Lets say i have an array of birth and death years of 100 people between the years 1900 - 2000.
ergo:
persons[0][birth] = 1900;
persons[0][death] = 1960;
.
.
.
persons[99][birth] = 1904;
persons[99][death] = 1982;
And i would want to find the year that most people lived in between those ranges.
The obvious way is by going through each person, calculate the years he lived (death - birth), and adding these years to an accumulating array using an inside loop. Runtime for that would be O(n^2).
Is there a more efficient way to do this?
The O notation may be a bit misleading sometimes. You do not have only one variable. Number of people is n. Length of life let be y, the whole interval Y. Precisely evaluated Your solution is not O(n^2), but O(MAX(n*y, Y)). If You say, that Y is 100 and always will be, Your solution is O(n), because y<=Y and they are both eliminated as constants. You do twice as much work with twice as much people, so it really is linear, just the constants are high. Marking it O(n^2) strictly speaking means, that length of life is proportional to number of people, that is probably nonsense. Let say, from now on, that y is equal to Y. Now Your solution simplifies to O(n*y).
You can improve Your algorithm. Store +1 for birth and -1 for death into the accumulating array. Only two values for every person. Now You can go through that array once and get number of people living in every year. The time would be O(MAX(n, y)). It is better than solution of #Aziuth when n*log(n) grows faster than y, but worse otherwise.
Create an array that stores a pair of data ([birth or death], year). Every person will create two sets of such data. This takes O(n).
Sort this array by year. This takes O(nlogn).
Process through that array and keep the current amount of people alive in a variable, decreasing and increasing accordingly. Store the maximal amount to that point. This takes again O(n).
Example:
A: born 1904, died 1960. B: born 1930, died 1960. C: born 1940, died 1950.
-> array of deaths and births starts as (birth, 1904), (death, 1960), ...
-> array is sorted: (birth, 1904), (birth, 1930), (birth 1940), (death, 1950), (death, 1960), (death, 1960)
-> processing array: At 1904, one person lives. At 1930, two persons live. At...
Related
You stand in an office by a door, with a measuring tape. Every time a person walks in you measure him or her and only keep tally of the “record” tallest. If the new person is taller than the preceding one, you count a record. If later another person is taller, you have another record, etc.
A 1000 persons pass through the door. How many records do you expect to have?
(Assume independence of height/arrival. Also note that the answer does not depend on any assumption about the probability distribution other than independence.)
PS - I'm able to come up with answer (~7.5) with a brute force approach. ( Running this scenario over 1000000 times and taking average ). But here I'm looking for a theoretical approach.
consider x_1 to x_1000 as the record, and max(i) as max of the sequence until i. The question is reduced to finding expected number of times the max(i) changes.
for i=0 to 999:
if x_i+1>max(i), then max(i) changes
Also, P(x_i+1>max(i))=1/i+1
answer=> summation of 1/1+i (i varies from 0 to 999) which is approx. 7.49
I have two sequences of intervals.
The first is fixed and non-overlapping, so something like:
[1..10], [12..15], [23..56], [72..89], ...
The second is not fixed, so it's just an ordered list of interval lengths:
[7, 2, 5, 26, ...]
The task at hand is to:
Place every interval from the second list at a given starting point, so that the second list becomes a list of fixed, non-overlapping intervals much like the first, while preserving its order
Find the alignment that minimizes the amount of integers that are in some interval from one of the lists but not in any interval from the other list
Very simple example:
[25..26], [58..68], [74..76], [78..86]
[10, 12]
The optimal solution is to place the interval of length 10 at [58..68] and the interval of length 12 at [74..86] which results in only the numbers 25, 26, and 77 being in one list but not the other.
The only thing I've come up with that seems mildly helpful is that if I lay down the intervals in order, I know how many 'penalties' the interval I've already created, so I have an upper bound for the score, which means I have an admissible heuristic and I can do A* search instead of looking at the entire tree. However, the total range of numbers spans from 0 to about 34M, so I'd like something better.
Any help would be hot!
OK, here's a half-thought-out answer. It should work in polynomial time, but I haven't bothered checking what the index is. It may well be possible to get a better index than the answer as outlined here. The details are left as an exercise to the reader :-) I hope it's not too unclear.
I'll define the score of a solution as the number of integers which appear in both lists of intervals. Let f(i,m) be the highest score it's possible to get using just the first i interval lengths, subject to the condition that none of your intervals goes above m. The function f, for fixed i, is essentially a (non-strictly) increasing function from the integers to a bounded subset of the integers. Therefore:
all values of f(i,m), for m > 0, are equal, with finitely many exceptions;
all values of f(i,m), for m < 0, are equal, with finitely many exceptions.
