I have an unordered list of n items, and I'm trying to find the most frequent item in that list. I wrote the following code:
def findFrequant(L):
count = int()
half = len(L)//2
for i in L:
for j in L:
if i == j:
count += 1
if count > half:
msg = "The majority vote is {0}".format(i)
return msg
else:
continue
count = 0
return "mixed list!"
Obviously this procedure with the two loops is O(n^2), and I'm trying to accomplish the same task in O(n log n) time. I'm not looking for a fix or someone to write the code for me, I'm simply looking for directions.
I don't recognize the language here so I'm treating it as pseudocode.
This depends on a hashtable with key of the same type of element of L and value type of int. Count each record in hashtable, then iterate the hashtable as a normal collection of key,value pairs applying normal maxlist algorithm.
O(n) is slightly worse than linear. We remember that expense of a good hash is not linear but may be approximated as linear. Linear space used.
def findFrequant(L):
hash = [,]
vc = 0
vv = null
for i in L
if hash.contains(i)
hash[i] = hash[i] + 1
else
hash[i] = 1
for (k,v) in hash
if v > vc
vv = k
vc = v
else if v == vc
vv = null
if (vv == null)
return "mixed list!"
else
return "The majority vote is {0}".format(v)
You could use a Merge Sort, which has a worst case time complexity of O(n log(n)) and a binary search which has a worst case time complexity of O(log(n)).
There are sorting algorithms with faster best case scenarios, like for example bubble sort which has a best case of O(n), but merge sort performs in O(n log(n)) at it's worst, while bubble sort has a worst case of O(n^2).
Pessimistic as we Computer Scientists are, we generally analyze based on the worst-case scenario. Therefore a combination of merge sort and binary search is probably best in your case.
Note that in certain circumstances a Radix sort might perform faster than merge sort, but this is really dependent on your data.
Related
I'm struggling to figure out what the time complexity for this code would be.
def under_ten(input_list : List[int]) -> List[int]:
res = []
for i in input_list:
if i < 10:
res.append(i)
res.sort()
return res
Since the loop iterates over every element of n, I think the best case should be O(n). What I'm not sure about is how sorting the result list affects the time complexity of the entire function. Is the worst case O(nlogn) (all numbers in n are under 10, so the result list is the same size as the input list)? And what would be the average case?
EDIT: Changed input name from n to input_list and added type hints, sorry if that caused some confusion (added type hints as well).
Your first observation is correct that iterating the input collection would be an O(N) operation, where N here is the length of the array called n. The running time of the sort operation at the end would depend on how large the res array is. In the worst case scenario, every number in n would be less than 10, and therefore would end up in res. The internal algorithm Python would be using for sort() would likely be either quicksort or mergesort (q.v. this SO question). Both of these algorithms use a divide-and-conquer approach which run in O(N*lgN). So, in the worst case, your under_ten() function would run in O(N*lgN).
Let N be the length of the list and K the number of elements smaller than 10.
The complexity is O(N + K log K), assuming that append is done in amortized constant time.
In the worst case, K=N, hence O(N Log N), provided the sort truly has a worst case O(N Log N). Otherwise, it could be O(N²).
Design an algorithm that sorts n integers where there are duplicates. The total number of different numbers is k. Your algorithm should have time complexity O(n + k*log(k)). The expected time is enough. For which values of k does the algorithm become linear?
I am not able to come up with a sorting algorithm for integers which satisfies the condition that it must be O(n + k*log(k)). I am not a very advanced programmer but I was in the problem before this one supposed to come up with an algorithm for all numbers xi in a list, 0 ≤ xi ≤ m such that the algorithm was O(n+m), where n was the number of elements in the list and m was the value of the biggest integer in the list. I solved that problem easily by using counting sort but I struggle with this problem. The condition that makes it the most difficult for me is the term k*log(k) under the ordo notation if that was n*log(n) instead I would be able to use merge sort, right? But that's not possible now so any ideas would be very helpful.
Thanks in advance!
Here is a possible solution:
Using a hash table, count the number of unique values and the number of duplicates of each value. This should have a complexity of O(n).
Enumerate the hashtable, storing the unique values into a temporary array. Complexity is O(k).
Sort this array with a standard algorithm such as mergesort: complexity is O(k.log(k)).
Create the resulting array by replicating the elements of the sorted array of unique values each the number of times stored in the hash table. complexity is O(n) + O(k).
Combined complexity is O(n + k.log(k)).
