return outside the function - python-3.x

Question

there is a syntax error that says that i am returning outside the function
I tried all the possible solutions that people used on the same problem with there code and i still get the same error
the error is in line 32 in the line return result
Problem Introduction
In this problem, we ask a simple question: how many times one string occurs as a substring of another? Recall that di erent occurrences of a substring can overlap with each other. For example, ATA occurs three times in CGATATATCCATAG.
Problem Description
Task. Find all occurrences of a pattern in a string.
Input Format. Strings 𝑃𝑎𝑡𝑡𝑒𝑟𝑛 and 𝐺𝑒𝑛𝑜𝑚𝑒.
Constraints. 1 ≤ |𝑃𝑎𝑡𝑡𝑒𝑟𝑛| ≤ 106; 1 ≤ |𝐺𝑒𝑛𝑜𝑚𝑒| ≤ 106; both strings are over A, C, G, T.
Sample 1.
Input:
Output:
1 12
4 1.5 2
4 4 24
TACG GT
Explanation:
The pattern is longer than the text and hence has no occurrences in the text.
Sample 2.
Input:
Output:
02
Explanation:
The pattern appears at positions 1 and 3 (and these two occurrences overlap each other).
Sample 3.
Input:
Output:
139
Explanation:
The pattern appears at positions 1, 3, and 9 in the text.
ATA ATATA
ATAT
GATATATGCATATACTT
# python3
import sys
def compute_prefix_function(pattern):
s = [0] * len(pattern)
border = 0
for i in range(1, len(pattern)):
while (border > 0) and (pattern[i] != pattern[border]):
border = s[border - 1]
if pattern[i] == pattern[border]:
border = border + 1
else:
border = 0
s[i] = border
return s
def find_pattern(pattern, text):
S = pattern + '$' + text
s = compute_prefix_function(S)
result = []
p = len(pattern)
for i in range(p + 1, len(S)):
if s[i] == p:
result.append(i - 2 * p)
return result
if __name__ == '__main__':
pattern = sys.stdin.readline().strip()
text = sys.stdin.readline().strip()
result = find_pattern(pattern, text)
print(" ".join(map(str, result)))

Answer 1

Realize that white space/indentation matters in python, and replaces the { }, used in most other language to define a scope. Something like this is a proper function (although, the return isn't really necessary):
def foo():
print("bar")
return
Doing this instead would result in an error message similar to your, though, because the return now is outside the scope of the function:
def foo():
print("bar")
return
For more info just Google for python indentation tutorial.