This means it's possible to represent all values of f(i,m) using a finite data structure (still considering a fixed value of i).
Now let F(i) be the value of this data structure representing all values of f(i,m). I claim that, given F(i), it is possible to calculate F(i+1). To do this, we only need to answer the following question for all x: If I place the new interval at x, how good is the best solution I can get? But we know what this is - it's just f(i,x) + the score we've got from this interval.
So if n is the number of intervals in the second list, the score of the best solution will be F(n).
To actually find the solution, you could work backwards from this.
You know what's the best score you can get. Say it's s_0. Then put the last interval as far left as possible, subject to the condition that it allows you to score s_0. That is, find the smallest m such that f(n,m) = s_0; and place the interval such that it only just stays inside the bound at m.
Then, let s_1 be the score you need to get from all the other intervals in order to get a total of s_0. Place the next-last interval as far left as possible, subject to the condition that you can still score s_1. That is, find the smallest m such that f(n,m) = s_1; and place the interval such that it only just stays inside the bound at m.
And so on...
this is a puzzle but i think it could be a classical algorithm which i am unaware of :
There are n people at the bottom of a mountain, and everyone wants to go up, then down the mountain. Person i takes u[i] time to climb this mountain, and d[i] time to descend it.
However, at same given time atmost 1 person can climb , and .atmost 1 person can descend the mountain. Find the least time to travel up and back down the mountain.
Update 1 :
well i tried with few examples and found that it's not reducible to sorting , or getting the fastest climbers first or vice versa . I think to get optimal solution we may have to try out all possible solutions , so seems to be NP complete.
My initial guess: (WRONG)
The solution i thought is greedy : sort n people by start time in ascending order. Then up jth person up and kth down where u[j]<= d[k] and d[k] is minimum from all k persons on top of mountain. I am not able to prove correctness of this .
Any other idea how to approach ?
A hint would suffice.
Try to think in the following manner: if the people are not sorted in ascending order of time it takes them to climb the mountain than what happens if you find a pair of adjacent people that are not in the correct order(i.e. first one climbs longer than second one) and swap them. Is it possible that the total time increases?
I think it is incorrect. Consider
u = [2,3]
d = [1,3]
Your algorithm gives ordering 0,1 whereas it should be 1,0.
I would suggest another greedy approach:
Create ordering list and add first person.
For current ordering keep track of two values:
mU - time of last person on the mountain - time of the end
mD - time of earliest time of first descending
From people who are not ordered choose the one which minimises abs(mD - d) and abs(mU - u). Then if abs(mD - d) < abs(mU - u) he should go at the beginning of ordering. Otherwise he goes at the end.
Some tweak may still be needed here, but this approach should minimise losses from cases like the one given in the example.
The following solution will only work with n <= 24.
This solution will require dynamic programming and bit-mask technique knowledge to be understood.
Observation: we can easily observe that the optimal total climb up time is fixed, which is equalled to the total climb up time of n people.
For the base case, if n = 1, the solution is obvious.
For n = 2, the solution is simple, just scan through all 4 possibilities and calculate the minimum down time.
For n = 3, we can see that this case will be equal to the case when one person climb up first, followed by two.
And the two person minimum down time can be easily pre-calculated. More important, this two person then can be treated as one person with up time is the total up time of the two, and down time is the minimum down time.
Storing all result for minimum down time for cases from n = 0 to n = 3 in array called 'dp', using bit-mask technique, we represent the state for 3 person as index 3 = 111b, so the result for case n = 3 will be:
for(int i = 0; i < 3; i++){
dp[3] = min(dp[(1<<i)] + dp[3^(1<<i)],dp[3]);
}
For n = 4... 24, the solution will be similar to case n = 3.
Note: The actual formula is not just simple as the code for case n = 3(and it requires similar approach to solve as case n = 2), but will be very similar,
Your approach looks sensible, but it may be over-simplified, could you describe it more precisely here?
From your description, I can't make out whether you are sorting or something else; these are the heuristics that I figured you are using:
Get the fastest climbers first, so the start using the Down path
asap.
Ensure there is always people at the top of the mountain, so
when the Down path becomes available, a person starts descending
immediately.The way you do that is to select first those people who
climb fast and descend slowly.
What if the fastest climber is also the fastest descender? That would leave the Down path idle until the second climber gets to the top, how does your algorithm ensures that this the best order?. I'm not sure that the problem reduces to a Sorting problem, it looks more like a knapsack or scheduling type.
We have a list of intervals of the form [ai, bi]. For each interval, we want to count the number of other intervals that are nested within it.
For example, if we had two intervals, A = [1,4] and B = [2,3]. Then the count for B would be 0 as there are no nested intervals for B; and the count for A would be 1 as B fits within A.
