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Interesting thought question for you guys. Given an array of length n, if I were to pick two random indexes in this array, a and b on average how far apart would they be? As in how many steps would I have to take to walk from a to b. There are no restrictions so there's a chance I pick the same index for both, and there's a chance a and b are at opposite ends of the array.
I've thought about this for a while, my initial idea being they're on average n/2(ish) apart, but I think this hunch is incorrect. An index chosen in the center of the array at most would have to walk n/2 places to find its corresponding second choice, whereas only at the ends of the array would the second choice ever be around n distance away.
Thanks!
After scribbling some grids of possible distances for the first few values of n, I think the exact result is in fact given by:
f(n) = (n² - 1) / 3n
Choosing two places in an array is equivalent to splitting the array up into 3 sections. The average size of each of those sections will be n/3 so the average distance between the two points is also n/3.
Using a monte carlo method in python:
from collections import defaultdict
import random
sample = [abs(random.choice(range(0,10)) - random.choice(range(0,10))) for i in range(0,10000)]
avg = float(sum(sample) / len(sample))
print ("Average: %f" % avg)
freq = defaultdict(int)
for s in sample:
freq[s] += 1
scale = 40.0 / max(freq.values())
for i in range(0,10):
print ("%d : %s" % (i, "#" * int(freq[i] * scale)))
Output:
Average: 3.293700
0 : ######################
1 : ########################################
2 : ####################################
3 : ###############################
4 : ##########################
5 : ######################
6 : #################
7 : #############
8 : #########
9 : ####
So, looks like it's n/3 - but it's not evenly distributed.
There is an easy way to know: for all the couples (a, b), computer their distance. Knowing that all the couples (a, b) have the same probability of appearance, you will just need to do the average of those distances in order to answer your question.
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Can you advise the solution for this..
ARRAY EQUALITY
Problem Statement
Amy has an array A, of N integers. She wants to make all the elements of the array equal.
On each day she can select a subarray of A, with length exactly M and perform following operation: ·
Pick any element of the selected subarray and increase or decrease it by 1. She can perform this operation any number of times she wants (possibly 0), on that day.
Find the minimum number of days required to make all the elements of the array A equal. NOTE: A subarray is the sequence of consecutive elements of the array. You are given T independent test cases.
Constraints
`
All input values are integers.
Input Format First-line contains T.
First line of each test case consists of two space separated integers N and M.
Second line of each test case consists of N space separated integers denoting the array A. Output Format Print in a newline for each test case single integer denoting the minimum number of days required to make all the elements of the array A equal.
Sample Input 1
1
5 3
1 2 2 3 1
Sample Output 1
1
I want to know the aproach to solve this problem?
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i have just wasted my 2 hours only for solving this programming question.if any one knows the trick of doing then please share it.question is given below.
You have given an array A having N integers. Let say G is the product of all elements of A.You have to find the number of distinct prime divisors of G.
Input Format
The first argument given is an Array A, having N integers.
Output Format
Return an Integer, i.e number of distinct prime divisors of
G.
Constraints
1 <= N <= 1e5
1 <= A[i] <= 1e5
For Example
Input:
A = [1, 2, 3, 4]
Output:
2
Explanation:
here G = 1 * 2 * 3 * 4 = 24
and distinct prime divisors of G are [2, 3]
Since this seems to be a homework question, I'll give some pushes in the right direction.
Theoretically, this is a very simple thing to do. Anyone can write code that loops through an array and multiplies its elements. It is also very easy to find pseudocode (or even real code) for factorizing a number into its prime factors.
However, this approach will not work here, since we will be dealing with HUGE numbers. The maximum value of G, given your constraints, is (10⁵)^(10⁵) = 10⁵⁰⁰⁰⁰⁰. This by far exceeds the number of electrons in the observable universe. We cannot factorize such huge numbers.
