def anagramSolution1(s1,s2): alist = list(s2) pos1 = 0 stillOK = True while pos1 < len(s1) and...

def anagramSolution1(s1,s2):
alist = list(s2)

pos1 = 0
stillOK = True

while pos1 < len(s1) and stillOK:
pos2 = 0
found = False
while pos2 < len(alist) and not found:
if s1[pos1] == alist[pos2]:
found = True
else:
pos2 = pos2 + 1

if found:
alist[pos2] = None
else:
stillOK = False

pos1 = pos1 + 1

return stillOK

include all operations and calculate the Big-O and the values for c and n₀.

Expert Solution

For the better understanfing of the complexity I will explain the complexity of each line one by one.

This operation is just an assignment operation which will take constant time means O(1)
similar to line this will also take O(1) time
Now let's consider the length of s1 is n, now because this loop is running n times the operation which are taking place inside this loop of O(1) time complexity will take O(n) time after n run.
Now let's consider the length of alist is m, since this is also a loop running m number of times inside a loop which is running n number of times time complexity for this loop will be O(m*n) time.
All the other statements inside this loop are either if or else are taking O(1) time. So overall complexity of this program will be O(m*n) time.

The order of times complexity of any algorithm is always the statement which is taking largest amount of time, because our focus is always on the worst case scenario.

We know that f(n)= O(g(n)) iff f(n)<=c*g(n) where c is a positive constant with c>=1

So time coopmlexity T(n) of the given code is

T(n)=O(m*n)., where m and n are length of strings list and s1 respectively. Value of c=1 and n₀=1

venereology answered 9 months ago

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