Palindrome Permutation

Description

null

Solution(javascript)

/**
 * @param {string} s
 * @return {boolean}
 */
const canPermutePalindrome = function (s) {
  const map = {}
  let oddCount = 0
  for (const c of s) {
    map[c] = map[c] || 0
    map[c] += 1
    if (map[c] % 2 === 0) {
      oddCount -= 1
    } else {
      oddCount += 1
    }
  }

  return oddCount <= 1
}