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How to reflect a number output in Python?

I have a programm which draws a picture from numbers in certain way
n = int(input(" Введіть ваше число "))
m = n * 2 - 1
pp = " "
i = 0
while m != 0:
l = []
while m > n:
while i < n:
i += 1
j = n - i
k = i
while j != 0:
l.append(pp)
j -= 1
while k != 0:
l.append(str(k))
k -= 1
m -= 1
a = " "
print(a.join(l))
l = []
i = 0
OUTPUT:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
But now I get a task to draw this picture
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Is there any hint how to reflect it without overwriting the whole code?

For the output you are expecting, a very simple way to do it is with this code :
n = 5
for i in range(n):
print(' '.join([str(x+1) for x in range(i+1)]))
Output :
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

You can try this. Its simple.
for i in range(1,n+1):
for j in range(1, i+1):
print(j, end="")
print()

How do you use the range function to count from 0 to 8 by increments of 1 for each iteration in a for loop

I have a triple for loop that creates a 1 row and 2 column collection of numbers starting at 0 0 and going up to 2 2. The third for loop counts from 0 to 8. The code looks as follows:
for N in range(0,3):
for K in range(0,3):
print(N,K)
for P in range(0,9):
print(P)
If you run this code you get the obvious output:
0 0
0
1
2
3
4
5
6
7
8
0 1
0
1
2
3
4
5
6
7
8
0 2
0
1
2
3
4
5
6
7
8
...
And so on. I want instead of the output of 0 to 8 after the N K printout, instead something that looks like:
0 0
0
0 1
1
0 2
2
1 0
3
1 1
4
1 2
5
2 0
6
2 1
7
2 2
8
My first guess was an if statement that said:
if P == Q:
break
where Q was several sets of sums and even the N,K array. However, I couldn't figure out the best way to get my
wanted output. I do think an if statement is the best way to achieve my wanted result, but I'm not quite sure of how to approach it. P is necessary for the rest of my code as it will be used in some subplots.

As this is just an increment by one at each print, you can just do compute the index with N * 3 + K
for N in range(0, 3):
for K in range(0, 3):
print(N, K)
print(N * 3 + K)
CODE DEMO

You can use zip to traverse two iterables in parallel. In this case, one of the iterables is the result of a nested list. That can be handled by using itertools.product, as follows:
import itertools
for (N, K), P in zip(itertools.product(range(3), range(3)), range(9)):
print(N, K)
print(P)

delete first/last occurrence in list python

I need to delete all first occurrence in list.
time limit = 2 s.
memory limit = 256 mb.
Given list a. Some elements in a repeated. If len(a) = 1 print 0.
Input: 1 1 5 2 4 3 3 4 2 5.
Output: 1 3 4 2 5
My solution: For 48 test 22 - good. Else - limited time. How it solve without .count
n = int(input())
A = list(map(int, input().split()))
C = []
if len(A) == 1:
print(0)
else:
for j in range(n):
if A[:j].count(A[j]):
C.append(A[j])
print(len(C))
print(*C)
Pls help

Another way than using sets
First
n = input().split(" ")
u = []
if(len(n) == 1): print(0, end="")
else:
d = {}
for a in n:
if a in d: u.append(int(a))
else: d[a] = 1
print(*u, end="")
# input: 1 1 5 2 4 3 3 4 2 5
# output: 1 3 4 2 5
# input: 5
# output: 0
# input: 1 1 1 1
# output: 1 1 1
Last
n = input().split(" ")
u = []
if(len(n) == 1): print(0, end="")
else:
n.reverse()
d = {}
for a in n:
if a in d: u.append(int(a))
else: d[a] = 1
u.reverse()
print(*u, end="")
# # input: 1 1 5 2 4 3 3 4 2 5
# # output: 1 5 2 4 3
# # input: 5
# # output: 0
# # input: 1 1 1 1
# # output: 1 1 1

To remove duplicates use set()
_ = input()
A = set(map(int, input().split()))
print(*list(A))

