I need to develop an algorithm which would accept two numbers m and n - dimensions of 2D array - as input and generate 2D array filled with numbers [1..m*n] with the following condition:
All (4) elements adjacent to a given element cannot be equal to currentElement + 1
Adjacent elements are located to the two/three/four sides (depending on position) of a given element
0 1 0
1 2 1
0 1 0
(E.g four 1s are adjacent to 2)
Example:
Input: m = 3, n = 3 (does not essentially have to be square matrix)
(Sample) output:
[
[7, 2, 5],
[1, 6, 9],
[3, 8, 4]
]
Note that there apparently may exist more than one possible output. In that case, numbers in the array have to be generated randomly (though still meeting the conditions), not following any preset sequence (e.g not [ [1, 3, 5], [4, 6, 2], [7, 9, 8] ] because it clearly uses a non-randomly generated sequence of numbers, odds first, then evens, etc)
Basically, for the same input, on two different occasions, two different arrays should be generated.
P.S: that was a coding interview question and I wonder how I could solve it, so, any help is highly appreciated.
Does anyone know an Algorithm that sorts k-approximately an array?
We were asked to find and Algorithm for k-approximate sorting, and it should run in O(n log(n/k)). but I can't seem to find any.
K-approx. sorting means that an array and any 1 <= i <= n-k such that sum a[j] <= sum a[j] i<=j<= i+k-1 i+1<=j<= i+k
I know I'm very late to the question ... But under the assumption that k is some approximation value between 0 and 1 (when 0 is completely unsorted and 1 is perfectly sorted) surely the answer to this is quicksort (or mergesort).
Consider the following array:
[4, 6, 9, 1, 10, 8, 2, 7, 5, 3]
Let's say this array is 'unsorted' - now apply one iteration of quicksort to this array with the (length[array]/2)th element as a pivot: length[array]/2 = 5. So the 5th element is our pivot (i.e. 8):
[4, 6, 2, 1, 3, 9, 7, 10, 8]
Now this is array is not sorted - but it is more sorted than one iteration ago, i.e. its approximately sorted but for a low approximation, i.e. a low value of k. Repeat this step again on the two halves of the array and it becomes more sorted. As k increases towards 1 - i.e. perfectly sorted - the complexity becomes O(N log(N/1)) = O(N log(N)).
Let's say we have an array like
[37, 20, 16, 8, 5, 5, 3, 0]
What algorithm can I use so that I can specify the number of partitions and have the array broken into them.
For 2 partitions, it should be
[37] and [20, 16, 8, 5, 5, 3, 0]
For 3, it should be
[37],[20, 16] and [8, 5, 5, 3, 0]
I am able to break them down by proximity by simply subtracting the element with right and left numbers but that doesn't ensure the correct number of partitions.
Any ideas?
My code is in ruby but any language/algo/pseudo-code will suffice.
Here's the ruby code by Vikram's algorithm
def partition(arr,clusters)
# Return same array if clusters are less than zero or more than array size
return arr if (clusters >= arr.size) || (clusters < 0)
edges = {}
# Get weights of edges
arr.each_with_index do |a,i|
break if i == (arr.length-1)
edges[i] = a - arr[i+1]
end
# Sort edge weights in ascending order
sorted_edges = edges.sort_by{|k,v| v}.collect{|k| k.first}
# Maintain counter for joins happening.
prev_edge = arr.size+1
joins = 0
sorted_edges.each do |edge|
# If join is on right of previous, subtract the number of previous joins that happened on left
if (edge > prev_edge)
edge -= joins
end
joins += 1
# Join the elements on the sides of edge.
arr[edge] = arr[edge,2].flatten
arr.delete_at(edge+1)
prev_edge = edge
# Get out when right clusters are done
break if arr.size == clusters
end
end
(assuming the array is sorted in descending order)
37, 20, 16, 8, 5, 5, 3, 0
Calculate the differences between adjacent numbers:
17, 4, 8, 3, 0, 2, 3
Then sort them in descending order:
17, 8, 4, 3, 3, 2, 0
Then take the first few numbers. For example, for 4 partitions, take 3 numbers:
17, 8, 4
Now look at the original array and find the elements with these given differences (you should attach the index in the original array to each element in the difference array to make this most easy).
