If we have an array of integers, then is there any efficient way other than O(n^2) by which one can find the number of pairs of integers which differ by a given value?
E.g for the array 4,2,6,7 the number of pairs of integers differing by 2 is 2 {(2,4),(4,6)}.
Thanks.
Create a set from your list. Create another set which has all the elements incremented by the delta. Intersect the two sets. These are the upper values of your pairs.
In Python:
>>> s = [4,2,6,7]
>>> d = 2
>>> s0 = set(s)
>>> sd = set(x+d for x in s0)
>>> set((x-d, x) for x in (s0 & sd))
set([(2, 4), (4, 6)])
Creating the sets is O(n). Intersecting the sets is also O(n), so this is a linear-time algorithm.
Store the elements in a multiset, implemented by a hash table. Then for each element n, check the number of occurences of n-2 in the multiset and sum them up. There is no need to check n+2 because that would cause you to count each pair twice.
The time efficiency is O(n) in the average case, and O(n*logn) or O(n^2) in the worst case (depending on the hash table implementation). It will be O(n*logn) if the multiset is implemented by a balanced tree.
Sort the array, then scan through with two pointers. Supposing the first one points to a, then step the second one forward until you've found where a+2 would be if it was present. Increment the total if it's there. Then increment the first pointer and repeat. At each step, the second pointer starts from the place it ended up on the previous step.
If duplicates are allowed in the array, then you need to remember how many duplicates the second one stepped over, so that you can add this number to the total if incrementing the first pointer yields the same integer again.
This is O(n log n) worst case (for the sort), since the scan is linear time.
It's O(n) worst case on the same basis that hashtable-based solutions for fixed-width integers can say that they're expected O(n) time, since sorting fixed-width integers can be done using radix sort in O(n). Which is actually faster is another matter -- hashtables are fast but might involve a lot of memory allocation (for nodes) and/or badly-localized memory access, depending on implementation.
Note that if the desired difference is 0 and all the elements in the array are identical, then the size of the output is O(n²), so the worst-case of any algorithm is necessarily O(n²). (On the other hand, average-case or expected-case behavior can be significantly better, as others have noted.)
Just hash the numbers in an array as you do in counting sort.Then take two variables, first pointing to index 0 and the other pointing to index 2(or index d in general case) initially.
Now check whether value at both indices are non-zero, if yes then increment the counter with larger of the two values else leave the counter unchanged as the pair does not exist. Now increment both the indices and continue until the second index reaches the end of the array.The total value of counter is the number of pairs with difference d.
Time complexity: O(n)
Space complexity: O(n)
Related
I have an unsorted array of n positive numbers and a parameter k, I need to find out if there is a pair of numbers in the array that the difference between than is less than k and I need to do so in time complexity of O(n) on probable average and in space complexity of O(n).
I believe it requires the use of a universal hash table but I'm not sure how, any ideas?
This answer works even on unbounded integers and floats (doing some assumptions on the nicety of the hashmap you'll be using - the java implementation should work for instance):
keep a hashmap<int, float> all_divided_values. For each key y,
if all_divided_values[y] exists, it will contain a value v that
is in the array such that floor(v/k) = y.
For each value v in the original array A, if v/k is in all_divided_values's keys, output (v, all_divided_values[v/k])
(they are distant by less than k). Else, store v in
all_divided_values[v/k]
Once all_divided_values is filled, go through A again. For each v, test whether all_divided_values[v/k - 1] exists, and if so,
output the pair (v, all_divided_values[v/k - 1]) if and only if abs(v-all_divided_values[v/k - 1])<=k
Inserting in a hashmap is usually (with Java hashmap for instance) O(1) in average, so the total time is O(n). But please note that technically this could be false, for instance if your language's implementation does not have a nice strategy about the hashmap.
Simple solution:
1- Sort the array
2- Calculate the difference between consecutive elements
a) If the difference is smaller than k return that pair
b) If no consecutive number difference yields a value smaller than k, then your array has no pair of numbers such that the difference is smaller than k.
Sorting is O(nlogn), but if you have only Integers of limited size, you can use Counting sort, that is O(n)
You can consider this way.
The problem can be modeled as this:-
consider each element (considering integer) now you convert them to a range (A[i]-K,A[i]+K)
Now you want to check if any of the two intervals overlap.
