iOS Interview - Leetcode 409. Longest Palindrome

May 25, 2024 • 1 min read • #swift, #interview, #leetcode

Primary idea: Count number of character with even and odd frequencies. Adjust the final result depending on if there is character with odd frequencies.
Time Complexity: O(n)
Space Complexity: O(n)

func longestPalindrome(_ s: String) -> Int {
    // Dictionary storing occurence frequencies of characters in the the string
    var frequencies = [Character: Int]()

    // Calculate the occurence frequencies for all characters
    for ch in s { frequencies[ch, default: 0] += 1 }

    var result = 0
    var hasOdd = false

    // Count characters with even and odd frequencies
    // adjust hasOdd flag if there is character with odd frequencies
    for (_, el) in frequencies.enumerated() {
        if el.value % 2 == 0 {
            result += el.value
        } else {
            hasOdd = true
            result += el.value - 1
        }
    }

    return result + (hasOdd ? 1 : 0)
}