---
tags: leetcode
---

# [266. Palindrome Permutation](https://leetcode.com/problems/palindrome-permutation/)
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# My Solution
## The Key Idea for Solving This Coding Question
If a string `s` can become a palindrome, there is exactly one character with odd number of occurences in `s`.
## C++ Code

class Solution {
public:
    bool canPermutePalindrome(string s) {
        vector<int> charCnt(26, 0);
        for (auto &c : s) {
            ++charCnt[c - 'a'];
        }

        int oddCnt = 0;
        for (int i = 0; i < charCnt.size(); ++i) {
            if (charCnt[i] & 0x1) {
                ++oddCnt;
                if (oddCnt > 1) {
                    return false;
                }
            }
        }

        return true;
    }
};

## Time Complexity
$O(n)$
$n$ is the length of `s`.
## Space Complexity
$O(1)$
# Miscellaneous
<!--
# Test Cases

"code"

"aab"

"carerac"

"as"

-->