Interval intersection problem without any sorted ness is not solvable in O(n) (worst case). You need to sort them and then inn O(n) you can check if hey intersect.
Same goes for your logic. Sort it and find it.
Given an array where number of occurrences of each number is odd except one number whose number of occurrences is even. Find the number with even occurrences.
e.g.
1, 1, 2, 3, 1, 2, 5, 3, 3
Output should be:
2
The below are the constraints:
Numbers are not in range.
Do it in-place.
Required time complexity is O(N).
Array may contain negative numbers.
Array is not sorted.
With the above constraints, all my thoughts failed: comparison based sorting, counting sort, BST's, hashing, brute-force.
I am curious to know: Will XORing work here? If yes, how?
This problem has been occupying my subway rides for several days. Here are my thoughts.
If A. Webb is right and this problem comes from an interview or is some sort of academic problem, we should think about the (wrong) assumptions we are making, and maybe try to explore some simple cases.
The two extreme subproblems that come to mind are the following:
The array contains two values: one of them is repeated an even number of times, and the other is repeated an odd number of times.
The array contains n-1 different values: all values are present once, except one value that is present twice.
Maybe we should split cases by complexity of number of different values.
If we suppose that the number of different values is O(1), each array would have m different values, with m independent from n. In this case, we could loop through the original array erasing and counting occurrences of each value. In the example it would give
1, 1, 2, 3, 1, 2, 5, 3, 3 -> First value is 1 so count and erase all 1
2, 3, 2, 5, 3, 3 -> Second value is 2, count and erase
-> Stop because 2 was found an even number of times.
This would solve the first extreme example with a complexity of O(mn), which evaluates to O(n).
There's better: if the number of different values is O(1), we could count value appearances inside a hash map, go through them after reading the whole array and return the one that appears an even number of times. This woud still be considered O(1) memory.
The second extreme case would consist in finding the only repeated value inside an array.
This seems impossible in O(n), but there are special cases where we can: if the array has n elements and values inside are {1, n-1} + repeated value (or some variant like all numbers between x and y). In this case, we sum all the values, substract n(n-1)/2 from the sum, and retrieve the repeated value.
Solving the second extreme case with random values inside the array, or the general case where m is not constant on n, in constant memory and O(n) time seems impossible to me.
Extra note: here, XORing doesn't work because the number we want appears an even number of times and others appear an odd number of times. If the problem was "give the number that appears an odd number of times, all other numbers appear an even number of times" we could XOR all the values and find the odd one at the end.
We could try to look for a method using this logic: we would need something like a function, that applied an odd number of times on a number would yield 0, and an even number of times would be identity. Don't think this is possible.
Introduction
Here is a possible solution. It is rather contrived and not practical, but then, so is the problem. I would appreciate any comments if I have holes in my analysis. If this was a homework or challenge problem with an “official” solution, I’d also love to see that if the original poster is still about, given that more than a month has passed since it was asked.
First, we need to flesh out a few ill-specified details of the problem. Time complexity required is O(N), but what is N? Most commentators appear to be assuming N is the number of elements in the array. This would be okay if the numbers in the array were of fixed maximum size, in which case Michael G’s solution of radix sort would solve the problem. But, I interpret constraint #1, in absence of clarification by the original poster, as saying the maximum number of digits need not be fixed. Therefore, if n (lowercase) is the number of elements in the array, and m the average length of the elements, then the total input size to contend with is mn. A lower bound on the solution time is O(mn) because this is the read-through time of the input needed to verify a solution. So, we want a solution that is linear with respect to total input size N = nm.
For example, we might have n = m, that is sqrt(N) elements of sqrt(N) average length. A comparison sort would take O( log(N) sqrt(N) ) < O(N) operations, but this is not a victory, because the operations themselves on average take O(m) = O(sqrt(N)) time, so we are back to O( N log(N) ).
Also, a radix sort would take O(mn) = O(N) if m were the maximum length instead of average length. The maximum and average length would be on the same order if the numbers were assumed to fall in some bounded range, but if not we might have a small percentage with a large and variable number of digits and a large percentage with a small number of digits. For example, 10% of the numbers could be of length m^1.1 and 90% of length m*(1-10%*m^0.1)/90%. The average length would be m, but the maximum length m^1.1, so the radix sort would be O(m^1.1 n) > O(N).
