Let's say I have an array with 5 elements. How can I calculate all possible repetitive permutations of this array in C.
Edit: What I mean is creating all possible arrays by using that 5 number. So the positon matters.
Example:
array = [1,2,3,4,5]
[1,1,1,1,1]
[1,1,1,1,2]
[1,1,1,2,3]
.
.
A common way to generate combinations or permutations is to use recursion: enumerate each of the possibilities for the first element, and prepend those to each of the combinations or permutations for the same set reduced by one element. So, if we say that you're looking for the number of permutations of n things taken k at a time and we use the notation perms(n, k), you get:
perms(5,5) = {
[1, perms(5,4)]
[2, perms(5,4)]
[3, perms(5,4)]
[4, perms(5,4)]
[5, perms(5,4)]
}
Likewise, for perms(5,4) you get:
perms(5,4) = {
[1, perms(5,3)]
[2, perms(5,3)]
[3, perms(5,3)]
[4, perms(5,3)]
[5, perms(5,3)]
}
So part of perms(5,5) looks like:
[1, 1, perms(5,3)]
[1, 2, perms(5,3)]
[1, 3, perms(5,3)]
[1, 4, perms(5,3)]
[1, 5, perms(5,3)]
[2, 1, perms(5,3)]
[2, 2, perms(5,3)]
...
Defining perms(n, k) is easy. As for any recursive definition, you need two things: a base case and a recursion step. The base case is where k = 0: perms(n, 0) is an empty array, []. For the recursive step, you generate elements by prepending each of the possible values in your set to all of the elements of perms(n, k-1).
If I get your question correctly, you need to generate all 5 digit numbers with digits 1,2,3,4 and 5. So there is a simple solution - generate all numbers base five up to 44444 and then map the 0 to 1, 1 to 2 and so on. Add leading zeros where needed - so 10 becomes 00010 or [1,1,1,2,1].
NOTE: you don't actually have to generate the numbers themselves, you may just iterate the numbers up to 5**5(excluding) and for each of them find the corresponing sequence by getting it's digits base 5.
int increment(size_t *dst, size_t len, size_t base) {
if (len == 0) return 0;
if (dst[len-1] != base-1) {
++dst[len-1];
return 1;
} else {
dst[len-1] = 0;
return increment(dst, len-1, base);
}
}
Armed with this function you can iterate over all repetitive permutations of (0 ... 4) starting from {0, 0, 0, 0, 0}. The function will return 0 when it runs out of repetitive permutations.
Then for each repetitive permutation in turn, use the contents as indexes into your array so as to get a repetitive permutation of the array rather than of (0 ... 4).
In your given example, each position could be occupied by either 1, 2, 3, 4, 5. As there are 5 positions, the total number of possibilities = 5 * 5 * 5 * 5 * 5 = 5 ^ 5 = 3125. In general, it would be N ^ N. (where ^ is the exponentiation operator).
To generate these possibilities, in each of the positions, put the numbers 1, 2, 3, 4, 5, one by one, and increment starting from the last position, similar to a 5 digit counter.
Hence, start with 11111. Increment the last position to get 11112 ... until 11115.
Then wrap back to 1, and increment the next digit 11121 continue with 11122 ... 11125, etc. Repeat this till you reach the first position, and you would end at 55555.
Related
For an numpy 1d array such as:
In [1]: A = np.array([2,5,1,3,9,0,7,4,1,2,0,11])
In [2]: A
Out[2]: array([2,5,1,3,9,0,7,4,1,2,0,11])
I need to split the array by using the values as a sub-array length.
For the example array:
The first index has a value of 2, so I need the first split to occur at index 0 + 2, so it would result in ([2,5,1]).
Skip to index 3 (since indices 0-2 were gobbled up in step 1).
The value at index 3 = 3, so the second split would occur at index 3 + 3, and result in ([3,9,0,7]).
Skip to index 7
The value at index 7 = 4, so the third and final split would occur at index 7 + 4, and result in ([4,1,2,0,11])
I'm using this simple array as an example, because I think it will help in my actual use case, which is reading data from binary files (either as bytes or unsigned shorts). I'm guessing that numpy will be the fastest way to do it, but I could also use struct/bytearray/lists or whatever would be best.
I hope this makes sense. I had a hard time trying to figure out how best to word the question.
Here is an approach using standard python lists and a while loop:
def custom_partition(arr):
partitions = []
i = 0
while i < len(arr):
pariton_size = arr[i]
next_i = i + pariton_size + 1
partitions.append(arr[i:next_i])
i = next_i
return partitions
a = [2, 5, 1, 3, 9, 0, 7, 4, 1, 2, 0, 11]
b = custom_partition(a)
print(b)
Output:
[[2, 5, 1], [3, 9, 0, 7], [4, 1, 2, 0, 11]]
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.
Say I have an array of numbers, e.g. [0, 1, 2, 3, 4, 5] and I want to end up with an array, e.g. [2, 1, 4, 0, 5, 3]. At my disposal, I have a single method that I can use:
move(fromIndex, toIndex)
Thus, to achieve my desired array, I could call the method a number of times:
move(2, 0); // [2, 0, 1, 3, 4, 5]
move(1, 2); // [2, 1, 0, 3, 4, 5] (swapped 2 with 0)
move(4, 2); // [2, 1, 4, 0, 3, 5]
move(3, 4); // [2, 1, 4, 3, 0, 5] (swapped 4 with 0)
move(4, 3); // [2, 1, 4, 0, 3, 5] (swapped 0 with 3)
move(5, 4); // [2, 1, 4, 0, 5, 3] (swapped 5 with 3)
Thus, I also have a list of move() operations to achieve my desired result. The list of move() operations can possibly be reduced in size by changing the order and the indexes, to end up with the same result.
