Suppose we have a string s, we have to check whether characters of the given string can be shuffled to make a palindrome or not.
So, if the input is like s = "raaecrc", then the output will be True as we can rearrange this to "racecar" which is a palindrome.
To solve this, we will follow these steps −
- freq := a map to store all characters and their frequencies in s
- odd_count := 0
- for each element i in the list of all values of freq, do
- if i is odd, then
- odd_count := odd_count + 1
- if odd_count > 1, then
- return False
- if i is odd, then
- return True
Let us see the following implementation to get better understanding −
Example
from collections import defaultdict def solve(st) : freq = defaultdict(int) for char in st : freq[char] += 1 odd_count = 0 for i in freq.values(): if i % 2 == 1: odd_count = odd_count + 1 if odd_count > 1: return False return True s = "raaecrc" print(solve(s))
Input
"raaecrc"
Output
True