Lest there be any concern that I have changed the problem definition too dramatically, my goal is still to describe an algorithm with time complexity linear to the number of elements, that is O(n). But, I will also need to perform operations of linear time complexity on the length of each element, so that on average over all the elements these operations will be O(m). Those operations will be multiplication and addition needed to compute hash functions on the elements and comparison. And if indeed this solution solves the problem in O(N) = O(nm), this should be optimal complexity as it takes the same time to verify an answer.
One other detail omitted from the problem definition is whether we are allowed to destroy the data as we process it. I am going to do so for the sake of simplicity, but I think with extra care it could be avoided.
Possible Solution
First, the constraint that there may be negative numbers is an empty one. With one pass through the data, we will record the minimum element, z, and the number of elements, n. On a second pass, we will add (3-z) to each element, so the smallest element is now 3. (Note that a constant number of numbers might overflow as a result, so we should do a constant number of additional passes through the data first to test these for solutions.) Once we have our solution, we simply subtract (3-z) to return it to its original form. Now we have available three special marker values 0, 1, and 2, which are not themselves elements.
Step 1
Use the median-of-medians selection algorithm to determine the 90th percentile element, p, of the array A and partition the array into set two sets S and T where S has the 10% of n elements greater than p and T has the elements less than p. This takes O(n) steps (with steps taking O(m) on average for O(N) total) time. Elements matching p could be placed either into S or T, but for the sake of simplicity, run through array once and test p and eliminate it by replacing it with 0. Set S originally spans indexes 0..s, where s is about 10% of n, and set T spans the remaining 90% of indexes s+1..n.
Step 2
Now we are going to loop through i in 0..s and for each element e_i we are going to compute a hash function h(e_i) into s+1..n. We’ll use universal hashing to get uniform distribution. So, our hashing function will do multiplication and addition and take linear time on each element with respect to its length.
We’ll use a modified linear probing strategy for collisions:
h(e_i) is occupied by a member of T (meaning A[ h(e_i) ] < p but is not a marker 1 or 2) or is 0. This is a hash table miss. Insert e_i by swapping elements from slots i and h(e_i).
h(e_i) is occupied by a member of S (meaning A[ h(e_i) ] > p) or markers 1 or 2. This is a hash table collision. Do linear probing until either encountering a duplicate of e_i or a member of T or 0.
If a member of T, this is a again a hash table miss, so insert e_i as in (1.) by swapping to slot i.
If a duplicate of e_i, this is a hash table hit. Examine the next element. If that element is 1 or 2, we’ve seen e_i more than once already, change 1s into 2s and vice versa to track its change in parity. If the next element is not 1 or 2, then we’ve only seen e_i once before. We want to store a 2 into the next element to indicate we’ve now seen e_i an even number of times. We look for the next “empty” slot, that is one occupied by a member of T which we’ll move to slot i, or a 0, and shift the elements back up to index h(e_i)+1 down so we have room next to h(e_i) to store our parity information. Note we do not need to store e_i itself again, so we’ve used up no extra space.
So basically we have a functional hash table with 9-fold the number of slots as elements we wish to hash. Once we start getting hits, we begin storing parity information as well, so we may end up with only 4.5-fold number of slots, still a very low load factor. There are several collision strategies that could work here, but since our load factor is low, the average number of collisions should be also be low and linear probing should resolve them with suitable time complexity on average.
Step 3
Once we finished hashing elements of 0..s into s+1..n, we traverse s+1..n. If we find an element of S followed by a 2, that is our goal element and we are done. Any element e of S followed by another element of S indicates e was encountered only once and can be zeroed out. Likewise e followed by a 1 means we saw e an odd number of times, and we can zero out the e and the marker 1.
Rinse and Repeat as Desired
If we have not found our goal element, we repeat the process. Our 90th percentile partition will move the 10% of n remaining largest elements to the beginning of A and the remaining elements, including the empty 0-marker slots to the end. We continue as before with the hashing. We have to do this at most 10 times as we process 10% of n each time.
