How do i print a pattern like this

Question

There is a question. How do i print a pattern like this

stackoverflow
stackoverflo
 tackoverflo
 tackoverfl
  ackoverfl
  ackoverf
   ckoverf
   ckover
    kover
    kove
     ove
     ov
      v

I have tried to use for loops but failed...

str = "stackoverflow"
k = len(str)
print(str)
print(str[:(k-1)])

And I don't know how to use for loops to finish it Are there any ways without using for loops to address this problem? Thanks...

Answer 1

Here is another way with recursive function printline(). For loop not needed.

# Recursive function
def printline(string, cutleft, padindex):
    # Print out the required line
    print(string.rjust(padindex+len(string)))

    # Last character to print out
    if len(string) == 1:
        return

    # Deciding to trim the left or right part of the string
    if cutleft:
        printline(string[1:], 0, padindex + 1)
    else:
        printline(string[:-1], 1, padindex)

# Calling the recursive function with initial value
printline('stackoverflow', 0, 0)

This is the output.

stackoverflow
stackoverflo
 tackoverflo
 tackoverfl
  ackoverfl
  ackoverf
   ckoverf
   ckover
    kover
    kove
     ove
     ov
      v

Answer 2

As OP stated that no loops. Does hardcoding count as valid answer?

print(“stackoverflow”)
print(“stackoverflo”)
print(“ tackoverflo”)
print(“ tackoverfl”)
print(“  ackoverfl”)
print(“  ackoverf”)
print(“   ckoverf”)
print(“   ckover”)
print(“    kover)
print(“    kove”)
print(“     ove”)
print(“     ov”)
print(“      v”)

Answer 3

You can use just simple slice operator :

a='stackoverflow'
print(a)

    #printing first whole string
    for i in range(len(a)):
        #loop range of the string

        if i%2==0:
            #here the logic see the problem you will find a pattern that it removing
            #last character when loop number is even and update the string with current
            #sliced string
            #print(i)
            # if you want use this print for understanding track of slicing
            print('{:^12s}'.format(a[:-1]))

            #Removing last character if loop index is even
            a=a[:-1]
            #update the string with current sliced string
        else:
            #print(i)
            #use this print for tracking of sliced string
            print('{:^14s}'.format(a[1:]))
            #remove first character if loop index is not even or loop index is odd.

            a=a[1:]
            #update the current string with sliced string

output:

stackoverflow
stackoverflo
 tackoverflo  
 tackoverfl 
  ackoverfl   
  ackoverf  
   ckoverf    
   ckover   
    kover     
    kove    
     ove      
     ov     
      v  

Answer 4

Well you could use 'some_string'.rjust(width, ' ') where width is an integer value and the second parameter is a string in my example used a blank space. Also you can use 'some_string'.ljust(width, ' '). For more information you should check this site https://www.programiz.com/python-programming/methods/string/rjust

for example:

def word_reduce(word):
   n = word.__len__()
   for i in range(n):
       left = i // 2 
       right = i - left
       result = word[left:n-right]
       print((' ').rjust(left + 1) + result)

s = 'stackoverflow'
word_reduce(s)

Answer 5

You can use the concept of recursion

def stackoverflow(pattern,alternate=0):
    if len(pattern) == 1:
        #base condition for recursion
        return 
    elif alternate == 0:
        #first time execution
        print(pattern)
        alternate = alternate + 1
        stackoverflow(pattern, alternate)
    elif alternate % 2 != 0:
        # truncate from right side
        pattern = pattern[:-1]
        print(pattern)
        alternate = alternate + 1
        stackoverflow(pattern, alternate)
    else:
        #truncate from left side
        pattern = pattern[1:]
        print(pattern)
        alternate = alternate + 1
        stackoverflow(pattern,alternate)

Answer 6

You could try this with only one counter

string = "stackoverflow"
loop=0
while string.replace(' ','')!='':
   print(' '*(loop//2)+string+' '*(loop//2))
   if loop%2==0:
       string=string[:-1]
   else:
       string=string[1:]
   loop=loop+1

Answer 7

Keep track of two index l and r that represent the slice of string to print. Then shorten that slice on each iteration.

s = 'stackoverflow'

l, r = 0, len(s)      # range of string to print
remove_left = True    # which side of the string to remove
space = 0             # how much space to print to the left

while l < r:
    print('%s%s' % (' ' * int(space/2), s[l:r]))

    if remove_left:
        r-= 1
    else:
        l+= 1

    remove_left = not remove_left
    space += 1

Output:

stackoverflow
stackoverflo
 tackoverflo
 tackoverfl
  ackoverfl
  ackoverf
   ckoverf
   ckover
    kover
    kove
     ove
     ov
      v

Answer 8

I suggest using a for loop. They're pretty easy to use once you get used to them. Here's a solution that uses a for loop:

def show(s):
    n = len(s)
    for i in range(n):
        n1 = i // 2
        n2 = i - n1
        print(" " * n1 + s[n1:n-n2])

s = "stackoverflow"
show(s)

The output is:

stackoverflow
stackoverflo
 tackoverflo
 tackoverfl
  ackoverfl
  ackoverf
   ckoverf
   ckover
    kover
    kove
     ove
     ov
      v

If you really don't want to use a for loop, you can replace it with a while loop as follows:

i = 0
while i < n:
    ...
    i += 1

Answer 9

Another possible solution is

s = "stackoverflow"
toggle = True # If true, remove first char. Else, cut last char.
left = 0 # number of spaces to prepend
right = 0 # number of spaces to append

while s: # while s is not empty
    print(' '*left + s + ' '*right)
    if toggle:
        s = s[1:] # remove first char
        left += 1
    else:
        s = s[:-1] # remove last char
        right += 1
    toggle = not toggle

which gives output

stackoverflow
 tackoverflow
 tackoverflo 
  ackoverflo 
  ackoverfl  
   ckoverfl  
   ckoverf   
    koverf   
    kover    
     over    
     ove     
      ve     
      v