The golang blog states :
"A slice can also be formed by "slicing" an existing slice or array. Slicing is done by specifying a half-open range with two indices separated by a colon. For example, the expression b[1:4] creates a slice including elements 1 through 3 of b (the indices of the resulting slice will be 0 through 2)."
Can someone please explain to me the logic in the above. IE. Why doesn't b[1:4] reference elements 1 through 4? Is this consistent with other array referencing?
Indexes point to the "start" of the element. This is shared by all languages using zero-based indexing:
| 0 | first | 1 | second | 2 | third | 3 | fourth | 4 | fifth | 5 |
[0] = ^
[0:1] = ^ --------> ^
[1:4] = ^-------------------------------------> ^
[0:5] = ^ ----------------------------------------------------------> ^
It's also common to support negative indexing, although Go doesn't allow this:
|-6 | |-5 | |-4 | |-3 | |-2 | |-1 |
| 0 | first | 1 | second | 2 | third | 3 | fourth | 4 | fifth | 5 |
The reason is given in the Go Language Specification section on Slices.
For a string, array, or slice a, the
primary expression
a[low : high]
constructs a substring or slice. The
index expressions low and high select
which elements appear in the result.
The result has indexes starting at 0
and length equal to high - low.
For convenience, any of the index
expressions may be omitted. A missing
low index defaults to zero; a missing
high index defaults to the length of
the sliced operand.
It's easy and efficient to calculate the length of the slice as high - low.
Half-open intervals make sense for many reasons, when you get down to it. For instance, with a half-open interval like this, the number of elements is:
n = end - start
which is a pretty nice and easy formula. For a closed interval, it would be:
n = (end - start) + 1
which is (not a lot, but still) more complicated.
It also means that for e.g. a string, the entire string is [1, len(s)] which also seems intuitive. If the interval was closed, to get the entire string you would need [1, len(s) + 1].
Go uses half-open intervals for slices like many other languages. In a more mathematical notation, the slice b[1:4] is the interval [1,4) which excludes the upper endpoint.
Related
Let's say I'm traversing through an array and I'm only going to traverse it 1 time. Now say I have a loop iterating through this array which stops once it hits the last index. The loop is searching for the next value that isn't a 0. The operation I am considering is the loop extracting the next non-zero value.
For example my array could be: | 1 | 0 | 0 | 7 | 0 | 0 | 0 | 6 |
In this example my loop would retrieve 1 and do something with it. Then 7 and 6. Once an index is visited it isn't visited again.
So, would this be a O(log n) operation since the extraction of an element is never going to take n iterations?
Big O is based on worst case scenarios, so we'd consider the case where your array looks like | 0 | 0 | 0 | 0 | 0 |.
In this case, it's O(n) because it would traverse each index once.
As example I have next arrays:
[100,192]
[235,280]
[129,267]
As intersect arrays we get:
[129,192]
[235,267]
Simple exercise for people but problem for creating algorithm that find second multidim array…
Any language, any ideas..
If somebody do not understand me:
I'll assume you wish to output any range that has 2 or more overlapping intervals.
So the output for [1,5], [2,4], [3,3] will be (only) [2,4].
The basic idea here is to use a sweep-line algorithm.
Split the ranges into start- and end-points.
Sort the points.
Now iterate through the points with a counter variable initialized to 0.
If you get a start-point:
Increase the counter.
If the counter's value is now 2, record that point as the start-point for a range in the output.
If you get an end-point
Decrease the counter.
If the counter's value is 1, record that point as the end-point for a range in the output.
Note:
If a start-point and an end-point have the same value, you'll need to process the end-point first if the counter is 1 and the start-point first if the counter is 2 or greater, otherwise you'll end up with a 0-size range or a 0-size gap between two ranges in the output.
This should be fairly simple to do by having a set of the following structure:
Element
int startCount
int endCount
int value
Then you combine all points with the same value into one such element, setting the counts appropriately.
Running time:
O(n log n)
Example:
Input:
[100, 192]
[235, 280]
[129, 267]
(S for start, E for end)
Points | | 100 | 129 | 192 | 235 | 267 | 280 |
Type | | Start | Start | End | Start | End | End |
Count | 0 | 1 | 2 | 1 | 2 | 1 | 0 |
Output | | | [129, | 192] | [235, | 267] | |
This is python implementation of intersection algorithm. Its computcomputational complexity O(n^2).
a = [[100,192],[235,280],[129,267]]
def get_intersections(diapasons):
intersections = []
for d in diapasons:
for check in diapasons:
if d == check:
continue
if d[0] >= check[0] and d[0] <= check[1]:
right = d[1]
if check[1] < d[1]:
right = check[1]
intersections.append([d[0], right])
return intersections
print get_intersections(a)
Here is the scenario:
We have several items that are shipped to many stores. We want to be able to allocate a certain quantity of each item to a store based on need. Each of these stores is also associated to a specific warehouse.
