I have to generate an ID using a macro and I'm facing some problems when there are 8 numeric values required.
The ID has to include numeric values for the 1st-8th charactera and the 9th must be alphabetic. From the 10th character onwards it has to be spaces.
These are my codes and I'm certain that there are no issues with the formula
Function GenerateRB()
strLastCharSelections = "X,W,M,L,K,J,E,D,C,B,A"
intNumber = GenerateRandomNumber(1000000, 9999999)
ReDim a(8)
For i = 1 To 8
a(i) = Mid(intNumber, i, 1)
Next
intTotal = a(1) * 9 + a(2) * 8 + a(3) * 7 + a(4) * 6 + a(5) * 5 + a(6) * 4 + a(7) * 3 + a(8) * 2
intRemainder = intTotal Mod 11
arrstrSplitLastCharSelections = Split(strLastCharSelections, ",")
strLastChar = arrstrSplitLastCharSelections(intRemainder)
GenerateRB = intNumber & strLastChar
End Function
The code works when its
ReDim a(7)
For i = 1 To 7
a(i) = Mid(intNumber, i, 1)
Next
intTotal = a(1) * 9 + a(2) * 8 + a(3) * 7 + a(4) * 6 + a(5) * 5 + a(6) * 4 + a(7) * 3
Any help will be appreciated as I'm very new to this, Thank you!
I'm assuming that GenerateRandomnNumber will return a numeric in the specified range - in this case, a seven digit number between 1000000 and 9999999.
So when selecting the ith digit in the statement a(i) = Mid(intNumber, i, 1), there are no issues when i is 1 to 7 - however, when it's 8 - there is no eighth digit, so the code will fail.
Related
I need to loop in a range (in my case column A) to get following results:
A B
1 = 2 - 1
2 = 3 - 1
3 = 3 - 2
4 = 4 - 1
5 = 4 - 2
= 4 - 3
= 5 - 1
= 5 - 2
= 5 - 3
= 5 - 4
I want number 5 to be deducted by 4, 3, 2 ,1 and then number 4 by 3, 2, 1 and so on.
I was somehow able to achieve this with collection but since dataset is rather big script is running 30+ minutes.
At this point I'm trying to figure out arrays but I don't know how to get desired result. My main concern is whether I can loop from bottom to top (From number 5 to 1, not 1 to 5) and how to fixate last row (fix number 5, conduct deductions and then fix number 4, do math magic and loop to 3 and so on).
My current code is:
Dim Arr As Variant
Dim lastc, lastr As Long
lastc = FindColNumber
lastr = ws.Cells(ws.Rows.count, lastc).End(xlUp).Row
Arr = ws.Range(ws.Cells(2, last), ws.Cells(lastr, lastc))
For i = LBound(Arr, 1) To UBound(Arr, 1) - 1
If (Arr(i, 1) > 0) And (Arr(i + 1, 1) > Arr(i, 1)) Then
Arr(i, 1) = Arr(i, 1) - Arr(i + 1, 1)
Code does deduction as following: 5-4, 4-3, 3-2, 2-1 and that's not what I need.
Any tips?
Thank you.