For example, if k is a small constant, sorting an array of n values converges toward linear time as n becomes larger and larger.
If during the first phase, where k is computed incrementally, it appears that k is not significantly smaller than n, drop the hash table and just sort the original array with a standard algorithm.
The runtime of O(n + k*log(k) indicates (like addition in runtimes often does) that you have 2 subroutines, one which runes in O(n) and the other that runs in O(k*log(k)).
You can first count the frequency of the elements in O(n) (for example in a Hashmap, look this up if youre not familiar with it, it's very useful).
Then you just sort the unique elements, from which there are k. This sorting runs in O(k*log(k)), use any sorting algorithm you want.
At the end replace the single unique elements by how often they actually appeared, by looking this up in the map you created in step 1.
A possible Java solution an be like this:
public List<Integer> sortArrayWithDuplicates(List<Integer> arr) {
// O(n)
Set<Integer> set = new HashSet<>(arr);
Map<Integer, Integer> freqMap = new HashMap<>();
for(Integer i: arr) {
freqMap.put(i, freqMap.getOrDefault(i, 0) + 1);
}
List<Integer> withoutDups = new ArrayList<>(set);
// Sorting => O(k(log(k)))
// as there are k different elements
Arrays.sort(withoutDups);
List<Integer> result = new ArrayList<>();
for(Integer i : withoutDups) {
int c = freqMap.get(i);
for(int j = 0; j < c; j++) {
result.add(i);
}
}
// return the result
return result;
}
The time complexity of the above code is O(n + k*log(k)) and solution is in the same line as answered above.
The question is rather simple, but I just can't find a good enough answer. On the most upvoted SO question regarding the big-O notation, it says that:
For example, sorting algorithms are typically compared based on comparison operations (comparing two nodes to determine their relative ordering).
Now let's consider the simple bubble sort algorithm:
for (int i = arr.length - 1; i > 0; i--) {
for (int j = 0; j < i; j++) {
if (arr[j] > arr[j+1]) {
switchPlaces(...)
}
}
}
I know that worst case is O(n²) and best case is O(n), but what is n exactly? If we attempt to sort an already sorted algorithm (best case), we would end up doing nothing, so why is it still O(n)? We are looping through 2 for-loops still, so if anything it should be O(n²). n can't be the number of comparison operations, because we still compare all the elements, right?
When analyzing the Big-O performance of sorting algorithms, n typically represents the number of elements that you're sorting.
So, for example, if you're sorting n items with Bubble Sort, the runtime performance in the worst case will be on the order of O(n2) operations. This is why Bubble Sort is considered to be an extremely poor sorting algorithm, because it doesn't scale well with increasing numbers of elements to sort. As the number of elements to sort increases linearly, the worst case runtime increases quadratically.
Here is an example graph demonstrating how various algorithms scale in terms of worst-case runtime as the problem size N increases. The dark-blue line represents an algorithm that scales linearly, while the magenta/purple line represents a quadratic algorithm.
Notice that for sufficiently large N, the quadratic algorithm eventually takes longer than the linear algorithm to solve the problem.
Graph taken from http://science.slc.edu/~jmarshall/courses/2002/spring/cs50/BigO/.
See Also
The formal definition of Big-O.
I think two things are getting confused here, n and the function of n that is being bounded by the Big-O analysis.
By convention, for any algorithm complexity analysis, n is the size of the input if nothing different is specified. For any given algorithm, there are several interesting functions of the input size for which one might calculate asymptotic bounds such as Big-O.
The commonest such function for a sorting algorithm is the worst case number of comparisons. If someone says a sorting algorithm is O(n^2), without specifying anything else, I would assume they mean the worst case comparison count is O(n^2), where n is the input size.
Another interesting function is the amount of work space, of space in addition to the array being sorted. Bubble sort's work space is O(1), constant space, because it only uses a few variables regardless of the array size.
Bubble sort can be coded to do only n-1 array element comparisons in the best case, by finishing after any pass that does no exchanges. See this pseudo code implementation, which uses swapped to remember whether there were any exchanges. If the array is already sorted the first pass does no exchanges, so the sort finishes after one pass.
n is usually the size of the input. For array, that would be the number of elements.
To see the different cases, you would need to change the algorithm:
for (int i = arr.length - 1; i > 0 ; i--) {
boolean swapped = false;
for (int j = 0; j<i; j++) {
if (arr[j] > arr[j+1]) {
switchPlaces(...);
swapped = true;
}
}
if(!swapped) {
break;
}
}
Your algorithm's best/worst cases are both O(n^2), but with the possibility of returning early, the best-case is now O(n).
n is array length. You want to find T(n) algorithm complexity.
