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By : Beach
Date : October 24 2020, 06:10 PM
seems to work fine This has a closed form solution where it is much easier to count the number of possibilities than to list all of them, so this feels more like a counting problem than a coding problem.
We are going to count how many possibilities we have at each position n and then multiply them together. code : ## how many different sequences that the difference between every adjacent pair of numbers is larger than 1

By : Malky Malk
Date : March 29 2020, 07:55 AM
will help you The answer to the question, "How many different sequences are there..." can be solved without enumerating every combination. Which is a good thing, because 25! is approximately 1.55 x 10^24, or way more than any existing computer is going to enumerate in a second. Or in a year.
This is a math problem, specifically combinatorics. See http://en.wikipedia.org/wiki/Combinatorics and possibly http://en.wikipedia.org/wiki/Combinations for information on how you would go about solving the problem. ## How to count sequences of repeated numbers in Python sequences?

By : Chuck IT
Date : March 29 2020, 07:55 AM
hop of those help? I want to count how many times small sequences of adjacent repeated number different from zero occurs in a given sequence. Let three sequences: , Use itertools.groupby:
code :
``````>>> from itertools import groupby
for k,g in groupby(seq1):
if k != 0:
print list(g)
...
[1, 1]
[2, 2]
[3, 3, 3]

for k,g in groupby(seq2):
if k != 0:
print list(g)
...
[1, 1]
[1, 1]
`````` ## How to count number of (weakly) decreasing sequences of numbers?

By : K.Ramya
Date : March 29 2020, 07:55 AM
this will help It's choose(q-p+N, N) where choose is the binomial coefficient.
Weakly decreasing sequences that are between p and q are in bijection with sequences of 0's and 1's of length q-p+N, where the sequences have exactly N ones. It's obvious that the number of such sequences is choose(q-p+N, N) because that's the number of ways of choosing N things from q-p+N things.
code :
`````` c = q
for x in xs
if x = 1 then output c
if x = 0 then c <- c - 1
``````
`````` c = q
while xs is not empty
output 1
xs <- tail(xs)
else
output 0
c <- c - 1
while c > p
output 0
c <- c - 1
`````` ## Getting a list of numbers from 0 to n, which has no adjacent consecutive numbers and the value of the number is not equa

By : PathFyndar
Date : March 29 2020, 07:55 AM
Does that help Here this works tested in python 3.6
This prints all odd numbers then all even numbers in reverse and if the last even number is equal to the index we interchange n-1 and n
code :
``````n=input("Enter ur number: ")
n = int(n)
l =*n
y=1
z=0
for x in range(n):
if y<=n and x+1<=n/2:
l[x]=y
y=y+2
else:
if(x==z):
l[x]= l[x-1]
l[x-1] = z
z=z+2
else:
l[x]=z
z=z+2
print(l)
`````` ## iterate over array count even numbers, print number of counts and even numbers in SWIFT

By : user2392924
Date : March 29 2020, 07:55 AM
To fix the issue you can do Your goal is to count how much elements matching condition which is in your case if number is even (number % 2 == 0) and then you need to print it.
To achieve this, first get number of these elements which matching condition and then print this number.
code :
``````var matching = 0

for myInt in myArray {

if myInt % 2 == 0 {
matching += 1 // if number matching condition, increase count of matching elements
}

}

print(number)
``````
``````let matching = myArray.filter({ \$0 % 2 == 0 }).count
``````
``````let matching = myArray.reduce(0) { \$1 % 2 == 0 ? \$0 + 1 : \$0 }
``````
``````let matching = myArray.count(where: { \$0 % 2 == 0 })
`````` 