As stated in the comments, you will need two loops and another output array:
Sub lkjlkjkdl()
Dim ws As Worksheet
Set ws = ActiveSheet
Dim Arr As Variant
Dim lastc As Long, lastr As Long
lastc = 1 'FindColNumber
lastr = ws.Cells(ws.Rows.Count, lastc).End(xlUp).Row
Arr = ws.Range(ws.Cells(2, lastc), ws.Cells(lastr, lastc))
Dim cnt As Long
cnt = ((UBound(Arr, 1) - 1) * UBound(Arr, 1)) / 2
Dim k As Long
k = 1
Dim outarr As Variant
ReDim outarr(1 To cnt, 1 To 1)
For i = LBound(Arr, 1) + 1 To UBound(Arr, 1)
Dim j As Long
For j = LBound(Arr, 1) To i - 1
outarr(k, 1) = Arr(i, 1) - Arr(j, 1)
k = k + 1
Next j
Next i
ws.Range("B2").Resize(cnt, 1).Value = outarr
End Sub
I have some data, stored in arrays like
Dim arrA, arrB, arrC, arrAi, arrBi
Dim i as integer, x as integer
for i = 1 to 100
if cells(i,1).value = "criteria" then ' just add value to array when it meets some criteria
x = x + 1
arrA(x) = cells(i,1).value
arrB(x) = cells(i,2).value
arrC(x) = cells(i,3).value
end if
next i
redim preserve arrA(1 to x)
redim preserve arrB(1 to x)
redim preserve arrC(1 to x)
And the data looks like
arrA: 26.1 40.2 80.3 26.0 41.3 78.7 25.8 40.8 80.0
arrB: 10 11 10 66 67 64 32 32 33
arrC: 1 2 3 1 2 3 1 2 3
Since the values in arrA 26.1, 26.0, 25.8 (position 1, 4, 7) belong to group 1 (referencing to values in arrC at same position), I would like to store 26.1 26.0 25.8 to arrAi and 10 66 32 to arrBi for subsequent calculations.
How can I loop through the 3 arrays and store values to another array as described above?
Thanks in advance.
Try the next way, please:
Sub handleArraysFromArrays()
'your existing code...
'but you fistly must declare
Dim arrA(1 To 100), arrB(1 To 100), arrC(1 To 100)
'....
'your existing code
'...
Dim k As Long, kk As Long
ReDim arrAi(1 To UBound(arrA))
ReDim arrBi(1 To UBound(arrA))
For i = 1 To UBound(arrC)
If arrC(i, 1) = 1 Then k = k + 1: arrAi(k, 1) = arrA(i, 1)
If arrC(i, 1) = 2 Then kk = kk + 1: arrBi(kk, 1) = arrA(i, 1)
Next i
ReDim Preserve arrAi(1 To k): ReDim Preserve arrBi(1 To kk)
Debug.Print UBound(arrAi), UBound(arrBi)
End Sub
This question already has answers here:
Find the missing and duplicate elements in an array in linear time and constant space
(8 answers)
Closed 4 years ago.
Given an array of first n natural numbers with one element missing and one duplicate element. figure out both numbers. I was asked this question in one of the interviews.
For example, the sample array is
5, 2, 1, 4, 5
where n=5
Answer for the given array should be:
missing element = 3
duplicate element = 5
Is it possible to come-up with the solution without iterating the array.
I was able to come-up with a solution having complexity O(nlogn), without extra space. Does any O(n) solution exist (without extra space)?
Let's say that you have numbers 1,...,n. Now, denote their sum 1 + 2 + ... + n as S. If we denote the missing number as j and the duplicit one as k, we will get for the modified sum S' = S - j + k, so that's one equation for two variables. We can repeat the same procedure, but this time, summing second powers, thus S_2 = 1 + 4 + ... + n^2. For the sequence with one missing and one duplicit number, the result will be S_2' = S_2 - j*j + k*k. Thus we get two equations for two variables.
In total, we have:
S' = S - j + k
S_2' = S_2 - j*j + k*k
therefore
k - j = S' - S =: a
k*k - j*j = S_2' - S_2 =: b
where we introduced the symbols a and b to simplify the notation.
Now, k*k - j*j = (k - j)*(k + j), thus:
k - j = a
k*k - j*j = a * (k + j) = b
summing both equations gives:
2*k = b/a + a
2*j = b/a - a
For your particular example:
S = 1 + 2 + 3 + 4 + 5 = 15
S_2 = 1 + 4 + 9 + 16 + 25 = 55
For the series with one missing and one duplicit element:
S' = 5 + 2 + 1 + 4 + 5 = 17
S_2' = 25 + 4 + 1 + 16 + 25 = 71
then:
a = S' - S = 2
b = S_2' - S_2 = 16
therefore:
2*k = 8 + 2 = 10
2*j = 8 - 2 = 6
which gives:
k = 5, j = 3
In practice (as noted by #HermanTheGermanHesse), one can obtain closed-form expressions for S as well as S_2 in terms of polynomials in n (so-called Faulhaber's formulae), namely: S = n*(n+1)/2 and S_2 = n*(n+1)*(2*n+1)/6. Therefore it is just sufficient to go once over the input data, accumulate S' and S_2', and use the formulae for k and j given above...