17 - difference between 37 and 20
8 - difference between 16 and 8
4 - difference between 20 and 16
Now print the stuff:
37 | 20 | 16 | 8, 5, 5, 3, 0
I think your problem can be solved using k-clustering using kruskal's algorithm . Kruskal algorithm is used to find the clusters such that there is maximum spacing between them.
Algorithm : -
Construct path graph from your data set like following : -
[37, 20, 16, 8, 5, 5, 3, 0]
path graph: - 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7
then weight for each edge will be difference between their values
edge(0,1) = abs(37-20) = 17
edge(1,2) = abs(20-16) = 4
edge(2,3) = abs(16-8) = 8
edge(3,4) = abs(8-5) = 3
edge(4,5) = abs(5-5) = 0
edge(5,6) = abs(5-3) = 2
edge(6,7) = abs(3-0) = 3
Use kruskal on this graph till there are only k clusters remaining : -
Sort the edges first according to weights in ascending order:-
(4,5),(5,6),(6,7),(3,4),(1,2),(2,3),(0,1)
Use krushkal on it find exactly k = 3 clusters : -
iteration 1 : join (4,5) clusters = 7 clusters: [37,20,16,8,(5,5),3,0]
iteration 2 : join (5,6) clusters = 6 clusters: [37,20,16,8,(5,5,3),0]
iteration 3 : join (6,7) clusters = 5 clusters: [37,20,16,8,(5,5,3,0)]
iteration 4 : join (3,4) clusters = 4 clusters: [37,20,16,(8,5,5,3,0)]
iteration 5 : join (1,2) clusters = 3 clusters: [37,(20,16),(8,5,5,3,0)]
stop as clusters = 3
reconstrusted solution : [(37), (20, 16), (8, 5, 5, 3, 0)] is what
u desired
While #anatolyg's solution may be fine, you should also look at k-means clustering. It's usually done in higher dimensions, but ought to work fine in 1d.
You pick k; your examples are k=2 and k=3. The algorithm seeks to put the inputs into k sets that minimize the sum of distances squared from the set's elements to the centroid (mean position) of the set. This adds a bit of rigor to your rather fuzzy definition of the right result.
While getting an optimal result is NP hard, there is a simple greedy solution.
It's an iteration. Take a guess to get started. Either pick k elements at random to be the initial means or put all the elements randomly into k sets and compute their means. Some care is needed here because each of the k sets must have at least one element.
Additionally, because your integer sets can have repeats, you'll have to ensure the initial k means are distinct. This is easy enough. Just pick from a set that has been "unqualified."
Now iterate. For each element find its closest mean. If it's already in the set corresponding to that mean, leave it there. Else move it. After all elements have been considered, recompute the means. Repeat until no elements need to move.
The Wikipedia page on this is pretty good.
Let's say I have a fixed sized array. I want to fill the array with either 1s or 2s so that all element sums up to X.
Example:
Required sum = 12
Array size = 7
Possible combinations:
array( 1, 2, 2, 2, 1, 2, 2 ) // sums to 12
array( 1, 1, 2, 2, 2, 2, 2 ) // sums to 12
Find the number of 2's in the array, this number is:
#2's = X - array_size
Chose random arbitrary #2's elements (for example the first elements), and
give them the value 2, the rest of the elements will get the value 1.
Note: it is easy to see that if X < array_size or X> 2*array_size there is no solution to the problem (and obviously the above algorithm will fail)
If I have several sets of numbers (just a 2D array where each row is a set):
[ 1, 3, -1, -1]
[ 2, 4, -1, -1]
[ 7, 8, 9, 10]
What would be an algorithm to create a list of sums (ignoring -1's)? the result for the above would be:
1+2+7,
1+2+8,
1+2+9,
1+2+10,
1+4+7,
1+4+8,
1+4+9,
1+4+10,
3+2+7,
3+2+8,
3+2+9,
3+2+10,
3+4+7,
3+4+8,
3+4+9,
3+4+10
For each number in the first list, generate all sums starting with that number and all sums recursively generated by applying the same method to all but the first list. When you have no lists left, that is the base case.
Pseudo-code:
function find_sums(lists):
if lists is empty:
return [""]
sums = []
for n in lists[0]:
if n != -1:
for sum in find_sums(lists from index 1 onwards):
sums.append(n + "+" + sum)
return sums
This is called the Cartesian product.