As per MSDN doc for Array.Sort,
If the number of partitions exceeds 2 * logN, where N is the range of the input array, it uses a Heapsort algorithm.
What I don't know is what are the "number of partitions" and the "range" of an array. What are they?
A partition in a sort is basically a section of the list based upon a pivot point. For example, using the quick sort algorithm to sort the following:
First Pass Second Pass
3 3 1
8 1 3
5 <- Pivot 5--------- 5
1 8 7
7 7 8
In the first pass, there are two partitions based off numbers that are less than or greater than 5
The range is the difference between the largest and smallest values, so in this example that is 7 (8 - 1)
So the line you are questioning works as
(2 * log(7)) > 2 == Use HeapSort
1.691 > 2 false
Related
With given 2d array[X][Y], i have to find the smallest possible sum of Y elements but:
the sum must be created by using just 2 rows,
each value must be from different index
Example:
for array
7 3 7 9
2 20 10 6
8 8 8 8
Result should be 18, as we get 3 + 7 from 1st row and 2 + 6 from 2nd.
I've been thinking about few hours but i can't figure out how to deal with it.
Try this one here.
Method 1 (Naive Approach): Check every possible submatrix in given 2D
array. This solution requires 4 nested loops and time complexity of
this solution would be O(n^4).
Method 2 (Efficient Approach): Kadane’s algorithm for 1D array can be
used to reduce the time complexity to O(n^3).
On the internet I only find code for the algorithm but I need understand in form of text first because I have trouble understand things from code only. And other description of the algorithm are very complicated for me (on Wikipedia and other sites).
Here is what I understand for far:
Let say we want search in array the element 10:
Index i 0 1 2 3 4
2 3 4 10 40
Some fibonacci number here:
Index j 0 1 2 3 4 5 6 7 8 9
0 1 1 2 3 5 8 13 21 34
First thing we do is find fibonacci number that is greater-equal to array length. Array length is 4 so we need take fibonacci number 5 that is in index position j=5.
But where we divide the array now and how continue? I really don't understand it.. Please help understand for exam...
The algorithm goes in the following way:
The length of the array is 5, so the fibonacci number which is greater than or equal to 5 is 5. The two numbers which are preceding in the Fibonacci sequence are 2 [n-2] and 3 [n-1] - (2, 3, 5).
So, arr[n-2] i.e. arr[2] is compared with the number to be searched which is 10.
If the element is smaller than the number, then the sequence is shifted 1 time to the left. Also, the previous index is saved for next iteration to give an offset for the index. In this case, since 4 is smaller, n-2 becomes 1 (1, 2, 3). arr[1 + 2(prev)] = arr[3] = 10. So, the index of the number is 3.
If the element is larger, the sequence is shifted 2 times to the left.
Always the min(n-2+offset,n)th element is compared with number to get the matching result.
You are giving array of length N and numbers in the array contain 1 to N no repetition. You need to check if the array can be divided into to list of equal sum.
I know it can be solved using subset sum problem whose time complexity is.
Is there an algorithm so that I can reduce the time complexity?
As per your requirements, we conclude the array will always contain numbers 1 to N.
So if Array.Sum()==Even the answer is YES, otherwise NO.
Since the sum of elements from 1 to n equals n*(n+1)/2, you have to check if n*(n+1) is a multiple of 4, which is equivalent to checking if n is a multiple of 4, or if n+1 is a multiple of 4. The complexity of it is O(1).
If this condition is met, the two subsets are :
if n is a multiple of 4: sum up the odd numbers of first half with even numbers of second half on one hand, and even numbers of first half with odd of second half on the other.
For instance, 1 3 5 8 10 12 , and 2 4 6 7 9 11.
if n = 3 modulo 4 : almost the same thing, just split the first 3 between 1 and 2 on one hand, 3 on the other, you have a remaining serie which has a size multiple of 4.
For instance : 1 2 4 7 , and 3 5 6 ; or if you prefer, 3 4 7, and 1 2 5 6.
This question already has answers here:
Finding out the duplicate element in an array
(2 answers)
Closed 6 years ago.
Can anyone could help me how to solve this code in C? I think that I have to use big O notation as a solution, but I have no idea about it.
The question: There is an array T sized N+1 where numbers from 1 to N are random. One number x is repeated twice (position is also random).
What should be the fastest way to find value of this number x?
For example:
N = 7
[6 3 5 1 3 7 4 2]
x=3
The sum of numbers 1..N is N*(N+1)/2.
So, the extra number is:
extra_number = sum(all N+1 numbers) - N*(N+1)/2
Everything is O(1) except the sum. The sum can be computed in O(N) time.
The overall algorithm is O(N).
Walk the array using the value as the next array index (minus 1), marking the ones visited with a special value (like 0 or the negation). O(n)
On average, only half the elements are visited.
v
6 3 5 1 3 7 4 2
v
. 3 5 1 3 7 4 2
v
. 3 5 1 3 7 . 2
v
. 3 5 1 . 7 . 2
v
. 3 5 . . 7 . 2
v !! all ready visited. Previous 3 is repeated.
. 3 5 . . 7 . 2
No overflow problem caused by adding up the sum. Of course the array needs to be modified (or a sibling bool array of flags is needed.)
This method works even if more than 1 value is repeated.
The algorithm given by Klaus has O(1) memory requirements, but requires to sum all the elements from the given array, which may be quite large to iterate (sum) all over them.
Another approach is to iterate over array and increment the occurence counter once per iteration, so the algorithm can be stopped instantly once it finds the duplicate, though the worst case scenario is to scan through all the elements. For example:
#define N 8
int T[N] = {6, 3, 5, 1, 3, 7, 4, 2};
int occurences[N+1] = {0};
int duplicate = -1;
for (int i = 0; i < N; i++) {
occurences[T[i]]++;
if (occurences[T[i]] == 2) {
duplicate = T[i];
break;
}
}
Note that this method is also immune to integer overflow, that is N*(N+1)/2. might be larger than integer data type can possibly hold.
I have to q sort an array but for a given range for ex.
Given array
4 5 3 7 2 1
and then i have the range 2 & 5 this means i have to sort from index 2 all the way through 5.
Resultant array
4 5 2 3 7 1
I know we can set one bound
like
qsort(array,4,sizeof(int),compa)
this will sort the array till 3rd index but will always start from index 0. I want to start the first bound by a desired value. Any Suggestions??
Just pass address to the middle of the array.
qsort(array + 2, 5 - 2, sizeof(int), compa);