Concluding Analysis
Partitioning via the median-of-medians algorithm has time complexity of O(N), which we do 10 times, still O(N). Each hash operation takes O(1) on average since the hash table load is low and there are O(n) hash operations in total performed (about 10% of n for each of the 10 repetitions). Each of the n elements have a hash function computed for them, with time complexity linear to their length, so on average over all the elements O(m). Thus, the hashing operations in aggregate are O(mn) = O(N). So, if I have analyzed this properly, then on whole this algorithm is O(N)+O(N)=O(N). (It is also O(n) if operations of addition, multiplication, comparison, and swapping are assumed to be constant time with respect to input.)
Note that this algorithm does not utilize the special nature of the problem definition that only one element has an even number of occurrences. That we did not utilize this special nature of the problem definition leaves open the possibility that a better (more clever) algorithm exists, but it would ultimately also have to be O(N).
See the following article: Sorting algorithm that runs in time O(n) and also sorts in place,
assuming that the maximum number of digits is constant, we can sort the array in-place in O(n) time.
After that it is a matter of counting each number's appearences, which will take in average n/2 time to find one number whose number of occurrences is even.
this is a homework question, and I'm not that at finding the complixity but I'm trying my best!
Three-way partitioning is a modification of quicksort that partitions elements into groups smaller than, equal to, and larger than the pivot. Only the groups of smaller and larger elements need to be recursively sorted. Show that if there are N items but only k unique values (in other words there are many duplicates), then the running time of this modification to quicksort is O(Nk).
my try:
on the average case:
the tree subroutines will be at these indices:
I assume that the subroutine that have duplicated items will equal (n-k)
first: from 0 - to(i-1)
Second: i - (i+(n-k-1))
third: (i+n-k) - (n-1)
number of comparisons = (n-k)-1
So,
T(n) = (n-k)-1 + Sigma from 0 until (n-k-1) [ T(i) + T (i-k)]
then I'm not sure how I'm gonna continue :S
It might be a very bad start though :$
Hope to find a help
First of all, you shouldn't look at the average case since the upper bound of O(nk) can be proved for the worst case, which is a stronger statement.
You should look at the maximum possible depth of recursion. In normal quicksort, the maximum depth is n. For each level, the total number of operations done is O(n), which gives O(n^2) total in the worst case.
Here, it's not hard to prove that the maximum possible depth is k (since one unique value will be removed at each level), which leads to O(nk) total.
I don't have a formal education in complexity. But if you think about it as a mathematical problem, you can prove it as a mathematical proof.
For all sorting algorithms, the best case scenario will always be O(n) for n elements because to sort n elements you have to consider each one atleast once. Now, for your particular optimisation of quicksort, what you have done is simplified the issue because now, you are only sorting unique values: All the values that are the same as the pivot are already considered sorted, and by virtue of its nature, quicksort will guarantee that every unique value will feature as the pivot at some point in the operation, so this eliminates duplicates.
This means for an N size list, quicksort must perform some operation N times (once for every position in the list), and because it is trying to sort the list, that operation is trying to find the position of that value in the list, but because you are effectively dealing with just unique values, and there are k of those, the quicksort algorithm must perform k comparisons for each element. So it performs Nk operations for an N sized list with k unique elements.
To summarise:
This algorithm eliminates checking against duplicate values.
But all sorting algorithms must look at every value in the list at least once. N operations
For every value in the list the operation is to find its position relative to other values in the list.
Because duplicates get removed, this leaves only k values to check against.
O(Nk)
Is it possible to compute the number of different elements in an array in linear time and constant space? Let us say it's an array of long integers, and you can not allocate an array of length sizeof(long).
P.S. Not homework, just curious. I've got a book that sort of implies that it is possible.
This is the Element uniqueness problem, for which the lower bound is Ω( n log n ), for comparison-based models. The obvious hashing or bucket sorting solution all requires linear space too, so I'm not sure this is possible.
You can't use constant space. You can use O(number of different elements) space; that's what a HashSet does.
You can use any sorting algorithm and count the number of different adjacent elements in the array.
I do not think this can be done in linear time. One algorithm to solve in O(n log n) requires first sorting the array (then the comparisons become trivial).
