Which of the following is not a feature of Hootsuite?
1 answer:
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Answer:
See explaination
Explanation:
def one_diff(s1,s2):#mthod return true if s1 and s2 has one difference
l1=list(s1)#store characters to list
l2=list(s2)
c=0
if len(l1)!=len(l2):
return False
for i in l1:
if i in l2:#for each character if it is in l1 and l2 continue
continue
else:
c+=1
if c==1:
return True#ret True if cis 1 else False
return False
def exchange(s,*k):#method that exchanges characters
l=list(k)
if len(l)!=0:#check validity of arguments
for i in l:
if i<0 and i+(2*i)>len(s):
res=s
elif i>len(s):
res=s
return res
if len(l)==0:#if no indexes are given
c1=s[0]
c2=s[-1]
res=c2
res=res+s[1:]
res=res[0]+res[1:].replace(c2,c1)
return (res)
elif len(l)==1:#if one index is given
c1=s[l[0]]
c2=s[-1]
res=''
for i in s:
if i==c1 or i==c2:
if i==c2:
res+=c1
else:
res+=c2
else:
res+=i
return (res)
else:#if two indexes are given
c1=s[l[0]]
c2=s[l[1]]
res=''
for i in s:
if i==c1 or i==c2:
if i==c2:
res+=c1
else:
res+=c2
else:
res+=i
return (res)
def is_chain(*k):#method that checks words for one difference
l=list(k)
if len(l)==0:
return False
if len(l)==1:
return True
if len(l)>1:
for i in range(len(l)-1):#for two words in l
if one_diff(l[i],l[i+1]):#call one_diff method
continue
else:
return False
return True
def scramble(s):#method that exchanges characters
i=0
while i<len(s)-1:
s=exchange(s,i,i+1)
i+=2
print(s)
please check out the screenshots in attachment.