It is much expensive to access memory then check condition if. So, you define T(n) to be number of access memory.
In the given algorithm BC and WC use O(n^2) accesses to memory because you check the if-condition O(n^2) times.
Make the complexity better: Hold a flag and if you don't do any swaps in the main-loop, it means your array is sorted and you can put a break.
Now, in BC the array is sorted and you access all elements once so O(n).
And in WC still O(n^2).
Given n, k and n number of integers. How would you find the pairs of integers for which their difference is k?
There is a n*log n solution, but I cannot figure it out.
You can do it like this:
Sort the array
For each item data[i], determine its two target pairs, i.e. data[i]+k and data[i]-k
Run a binary search on the sorted array for these two targets; if found, add both data[i] and data[targetPos] to the output.
Sorting is done in O(n*log n). Each of the n search steps take 2 * log n time to look for the targets, for the overall time of O(n*log n)
For this problem exists the linear solution! Just ask yourself one question. If you have a what number should be in the array? Of course a+k or a-k (A special case: k = 0, required an alternative solution). So, what now?
You are creating a hash-set (for example unordered_set in C++11) with all values from the array. O(1) - Average complexity for each element, so it's O(n).
You are iterating through the array, and check for each element Is present in the array (x+k) or (x-k)?. You check it for each element, in set in O(1), You check each element once, so it's linear (O(n)).
If you found x with pair (x+k / x-k), it is what you are looking for.
So it's linear (O(n)). If you really want O(n lg n) you should use a set on tree, with checking is_exist in (lg n), then you have O(n lg n) algorithm.
Apposition: No need to check x+k and x-k, just x+k is sufficient. Cause if a and b are good pair then:
if a < b then
a + k == b
else
b + k == a
Improvement: If you know a range, you can guarantee linear complexity, by using bool table (set_tab[i] == true, when i is in table.).
Solution similar to one above:
Sort the array
set variables i = 0; j = 1;
check the difference between array[i] and array[j]
if the difference is too small, increase j
if the difference is too big, increase i
if the difference is the one you're looking for, add it to results and increase j
repeat 3 and 4 until the end of array
Sorting is O(n*lg n), the next step is, if I'm correct, O(n) (at most 2*n comparisons), so the whole algorithm is O(n*lg n)
Here is an algorithm for finding kth smallest number in n element array using partition algorithm of Quicksort.
small(a,i,j,k)
{
if(i==j) return(a[i]);
else
{
m=partition(a,i,j);
if(m==k) return(a[m]);
else
{
if(m>k) small(a,i,m-1,k);
else small(a,m+1,j,k);
}
}
}
Where i,j are starting and ending indices of array(j-i=n(no of elements in array)) and k is kth smallest no to be found.
I want to know what is the best case,and average case of above algorithm and how in brief. I know we should not calculate termination condition in best case and also partition algorithm takes O(n). I do not want asymptotic notation but exact mathematical result if possible.
First of all, I'm assuming the array is sorted - something you didn't mention - because that code wouldn't otherwise work. And, well, this looks to me like a regular binary search.
Anyway...
The best case scenario is when either the array is one element long (you return immediately because i == j), or, for large values of n, if the middle position, m, is the same as k; in that case, no recursive calls are made and it returns immediately as well. That makes it O(1) in best case.
For the general case, consider that T(n) denotes the time taken to solve a problem of size n using your algorithm. We know that:
T(1) = c
T(n) = T(n/2) + c
Where c is a constant time operation (for example, the time to compare if i is the same as j, etc.). The general idea is that to solve a problem of size n, we consume some constant time c (to decide if m == k, if m > k, to calculate m, etc.), and then we consume the time taken to solve a problem of half the size.
Expanding the recurrence can help you derive a general formula, although it is pretty intuitive that this is O(log(n)):
T(n) = T(n/2) + c = T(n/4) + c + c = T(n/8) + c + c + c = ... = T(1) + c*log(n) = c*(log(n) + 1)
That should be the exact mathematical result. The algorithm runs in O(log(n)) time. An average case analysis is harder because you need to know the conditions in which the algorithm will be used. What is the typical size of the array? The typical size of k? What is the mos likely position for k in the array? If it's in the middle, for example, the average case may be O(1). It really depends on how you use this.