But luckily we don't need to know the value of G. We are only required to calculate it's prime factors, but we don't need to know the value of G to do so. So instead you will have to factorize the individual numbers in the array. I would recommend something like this code:
factors = set()
for num in A:
f = factorize(num) # Function that returns the set of prime factors in num
factors |= f # Add all elements in f to factors
You were only interested in the distinct prime factors, so using a set will take care of that. Just add everything, and it will automatically throw away any duplicates. factorize(x) is a function you will need to write that takes a number as argument and return the set of prime factors.
Good luck!
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We have for example these matrices:
M = rand(30,45,4);
U = rand(4,4);
In first step I want to multiply each element of the matrix M, which is a 4 element vector, with the U matrix. This will result in in a new M1 matrix whose elements are also 1x4. For example, M1(1,1,:) = M(1,1,:)*U; This in matrix dimensions, for the specific example, corresponds in 1x4 = (1x4)(4x4). The whole M1 matrix has the same dimensions as M.
In second step, the M1 must be multiplied with the M' in the same way and obtain the final F matrix which is a 2D matrix with 1D elements. For example, F(1,1) = M1(1,1,:)*M(1,1,:)';
In other words, each element of the F matrix is where is each of the 1x4 elements of the M matrix.
Attached example:
M = rand(600, 800, 4);
U = rand(4,4);
F = nan(size(M,1),size(M,2));
for i=1:size(M,1)*size(M,2)
[r,c]=ind2sub(size(M),i);
F(i) = squeeze(M(r,c,:))'*U*squeeze(M(r,c,:));
end
Elapsed time is 58.627296 seconds.
#Luis Mendo answer:
tic
Mr = reshape(M, [], size(M,3));
result = reshape(sum(Mr*U.*Mr,2), size(M,1), size(M,2));
toc
Elapsed time is 0.062898 seconds.
Could you please somebody suggest a way without using for loops?
Thank you for you time!
PS: Matlab 2013a 64x, Intel(R) Core(TM) i3 CPU 2.40 GHz, 4GB memory
I'm still not sure I understood correctly, but maybe this is what you want:
Mr = reshape(M, [], size(M,4));
result = reshape(sum(Mr*U.*Mr,2), size(M,1), size(M,2));
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Im truggling to find a closed form for the number of k-permutations of a set S of cardinality n.
The combinations should regard ordering, but no repitions.
Example:
|S| = n = 3
S = {a,b,c}
k = 2
{a,b}
{b,a}
{b,c}
{c,b}
{a,c}
{c,a}
Anyone could help me out how to compute the number of viable permutations (and not the permutations itself)?
What I've tried:
I've read through different material and found out, that including repitititions it is
O(n) = n^k
My initial though was, that I need to eliminiate the permutations like
{a,a}
{b,b}
{c,c}
But I struggle finding a closed form for the number of perceivable repititions.
You are looking for the number of k-permutations of a set S of cardinality n.
The formulae is well known : n!/(n-k)!
Pseudo-proof :
for the 1st element, you are able to choose among the n elements of S ;
for the 2nd, only among : n-1, because you don't want doublons ;
...
for the ith, only among : n-(i-1) ;
...
for the kth, only among : n-(k-1) ;
So, finally :
n * (n-1) * ... * (n-i) * ... * (n-k+1)
= n! / (n-k)!
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Possible Duplicate:
Determining the number of occurrences of each unique element in a vector
I've the following array:
v = [ 1 5 1 6 7 1 5 5 1 1]
And I need to count the values and show the number that has more appearances.
From the example on the top, the solution would be 1 (there are five 1's)
Thanks in advance
Use mode.
If you need to return the number of elements as well, do the following:
m = mode(v);
n = sum(v==m);
fprintf('%d appears %d times\n',m,n);
Another method is using the hist function, if you're dealing with integers.
numbers=unique(v); %#provides sorted unique list of elements
count=hist(v,numbers); %#provides a count of each element's occurrence
Just make sure you specify an output value for the hist function, or you'll end up with a bar graph.
#Jacob is right: mode(v) will give you the answer you need.
I just wanted to add a nice way to represent the frequencies of each value:
bar(accumarray(v', 1))
will show a nice bar diagram with the count of each value in v.