This will return on the first True, meaning iteration will not be complete. Be sure to completely compare little_string.

Think in terms of *slices*. We can measure the length of little_string and then iterate over big_string and compare equal length slices when the first letter of little_string is encountered.

What’s more, the above is comparing a character to a string, not a character.

While this is not the recommended solution, since it is not an algorithm (the preferred solution at this level)…

```
return little_string in big_string
```

The above uses *membership testing* with the `in`

operator. Our algorithm will want to emulate this, but in a loop construct.

```
def contains(haystack, needle):
n = len(needle)
f = needle[0]
for i, x in enumerate(haystack):
if x == f:
if haystack[i:n + i] == needle:
return True
return False
print (contains('superimposed', 'impose')) # True
```

Now if we assume that the `enumerate`

function is as yet unknown to you, then we can revert to `range`

to obtain the index, and the current character.

```
for i in range(len(big_string)):
if big_string[i] == f:
if big_string[i:n + i] == little_string:
```

Of course, the above solutions both assume that you are familiar with slicing. If not, then we need to get more creative using only the concepts currently in your tool belt.

We’re now eight hours later, and have this to consider that uses none of the above, only the `range`

, as suggested.

```
# without slicing or enumerate
def has_in(haystack, needle):
k = range(len(needle))
h = range(len(haystack))
f = needle[0]
r = False
for i in h:
if haystack[i] == f:
for j in k:
if haystack[i + j] == needle[j]:
r = True
continue
else:
r = False
break
if r: return r
return r
print (has_in('superimposed self-imposed', 'impose')) # True
print (has_in('superimposed self-imposed', 'imposter')) # False
```

https://repl.it/@mtf/needle-haystack-found-string-in-string

Above we could technically return the index, as well, but it would be to the first match, not subsequent matches. That will take more logic.

Such as,

```
def has_at(haystack, needle):
k = range(len(needle))
h = range(len(haystack) - k[-1])
f = needle[0]
r = False
s = []
for i in h:
if haystack[i] == f:
for j in k:
if haystack[i + j] == needle[j]:
r = True
continue
else:
r = False
break
if r: s.append(i)
return s or r
print (has_at('superimposed self-imposed impose', 'impose')) # [5, 18, 26]
print (has_at('superimposed self-imposed impose', 'imposter')) # False
```