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For a string let the unique set of characters that
occur in it one or more times be C. Considers a
permutation of the elements of C as (c1,c2,c3...). Let
f(c) be the number of times c occurs in s.
if any such permutation of the elements of the elements of
C satifies f(c) = f(ci-1)+ f(ci-2) for all i>=3, the
string is said to be dynamic string.
Note that if the number of distinct characters in the
characters in the sting is less than 3, i.e if C|<3, then
the is always dynamic.
occur in it one or more times be C. Considers a
permutation of the elements of C as (c1,c2,c3...). Let
f(c) be the number of times c occurs in s.
if any such permutation of the elements of the elements of
C satifies f(c) = f(ci-1)+ f(ci-2) for all i>=3, the
string is said to be dynamic string.
Note that if the number of distinct characters in the
characters in the sting is less than 3, i.e if C|<3, then
the is always dynamic.
1 expert answers available & 3 regular answers available
This questions look tricky, but its very simple
REMEMBER, you no need to use the full question for getting the output ,
Explaination :- In this question , they said that 'C' stores number of unique set of characters in the string.
For output the condition is given , if C less than 3 then print "Dynamic"
else print "Not"
Basically , you need to count the number of unique set of characters and check/compare with the conditions
Take a look at the python code for better understanding
** variable 't' in code is number of test case **
Certified
python code:-
t=int(input())
a=[]
for i in range(t):
s=str(input())
for j in s:
if j not in a:
a.append(s)
if len(a)<3:
print('Dynamic')
else:
print('Not')
t=int(input())
a=[]
for i in range(t):
s=str(input())
for j in s:
if j not in a:
a.append(s)
if len(a)<3:
print('Dynamic')
else:
print('Not')
t=int(input())
a=[]
for i in range(t):
s=str(input())
for j in s:
if j not in a:
a.append(s)
if len(a)<3:
print('Dynamic')
else:
print('Not')