The catch is that at the warehouse level, the total quantity of each item must be a multiple of a number (6 for example).
I have already calculated out the quantity needed by each store at store level, but they do not sum up to a multiple of 6 at the warehouse level.
My solution was this using Excel:
Using a SUMIFS formula to keep track of the sum of each item allocated at the warehouse level. Then another MOD(6) formula that calculates the remaining until a multiple of 6. Then my actually VBA code loops through and subtracts 1 (if MOD <= 3) or adds (if MOD > 3) from the store level units needed until MOD = 0 for all rows.
Now this works for me, but is extremely slow even when I have just ~5000 rows.
I am looking for a faster solution, because everytime I subtract/add to units needed, the SUMIFS and MOD need to be calculated again.
EDIT: (trying to be clearer)
I have a template file that I paste my data into with the following setup:
+------+-------+-----------+----------+--------------+--------+
| Item | Store | Warehouse | StoreQty | WarehouseQty | Mod(6) |
+------+-------+-----------+----------+--------------+--------+
| 1 | 1 | 1 | 2 | 8 | 2 |
| 1 | 2 | 1 | 3 | 8 | 2 |
| 1 | 3 | 1 | 1 | 8 | 2 |
| 1 | 4 | 1 | 2 | 8 | 2 |
| 2 | 1 | 2 | 1 | 4 | 2 |
| 2 | 2 | 2 | 3 | 4 | 2 |
+------+-------+-----------+----------+--------------+--------+
Currently the WarehouseQty column is the SUMIFS formula summing up the StoreQty for each Item-Store combo that is associated to the Warehouse. So I guess the Warehouse/WarehouseQty columns is actually duplicated several times every time an Item-Store combo shows up. The WarehouseQty is the one that needs to be a multiple of 6.
Screen updating can be turned OFF to speed up length computations like this:
Application.ScreenUpdating = FALSE
The opposite assignment turns screen updating back on again.
put the data into an array first, rather than cells, then put the data back after you have manipulated it - this will be much faster.
an example which uses your criteria:
Option Explicit
Sub test()
Dim q() 'this is what will be used for the range
Dim i As Long
q = Range("C2:C41") 'put the data into the array - *ALWAYS* 2 dimensions, even if a single column
For i = LBound(q) To UBound(q) ' use this, in case it's a dynamic array - 1 to 40 would have worked here
Select Case q(i, 1) Mod 6 ' calculate remander
Case 0 To 3
q(i, 1) = q(i, 1) - (q(i, 1) Mod 6) 'make a multiple of 6
Case 4 To 5
q(i, 1) = q(i, 1) - (q(i, 1) Mod 6) + 6 ' and go higher in the later numbers
End Select
Next i
Range("D2:D41") = q ' drop the data back
End Sub
Guessing you may find that stopping the screen refresh may help quite a chunk and therefore not need any more suggestions.
Another option would be to reduce your adjustment to a quantity which is divisible by 6 to a number of if statements, depending on the value of mod(6).
You could also address how you sum up the number of a particular item across all stores, using a pivot table and reading the sum totals from there is a lot quicker than using sumifs in a macro
Based on your modifications to the question:
You're correct that you could have huge amounts of replication doing the calculation row by row, as well as adjusting the quantity by a single unit at a time even though you know exactly how many units you need to add / remove from the mod(6) formula.
Could you not create a new sheet with all your possible combinations of product Id and store. You could then use sumifs() for each of these unique combinations and in a final step round up/down at a warehouse level?
So here is an interesting problem in C#. I'm looking for a better way of solving it:
Given a matrix M (not necesarily square) of matches, find the best matching elements. Element i matches elem j by value M(i,j). M(i,j) != M(j,i).
Since #rows != #columns, find the best min(#rows,#columns) matching pairs (i,j).
Basically the problem is to pick the maximum from each row/column such that no row/column is picked twice.
Example:
1 2 3
+---------
a | 10 3 1
b | 12 99 2
c | 20 5 3
d | 5 7 4
The maximum value in this matrix is 99 so the best match is (b,2). For the next selection we cannot use anymore row b and column 2. Is like cutting them
1 2 3 or, if you prefer, 1 3
+--------- a smaller matrix: +------
a | 10 || 1 a | 10 1
b | ===++=== c | 20 3
c | 20 || 3 d | 5 4
d | 5 || 4
The max is now 20 and the match is (c, 1). The remaining matrix has only one column.
