Suppose we have a string s, we have to check whether any permutation of s is a palindrome or not.
So, if the input is like s = "admma", then the output will be True, as we can rearrange "admma" to "madam" which is a palindrome.
To solve this, we will follow these steps −
- c := a map holding each individual character count of s
- count := 0
- for each i in list of all values of c, do
- if i is odd, then
- if count is same as 0, then
- count := count + 1
- come out from the loop
- return False
- if count is same as 0, then
- if i is odd, then
- return True
Let us see the following implementation to get better understanding −
Example
from collections import Counter class Solution: def solve(self, s): c = Counter(s) count = 0 for i in c.values(): if i % 2 != 0: if count == 0: count += 1 continue return False return True ob = Solution() s = "admma" print(ob.solve(s))
Input
"admma"
Output
True