My question is, does there exist a sub- O(n2) algorithm for this problem where n is the number of intervals?
EDIT: Here are the conditions the intervals meet. The end points of the intervals are floating point numbers. The lower limit for the ai's/bi's is 0 and the upper limit is whatever max float is. Also, there is the condition that ai < bi, so no intervals of length 0.
Yes, it is possible.
We will borrow the typical computational geometry "scan line" trick.
First, let's answer an easier (but closely related) question. Instead of reporting how many other intervals each interval contains, let's report how many intervals each is contained in. So for your example with only two intervals, interval I0 = [1,4] has value zero because it is contained in zero intervals, while I1 = [2,3] has value one because it is contained in one interval.
You will see in a minute (a) why this question is easier and (b) how it leads to the answer for the original question.
To solve this easier question: Take all starting and ending points -- all of the ai and bi -- and put them into a master list. Call each element of this list an "event". So an event would be something like "interval I37 started" or "interval I23 ended".
Sort this list of events and process it in order.
As you process the list of events, maintain a set S of "active intervals". An interval is "active" if we have encountered its start event but not its ending event; that is, if we are within that interval.
Now, whenever we see an ending event bj, we are ready to compute how many intervals contain Ij (= [aj, bj]). All we need to do is examine the set S of active intervals and determine how many of them started before aj. That is our answer for how many intervals contain interval Ij.
To do this efficiently, keep S itself sorted by starting point; e.g., by using a self-balancing binary tree.
Sorting the list of events is O(2n log 2n) = O(n log n). Adding or removing an element from a self-balancing binary tree is O(log n). Asking "how many elements of the self-balancing binary tree are less than x?" is also O(log n). Therefore this entire algorithm is O(n log n).
So, that solves the easy question. Call that the "easy algorithm". Now for what you actually asked.
Think of the number line as extending to infinity and wrapping around to -infinity, and define an interval with bi < ai to start at ai, stretch to infinity, wrap to minus infinity, and end at bi.
For any interval Ij = [aj, bj], define Complement(Ij) as the interval [bj, aj]. (For example, the interval [2, 3] starts at 2 and ends at 3; so Complement([2,3]) = [3,2] starts at 3, stretches to infinity, wraps to -infinity, and ends at 2.)
Observe that interval I contains interval J if and only if Complement(J) contains Complement(I). (Prove this.)
So, we can answer your original question simply by running the "easy algorithm" on the set of complements of all of the intervals. That is, start your scan at -infinity with the set S of "active intervals" containing all intervals (because all complements contain infinity/-infinity). Keep S sorted by end point (i.e. start point of complement).
Sort all start points and end points and process them in order. When you encounter a starting point for interval Ij (= [aj, bj]), you are actually hitting the end point of its complement... So remove Ij from S, query S to see how many of its endpoints (i.e. complement start points) come before bj, and report that as the answer for Ij. If you later encounter the end point of Ij, you are encountering the start point of its complement, so you need to add it back into the set S of active intervals.
This final algorithm is O(n log n) for the same reasons the "easy algorithm" was.
[Update]
One clarification, one correction, one comment...
Clarification: Of course, the "self-balancing binary tree" has to be augmented such that each sub-tree knows how many elements it contains. Otherwise, you cannot answer "how many elements are less than x?" This augmentation is straightforward to maintain, but it is not something that every implementation provides; e.g. the C++ std::set does not, to my knowledge.
Correction: You do not want to add any elements back in to the set S of active intervals; in fact, doing so can result in the wrong answer. For example, if the intervals are just [1,2] and [3,4], you would hit 1 (and remove [1,2] from the set), then 2 (and add it back in again), then 3... And since 2<4, you would conclude that [3,4] contains [1,2]. Which is wrong.
Conceptually, you already processed all of the "start events" for the complement intervals; that is why S begins will all intervals inside of it. So all you need to worry about are the ending points; you do not want to add any elements to S, ever.
Put another way, instead of having the intervals wrap around, you can think of [bi,ai] (where bi > ai) as meaning [bi - infinity, ai] with no wrap-around. The logic still works, but the processing is more clear: First you process all of the "whatever - infinity" terms (i.e. the end points), then you process the others (i.e. the start points).
With this correction, I am pretty sure my solution actually works. This formulation also extends -- I think -- to the case where you have both normal and "backward" intervals together in one input.
Comment: This problem is tricky because if you have to enumerate the set of all intervals contained within every interval, the output itself can be O(n^2). So any working approach has to somehow count the intervals without even being able to identify them :-).
Here is a O(N*LOG(N)):
let Ii = Interval i = (ai, bi)
let L = list of intervals I
sort L by ai
divide L in half into L1a and L2a.
sort L1a and L2a by bi to get L1b and L2b
merge sort L1b and L2b keeping track of the count of nestings (e.g. because all intervals in L1b start before intervals in L2b, when we find an endpoint in L1b that is higher than an endpoint in l2b, we know everything between them is nested inside - think about it)..
