String separation in space areas

Question

we have a string with spaces and we have max string length, i.e.

str1 = 'we have a string with spaces and we have max string length'
b=8

And divide string in rows with max b length in row, but we can divide only in space area.

Expected Output:

x='
we have
a string
with
spaces
and we
have max
string
length'

and it should work for any b.
I have tried to make a list from a string and check can two elements of list be less than b(but the can be 3 or 4 etc, so dont know what to do in this case).
!And very important that by the task description b is actually named as ‘len’ , so i believe this is made to prevent using this function…

str1 = 'we have a string with spaces and we have max string length'
def WordSearch(len:int, a:str->int:
    k=[]
    t=str1.split(sep=' ')
    z=str()
    lst1 = ['wp' + str(i) for i in range(len(t))]
    for j in lst1:
        for i in range(len(t)):
            if len(j)+len(t[i])<lenn:
                'wp0'+t[0]+' '

    print(lst1)
WordSearch(len=8,a=str1)

Is it possible to fix code? or i am going completely wrong way….

Asked By: Netbek

||

Source

Answer 1

No need to reinvent the wheel:

import textwrap
output = textwrap.wrap('we have a string with spaces and we have max string length', 8)
output == ['we have', 'a string', 'with', 'spaces', 'and we', 'have max', 'string', 'length']

You can combine the items with "n".join(output), print them, or do anything else that you wish.

Answered By: Bharel

Answer 2

It looks like some algorithm exercise, anyway it just need to take care of corner cases.

str1 = 'we have a string with spaces and we have max string length'
def WordSearch(len:int, a:str):
    k=[]
    first_str = ''
    first_len = 0
    second_str = ''
    second_len = 0
    is_first = True
    for c in a:
        if c == ' ':
            if is_first:
                is_first = False
                continue
            if first_len + second_len + 1 > len:
                k.append(first_str)
                first_str = second_str
                first_len = second_len
                second_str = ''
                second_len = 0
            else:
                first_str = first_str + ' ' + second_str
                first_len += second_len + 1
                second_str = ''
                second_len = 0
        else:
            if is_first:
                first_str += c
                first_len += 1
            else:
                second_str += c
                second_len += 1

    if second_len != 0:
        if first_len != 0:
            if (first_len + second_len + 1 > len):
                k.append(first_str)
                k.append(second_str)
            else:
                k.append(first_str + ' ' + second_str)
        else:
            k.append(second_str)
    elif first_len != 0:
        k.append(first_str)
    result = 'rn'.join(k)
    print(result)
WordSearch(len=8,a=str1)

Answered By: Lamp

Answer 3

If you really want to write it yourself (rather than using textwrap) then:

def reformat(s, b):
    tokens = s.split()
    if len(tokens) > 0 and max(map(len, tokens)) <= b:
        result = [[tokens[0]]]
        for token in tokens[1:]:
            e = result[-1]
            if len(token) + sum(map(len, e)) + len(e) <= b:
                e.append(token)
            else:
                result.append([token])
        return [' '.join(e) for e in result]
s = 'we have a string with spaces and we have max string length'
print(*reformat(s, 8), sep='n')

Output:

we have
a string
with
spaces
and we
have max
string
length

Answered By: Stuart

String separation in space areas

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