Python help please! Again, the solution has a number of iterations. However, unl

Question

Python help please!

Again, the solution has a number of iterations. However, unlike the first solution, none of them are nested. The first two iterations used to count the characters are both based on n. The third iteration, comparing the two lists of counts, always takes 26 steps since there are 26 possible characters in the strings. Adding it all up gives us T(n) = 2n + 26 steps. That is O(n). For this part you need to define three variables that are called: number_of_vists 1, number_of_vists 2, number_of_vists 3. Each of these variables should be added to the written function in order to count the number of iterations for each of the loops. In the end, you need to print sum of these variables as the number of steps that you need to execute in order to find two strings are anagrams or not.

Explanation / Answer

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

print(anagramSolution1('abcd','dcba'))
def anagramSolution2(s1,s2):
alist1 = list(s1)
alist2 = list(s2)

alist1.sort()
alist2.sort()

pos = 0
matches = True

while pos < len(s1) and matches:
if alist1[pos]==alist2[pos]:
pos = pos + 1
else:
matches = False

return matches

print(anagramSolution2('abcde','edcba'))

def anagramSolution4(s1,s2):
c1 = [0]*26
c2 = [0]*26

for i in range(len(s1)):
pos = ord(s1[i])-ord('a')
c1[pos] = c1[pos] + 1

for i in range(len(s2)):
pos = ord(s2[i])-ord('a')
c2[pos] = c2[pos] + 1

j = 0
stillOK = True
while j<26 and stillOK:
if c1[j]==c2[j]:
j = j + 1
else:
stillOK = False

return stillOK

print(anagramSolution4('apple','pleap'))
def anagramSolution4(s1,s2):
c1 = [0]*26
c2 = [0]*26

for i in range(len(s1)):
pos = ord(s1[i])-ord('a')
c1[pos] = c1[pos] + 1

for i in range(len(s2)):
pos = ord(s2[i])-ord('a')
c2[pos] = c2[pos] + 1

j = 0
stillOK = True
while j<26 and stillOK:
if c1[j]==c2[j]:
j = j + 1
else:
stillOK = False

return stillOK

print(anagramSolution4('apple','pleap'))

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