The time complexity of the following code is O(2^n), could you please explain to me why?
int f(int n)
{
if (n == 1)
return 1;
return 1 + f(f(n-1));
}
Thanks in advance
f(n) = n is a solution for this recurence relation
Proof:
f(n) = f(f(n-1)) + 1 = f(n-1) + 1 = (n-1) + 1 = n
Sample VBA code that verifies it:
Sub test()
For i = 1 To 20
Debug.Print i, f(i)
Next
End Sub
Function f(ByVal n As Long) As Long
If n = 1 Then
f = 1
Else
f = 1 + f(f(n - 1))
End If
End Function
Since we have established that f(n) = n, we can conclude that
f(n-1) = n-1
Assuming that it takes An calls to get f(n), An being a recurrence relation:
An = 1 + 2 * An-1
The call to calculate f(n)
The numbers of calls necessary to calculate f(n-1) which will return n-1
then the same number of calls to calculate f(f(n-1)) since we are calling f with n-1 again.
The solution of the recurrence relation is 2^n - 1
Cost complexity is 2^n -1
VBA Code:
Option Explicit
Dim count As Long
Sub test()
Dim i As Integer
For i = 1 To 20
count = 0
f i
Debug.Print i, count
Next
End Sub
Function f(ByVal n As Long) As Long
count = count + 1
If n = 1 Then
f = 1
Else
f = 1 + f(f(n - 1))
End If
End Function
Output:
1 1
2 3
3 7
4 15
5 31
6 63
7 127
8 255
9 511
10 1023
11 2047
12 4095
13 8191
14 16383
15 32767
16 65535
17 131071
18 262143
19 524287
20 1048575
how to multiply two dimensional a(5,3) and b (3,5)
so the arry c( , )= a(5,3) and b (3,5)
the first row of arry a * the columns of arry b then
the second row of arry a * the columns of arry b then
[EDIT] the code of your comment:
Dim arry1(5, 3) As Integer
Dim arry2(3, 5) As Integer
Dim i, j As Integer
For i = 1 To 5
For j = 1 To 3
arry1(i, j) = Int(9 * Rnd + 1)
Next i, j
For i = 1 To 3
For j = 1 To 5
arry2(i, j) = Int(9 * Rnd + 1)
Next i, j
Congrats! it works :)
Some comments:
when you use Dim a, b As Integer then b will be an Integer, but a will be a Variant. It is better to use Dim a As Integer, b As Integer so that both a and b will be Integers
why do you ReDim c(5, 5) As Integer ? You can use Dim c(5, 5) As Integer
Instead of printing to the form I would show the data in a textbox or label control
Dim a, b As Integer
Dim arry1(5, 3) As Integer
Dim arry2(3, 5) As Integer
ReDim c(5, 5) As Integer
Dim i, j, s As Integer
Print ""
Print Space(15); "the first arry"
Print ""
For i = 1 To 5
For j = 1 To 3
arry1(i, j) = Int(9 * Rnd + 1)
Print Space(3); arry1(i, j);
Next j
Print ""
Next i
Print ""
Print Space(15); " the second arry"
Print ""
For i = 1 To 3
For j = 1 To 5
arry2(i, j) = Int(9 * Rnd + 1)
Print Space(3); arry2(i, j);
Next j
Print ""
Next i
Print ""
Print Space(15); " the result"
Print ""
cmd1.Visible = False
For a = 1 To 5
For b = 1 To 5
For j = 1 To 3
s = s + (arry1(b, j) * arry2(j, a))
c(a, b) = s
Next j
s=0
Next b
Next a
For i = 1 To 5
For j = 1 To 5
Print Space(3); c(i, j);
Next j
Print ""
Next i
thank you for your answers i solved the problem by may self and its pleasure me if you tel your openion in my code ... thank you