Question

Consider that two strings are perms of one another if the same number of every character is present in both. Thus the two strings ‘aabcc’ and ‘ccbaa’ are perms, for each has two a’s, one b and 2 c’s. Another example: the strings ‘xyyz’ and ‘yxz’ are not perms, for one has two y’s and the other has only one. Take two lists of strings. Find how many members of the first have a perm in the second using Phyton program.

Answer 1

Source Code::-

def perms(str1,str2):

#perms() which takes two string as an argument.

if len(str1) == len(str2):

#First will check the length of two string, if length is equal then only will match character

for char in str1:

#Looping one by one character

if (str1.count(char)) != (str2.count(char)):

#If character count is not equal then returns False

return False

return True

else:

return False

if __name__=="__main__":

list1 = ["aabbcc","abc","xyz","pqrst","aabcc"] #Created list1 which has five items

list2 = ["aabbcc","xxyz","cba","pqqrst","ccbaa"] ##Created list2 which has five items

count = 0 #Initalize the count variable

for item1 in list1:

   #looping item from list1

   for item2 in list2:

   #Looping item from list2

   if not(perms(item1,item2)):

#calling perms() method which will find the perms between two words.

#If method is returning False, means that two words are not perms.

pass

   else:

   #If perms() method returns True, means that two words are perms.

   count+=1 #Increament the count variable.

print("No of perms found between two list is::%d"%count)

Output::-

No of perms found between two list is::3

Screen Shot::-

Output screen shot::-