After another pick we'll get the match (d, 3) with match = 4
In the end "a" has no match.
My current implementation uses 2 array to store the already matched rows/columns and for each match goes through the entire matrix, picking the first maximum that belongs to row/col not match.
PS: in case of value multiple matches having the same value, just pick one of them
PS2: The array is stored as int [,]
How would you approach this problem in a more optimal/beautiful way?
If you are trying to maximise the sum of the cells chosen, such that exactly one cell is picked from each row and from each column, then this is http://en.wikipedia.org/wiki/Assignment_problem. If your matrix is not square, you can make it square by adding rows or columns to them, with values in the new cells which mean that they won't be picked unless there is no other way to fill out the solution.
(If you are not maximising the sum, you need to say what function of the values chosen you are maximising - is (1,3) better than (2,2)?. Otherwise you are into http://en.wikipedia.org/wiki/Multi-objective_optimization, which is possible, but more complicated).
You could first sort all of the entries of the matrix in descending order, and then process the sorted list. Whenever you see an entry that isn't in an already-picked row/col, it means that entry should be picked, so you mark the corresponding row/column and continue further down the list until either all rows or all columns have been picked.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Given a 2d array sorted in increasing order from left to right and top to bottom, what is the best way to search for a target number?
The following was asked in a Google interview:
You are given a 2D array storing integers, sorted vertically and horizontally.
Write a method that takes as input an integer and outputs a bool saying whether or not the integer is in the array.
What is the best way to do this? And what is its time complexity?
Start at the Bottom-Left corner of the Matrix and follow the rules stated below to traverse the matrix:
The matrix traversal is based on these conditions:
If the input number is greater than current number: Move Right
If the input number is less than current number: Move Up.
If the input number is equal to current number: Return Success
If the input number is not equal to current number and no transition is possible: Return Fail
Time Complexity: (Thanks to Martinho Fernandes)
The time complexity is O(N+M). In the worst case, the element searched for is in the upper-left corner, meaning you'll go up N times, and left M times.
Example
Input matrix:
--------------
| 1 | 4 | 6 |
--------------
| 2 | 5 | 9 |
--------------
| *3* | 8 | 10 |
--------------
Number to search: 4
Step 1:
Start at the cell where you have 3 (Bottom-Left).
3 < 4: Move Right
| 1 | 4 | 6 |
--------------
| 2 | 5 | 9 |
--------------
| 3 | *8* | 10 |
--------------
Step 2:
8 > 4: Move Up
| 1 | 4 | 6 |
--------------
| 2 | *5* | 9 |
--------------
| 3 | 8 | 10 |
--------------
Step 3:
5 > 4: Move Up
| 1 | *4* | 6 |
--------------
| 2 | 5 | 9 |
--------------
| 3 | 8 | 10 |
--------------
Step 4:
4=4: Return the index of the number
I would start by asking details about what it means to be "sorted vertically and horizontally"
If the matrix is sorted in a way that the last element of each row is less than the first element of the next row, you can run a binary search on the first column to find out in what row that number is, and then run another binary search on the row. This algorithm will take O(log C + log R) time, where C and R are, respectively the number of rows and columns. Using a property of the logarithm, one can write that as O(log(C*R)), which is the same as O(log N), if N is the number of elements in the array. This is almost the same as treating the array as 1D and running a binary search on it.
But the matrix could be sorted in a way that the last element of each row is not less than the first element of the next row:
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 10
3 4 5 6 7 8 9 10 11
In this case, you could run some sort of horizontal an vertical binary search simultaneously:
Test the middle number of the first column. If it's less than the target, consider the lines above it. If it's greater, consider those below;
Test the middle number of the first considered line. If it's less, consider the columns left of it. If it's greater, consider those to the right;
Lathe, rinse, repeat until you find one, or you're left with no more elements to consider;
This method is also logarithmic on the number of elements.
The first method that comes to mind is a vertical binary search, followed by a horizontal one when you find the row it should be in. Complexity will be O(log NM) where N and M are the dimensions of the array.
Further explanation:
Consider just the first number of every row. When you perform a binary search of these first numbers for the specified number, the result will be either the specified number if you're lucky, otherwise it will be the position before or after where the specified number would go depending on the binary search implementation. Once you find the two of the first numbers that the specified number should go between, you know that the number is in that row, and a second binary search will find the number if it is in the row.