Now you have updated the counts on how often an interval in L2 is nested inside an interval in L1.
after merging L1 and L2, we repeat the process (recursion) by dividing L1 into L11a and l12a, also dividing L2 into L21a and L21a..
I was recently asked this question in an interview. Even though I was able to come up the O(n²) solution, the interviewer was obsessed with an O(n) solution. I also checked few other solutions of O(n logn) which I understood, but O(n) solution is still not my cup of tea which assumes appointments sorted by start-time.
Can anyone explain this?
Problem Statement: You are given n appointments. Each appointment contains a start time and an end time. You have to retun all conflicting appointments efficiently.
Person: 1,2, 3, 4, 5
App Start: 2, 4, 29, 10, 22
App End: 5, 7, 34, 11, 36
Answer: 2x1 5x3
O(n logn) algorithm: separate start and end point like this:
2s, 4s, 29s, 10s, 22s, 5e, 7e, 34e, 11e, 36e
then sort all of this points (for simplicity let's assume each point is unique):
2s, 4s, 5e, 7e, 10s, 11e, 22s, 29s, 34e, 36e
if we have consecutive starts without ends then it is overlapping:
2s, 4s are adjacent so overlapping is there
We will keep a count of "s" and each time we encounter it will +1, and when e is encountered we decrease count by 1.
The general solution to this problem is not possible in O(n).
At a minimum you need to sort by appointment start time, which requires O(n log n).
There is an O(n) solution if the list is already sorted. The algorithm basically involves checking whether the next appointment is overlapped by any previous ones. There is a bit of subtlety to this one as you actually need two pointers into the list as you run through it:
The current appointment being checked
The appointment with the latest end time encountered so far (which might not be the previous appointment)
O(n) solutions for the unsorted case could only exist if you have other constraints, e.g. a fixed number of appointment timeslots. If this was the case, then you can use HashSets to determine which appointment(s) cover each timeslot, algorithm roughly as follows:
Create a HashSet for each timeslot - O(1) since timeslot number is a fixed constant
For each appointment, store its ID number in HashSets of slot(s) that it covers - O(n) since updating a constant number of timeslots is O(1) for each appointment
Run through the slots, checking for overlaps - O(1) (or O(n) if you want to iterate over the overlapping appointments to return them as results)
Assuming you have some constraint on the start and end times, and on the resolution at which you do the scheduling, it seems like it would be fairly easy to turn each appointment into a bitmap of times it does/doesn't use, then do a counting sort (aka bucket sort) on the in-use slots. Since both of those are linear, the result should be linear (though if I'm thinking correctly, it should be linear on the number of time slots rather than the number of appointments).
At least if I asked this as an interview question, the main thing I'd be hoping for is the candidate to ask about those constraints (i.e., whether those constraints are allowed). Given the degree to which it's unrealistic to schedule appointments for 1000 years from now, or schedule to a precision of even a minute (not to mention something like a nanosecond), they strike me as reasonable constraints, but you should ask before assuming them.
A naive approach might be to build two parallel trees, one ordered by the beginning point, and one ordered by the ending point of each interval. This allows discarding half of each tree in O(log n) time, but the results must be merged, requiring O(n) time. This gives us queries in O(n + log n) = O(n).
This is the best I can think of, in horrible pseudocode. I attempted to reduce the problem as much as possible. This is only less than On^2 (I think).
Note that the output at the end will not show every appointment that a given appointment will conflict with on that appointment's specific output line...but at some point every conflict is displayed.
Also note that I renamed the appointments numerically in order of starting time.
output would be something like the following:
Appointment 1 conflicts with 2
Appointment 2 conflicts with
Appointment 3 conflicts with
Appointment 4 conflicts with 5
Appointment 5 conflicts with
appt{1},appt{2},appt{3} ,appt{4} ,appt{5}
2 4 10 22 29
5 7 11 36 34
pseudocode
list=(1,2,3,4,5)
for (i=1,i<=5,i++)
list.shift() **removes first element
appt{i}.conflictswith()=list
for (i=1,i<=n,i++)
{ number=n
done=false
while(done=false)
{if (number>i)
{if (appt(i).endtime() < appt(number).startime())
{appt{i}.conflictswith().pop()}
else
{done=true}
number--
}
else
{done=true}
}
}
for (i=1,i<=n,i++)
print "Appointment ",i," conflicts with:",appt{i}.conflictswith()
I came across a Data Structure called Interval tree, with the help of which we can find intervals in less than O(n log (n)) time, depending upon the data provided