Problem Description:

Given a string, determine if a permutation of the string could form a palindrome.

For example,

"code" -> False, "aab" -> True, "carerac" -> True.

Hint:

1. 1. 1. Consider the palindromes of odd vs even length. What difference do you notice?
      2. Count the frequency of each character.
      3. If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times?

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Just check there are no more than 2 characters that appear an odd number of times in the string.

My C++ code using an array as a hash map is as follows.

1 class Solution {2 public:3     bool canPermutePalindrome(string s) {4         int odd = 0, counts[256] = {
   0};5         for (char c : s)6             odd += ++counts[c] & 1 ? 1 : -1;7         return odd <= 1;8     }9 };

BTW, Stefan has posted many nice solutions , including the following one that uses bitset.

1 class Solution {2 public:3     bool canPermutePalindrome(string s) {4         bitset<256> b;5         for (char c : s) b.flip(c);6         return b.count() < 2;7     }8 };