Is there an algorithm that I can use on my list of move() operations to reduce the number of operations to a minimum?
We can create a graph with an element pointing towards the number it needs to be swapped with to get to the desired position. Hence we will get multiple graphs with possible cycles. In your particular case,we will get
2->0->3->5->4->2 (first and last elements denote a cycle)
This means that 2 wants to be swapped with 0 to get to the desired position. Similarly,0 wants to be swapped with 3 to get to the desired position. Notice that 1 does not want to be swapped.
Now, we can swap two adjacent elements of the graph to reduce the graph size by 1. Say we swap 3 and 5 so now the arr = [0,1,2,5,4,3]. Now 3 is in desired state so we can remove it from graph
2->0->5->4->2
We need to repeat this process (m-1) times to remove the graph completely. Here m represents the number of edges of the graph. We can have multiple disconnected graphs or graphs without cycles. We need to make sure that we are swapping elements from the same graph. The final answer would be the sum of all steps (that is m-1 for each component) of the graph.
I have a numpy array, I'd like to take the 3 numbers in each row, minus them from the next row and store those values in another array.
something like
for i in array:
a = i - i+1
I know this is very wrong, but at least this gives the idea of what I want.
Obviously i+1 will just result in the value + 1 and then all I have is a = 1,1,1
When I say i+1 I mean the next in line.
So for example:
input = np.array([[4,4,5], [2,3,1],[1,2,0]])
output = np.array([2,1,4],[1,1,1]) etc....
What would be the best way to do this iteratively on thousands of rows?
IIUC, instead of looping, you can just shift your arrays 1 up using np.roll, subtract that from your original input, and take all the resulting arrays except the last (because there will be nothing to subtract from the last array):
>>> inp = np.array([[4,4,5], [2,3,1],[1,2,0]])
>>> inp
array([[4, 4, 5],
[2, 3, 1],
[1, 2, 0]])
>>> (inp - np.roll(inp,-1,axis=0))[:-1]
array([[2, 1, 4],
[1, 1, 1]])
Or, a more straightforward way would just be to use numpy indexing:
>>> inp[:-1] - inp[1:]
array([[2, 1, 4],
[1, 1, 1]])
Given an array, the output array consecutive elements where total sum is 0.
Eg:
For input [2, 3, -3, 4, -4, 5, 6, -6, -5, 10],
Output is [3, -3, 4, -4, 5, 6, -6, -5]
I just can't find an optimal solution.
Clarification 1: For any element in the output subarray, there should a subset in the subarray which adds with the element to zero.
Eg: For -5, either one of subsets {[-2, -3], [-1, -4], [-5], ....} should be present in output subarray.
Clarification 2: Output subarray should be all consecutive elements.
Here is a python solution that runs in O(n³):
def conSumZero(input):
take = [False] * len(input)
for i in range(len(input)):
for j in range(i+1, len(input)):
if sum(input[i:j]) == 0:
for k in range(i, j):
take[k] = True;
return numpy.where(take, input)
EDIT: Now more efficient! (Not sure if it's quite O(n²); will update once I finish calculating the complexity.)
def conSumZero(input):
take = [False] * len(input)
cs = numpy.cumsum(input)
cs.insert(0,0)
for i in range(len(input)):
for j in range(i+1, len(input)):
if cs[j] - cs[i] == 0:
for k in range(i, j):
take[k] = True;
return numpy.where(take, input)
The difference here is that I precompute the partial sums of the sequence, and use them to calculate subsequence sums - since sum(a[i:j]) = sum(a[0:j]) - sum(a[0:i]) - rather than iterating each time.
Why not just hash the incremental sum totals and update their indexes as you traverse the array, the winner being the one with largest index range. O(n) time complexity (assuming average hash table complexity).
[2, 3, -3, 4, -4, 5, 6, -6, -5, 10]
sum 0 2 5 2 6 2 7 13 7 2 12
The winner is 2, indexed 1 to 8!
To also guarantee an exact counterpart contiguous-subarray for each number in the output array, I don't yet see a way around checking/hashing all the sum subsequences in the candidate subarrays, which would raise the time complexity to O(n^2).
Based on the example, I assumed that you wanted to find only the ones where 2 values together added up to 0, if you want to include ones that add up to 0 if you add more of them together (like 5 + -2 + -3), then you would need to clarify your parameters a bit more.
The implementation is different based on language, but here is a javascript example that shows the algorithm, which you can implement in any language:
var inputArray = [2, 3, -3, 4, -4, 5, 6, -6, -5, 10];
var ouputArray = [];
for (var i=0;i<inputArray.length;i++){
var num1 = inputArray[i];
for (var x=0;x<inputArray.length;x++){
var num2 = inputArray[x];
var sumVal = num1+num2;
if (sumVal == 0){
outputArray.push(num1);
outputArray.push(num2);
}
}
}
Is this the problem you are trying to solve?
Given a sequence , find maximizing such that
If so, here is the algorithm for solving it:
let $U$ be a set of contiguous integers
for each contiguous $S\in\Bbb Z^+_{\le n}$
for each $\T in \wp\left([i,j)\right)$
if $\sum_{n\in T}a_n = 0$
if $\left|S\right| < \left|U\left$
$S \to u$
return $U$
(Will update with full latex once I get the chance.)