if (canByEnd(requests, m, k + 1))
h = m - 1;
else
l = m + 1;
}
int end = (int) l;
l = 0;
h = Integer.MAX_VALUE;
while (l <= h) {
long m = (l + h) >>> 1;
if (canByStart(requests, m, k + 1))
h = m - 1;
else
l = m + 1;
}
int start = (int) l;
List<Integer> poss = new ArrayList<>();
for (int i = 0; i != requests.length; i++) {
int[] r = requests[i];
if (r[0] > end || r[1] < start)
continue;
// if ((r[0] <= start && r[1] >= start) || (r[0] <= end && r[1] >= end))
poss.add(i);
}
Collections.sort(poss);
int[] res = new int[poss.size()];
for (int i = 0; i != poss.size(); i++) res[i] = poss.get(i);
return res;
}
boolean canByStart(int[][] reqs, long time, final int K) {
int count = 0;
for (int[] r : reqs)
if (r[0] <= time)
count++;
return count >= K;
}
boolean canByEnd(int[][] reqs, long time, final int K) {
int count = 0;
for (int[] r : reqs) {
if (r[1] <= time)
count++;
}
return count >= K;
}
}
Solution in Python :
class Solution:
def solve(self, requests, k):
# calculate the earliest and latest visit times for user k.
# return all requests that overlap with this interval.
ks = sorted(s for s, e in requests)[k]
ke = sorted(e for s, e in requests)[k]
return [i for i, (s, e) in enumerate(requests) if not (ke < s or e < ks)]
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