If you are guaranteed that the numbers in the array are bounded above and below, by say a and b, then you could allocate an array of size b - a, and use it to keep track of which numbers have been seen.
i.e., you would move through your input array take each number, and mark a true in your target array at that spot. You would increment a counter of distinct numbers only when you encounter a number whose position in your storage array is false.
Assuming we can partially destroy the input, here's an algorithm for n words of O(log n) bits.
Find the element of order sqrt(n) via linear-time selection. Partition the array using this element as a pivot (O(n)). Using brute force, count the number of different elements in the partition of length sqrt(n). (This is O(sqrt(n)^2) = O(n).) Now use an in-place radix sort on the rest, where each "digit" is log(sqrt(n)) = log(n)/2 bits and we use the first partition to store the digit counts.
If you consider streaming algorithms only ( http://en.wikipedia.org/wiki/Streaming_algorithm ), then it's impossible to get an exact answer with o(n) bits of storage via a communication complexity lower bound ( http://en.wikipedia.org/wiki/Communication_complexity ), but possible to approximate the answer using randomness and little space (Alon, Matias, and Szegedy).
This can be done with a bucket approach when assuming that there are only a constant number of different values. Make a flag for each value (still constant space). Traverse the list and flag the occured values. If you happen to flag an already flagged value, you've found a duplicate. You have to traverse the buckets for each element in the list. But that's still linear time.
Given an unsorted integer array, and without making any assumptions on
the numbers in the array:
Is it possible to find two numbers whose
difference is minimum in O(n) time?
Edit: Difference between two numbers a, b is defined as abs(a-b)
Find smallest and largest element in the list. The difference smallest-largest will be minimum.
If you're looking for nonnegative difference, then this is of course at least as hard as checking if the array has two same elements. This is called element uniqueness problem and without any additional assumptions (like limiting size of integers, allowing other operations than comparison) requires >= n log n time. It is the 1-dimensional case of finding the closest pair of points.
I don't think you can to it in O(n). The best I can come up with off the top of my head is to sort them (which is O(n * log n)) and find the minimum difference of adjacent pairs in the sorted list (which adds another O(n)).
I think it is possible. The secret is that you don't actually have to sort the list, you just need to create a tally of which numbers exist. This may count as "making an assumption" from an algorithmic perspective, but not from a practical perspective. We know the ints are bounded by a min and a max.
So, create an array of 2 bit elements, 1 pair for each int from INT_MIN to INT_MAX inclusive, set all of them to 00.
Iterate through the entire list of numbers. For each number in the list, if the corresponding 2 bits are 00 set them to 01. If they're 01 set them to 10. Otherwise ignore. This is obviously O(n).
Next, if any of the 2 bits is set to 10, that is your answer. The minimum distance is 0 because the list contains a repeated number. If not, scan through the list and find the minimum distance. Many people have already pointed out there are simple O(n) algorithms for this.
So O(n) + O(n) = O(n).
Edit: responding to comments.
Interesting points. I think you could achieve the same results without making any assumptions by finding the min/max of the list first and using a sparse array ranging from min to max to hold the data. Takes care of the INT_MIN/MAX assumption, the space complexity and the O(m) time complexity of scanning the array.
The best I can think of is to counting sort the array (possibly combining equal values) and then do the sorted comparisons -- bin sort is O(n + M) (M being the number of distinct values). This has a heavy memory requirement, however. Some form of bucket or radix sort would be intermediate in time and more efficient in space.
Sort the list with radixsort (which is O(n) for integers), then iterate and keep track of the smallest distance so far.
(I assume your integer is a fixed-bit type. If they can hold arbitrarily large mathematical integers, radixsort will be O(n log n) as well.)
It seems to be possible to sort unbounded set of integers in O(n*sqrt(log(log(n))) time. After sorting it is of course trivial to find the minimal difference in linear time.
But I can't think of any algorithm to make it faster than this.
No, not without making assumptions about the numbers/ordering.
It would be possible given a sorted list though.
I think the answer is no and the proof is similar to the proof that you can not sort faster than n lg n: you have to compare all of the elements, i.e create a comparison tree, which implies omega(n lg n) algorithm.
EDIT. OK, if you really want to argue, then the question does not say whether it should be a Turing machine or not. With quantum computers, you can do it in linear time :)