Splitting a string of integers with spaces without converting to a list in python

Basically, I need a code that takes an integer and then prints out the string of numbers with the certain range.
For example:
n = 11
1 2 2 3 3 3 4 4 4 4
n = 7
1 2 2 3 3 3 4
a = []
n = int(input())
if n == 0:
print(n)
for i in range(int(n) + 1):
a += [i] * i
a = ' '.join(map(str, a))
print(a[:n])
This does the job but it counts spaces as characters, so I tried to convert it to an int
n = int(n)
print(' '.join(a[:n]))
But when the n >= 47, it starts to print out 10 as 1 0 which is incorrect
I also tried this code
n = int(input())
for i in range(n):
b = (str(i) * i)
print(b, end = ' ')
But I don't understand how to separate the b with spaces without converting the string to a list and printing it in one line either.
I am not sure if it is even possible.

Maybe something like this?
# initialize variables:
number = 11
counter = 0
for i in range(number):
for j in range(i):
# print number:
print('%d ' %i, end='')
# increment counter (tracks how many numbres to print):
counter += 1
# exit inner loop if counter reaches number:
if counter >= number-1: break
# exit outer loop if counter reaches number:
if counter >= number-1: break
Output:
1 2 2 3 3 3 4 4 4 4

Here's a solution using itertools. Generators, chain, repeat and islice are all lazily evaluated, so this uses O(1) space.
>>> n = 7
>>> from itertools import chain, repeat, islice
>>> gen = chain.from_iterable(repeat(i, i) for i in range(1, n+1))
>>> for i in islice(gen, n):
... print(i, end=' ')
...
1 2 2 3 3 3 4

This seems simple. But this does solve the problem?
>>> for i in t:
... if i!=" ": print i
...
1
2
2
3
3
3
4
4
4
4
or even string replace like,
print t.replace(" ","\n")
1
2
2
3
3
3
4
4
4
4

Conway sequence in python

I have an assignement for which I have to write an algorithm who generates the l lines of the Conway sequence of a chosen r integer.
Example for l=8 and r=1 :
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
3 1 2 2 1 1
1 3 1 1 2 2 2 1
1 1 1 3 2 1 3 2 1 1
I have tried to write an algorithm in Python3, but I can't manage to get the correct output. I'd appreciate if you could help me to figure out what's wrong... Anyway, here's what I've wrote so far :
r = int(input())
l = int(input())
cur=str(r).strip()
print(r)
for j in range(l-1):
r=cur
cur=""
i=0
while i<len(r) :
c=1 #counts the consecutive same "numbers"
while ((i+c*2)<len(r)) and (r[i]==r[i+c*2]):
c+=1
cur+=str(c)+" "+r[i]+" "
i+=c+1
cur=cur.strip()
print(cur)
And that's the output I'm getting for l=8 and r=1 :
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
3 1 1 1 2 2 1
1 3 3 1 1 1 2 2 1
1 1 2 3 6 2 2 1
I also feel that my code is pretty ugly so feel free to give your remarks

Really, groupby is built for this:
from itertools import groupby
def lookandsay(i):
digits = str(i) #convert to string
newdigits = []
for k, g in groupby(digits):
newdigits += [sum(1 for _ in g), k]
return int("".join(map(str, newdigits)))
def conway(i):
yield i
while True:
i = lookandsay(i)
yield i
cw01 = conway(1)
for _ in range(8):
print(" ".join(str(next(cw01))))

Look at this logic,
I got below code from: HERE
r = [input()]
l = int(input())
# Write an answer using print
# To debug: print("Debug messages...", file=sys.stderr, flush=True)
for i in range(l - 1):
temp = list()
lastChar = r[0]
counter = 0
for char in r:
if lastChar == char:
counter += 1
else:
temp.extend([str(counter), lastChar])
lastChar = char
counter = 1
temp.extend([str(counter), lastChar])
r = temp
print(r)
print(" ".join(r))