Sherlock anagrams - Swift solution

sherlock.swift

// Given a string return a number of all possible anagrams

func sherlockAndAnagrams(s: String) -> Int {
    var anagramCount = 0
    var frequencyMap = [String: Int]()

    // Iterate through all possible substrings
    for i in 0..<s.count {
        for j in i+1..<s.count + 1 {
            let substring = String(s[s.index(s.startIndex, offsetBy: i)..<s.index(s.startIndex, offsetBy: j)])
            let sortedSubstring = String(substring.sorted())

            // Increment the count for the sorted substring in the frequency map
            frequencyMap[sortedSubstring, default: 0] += 1
        }
    }

    // Count the number of anagram pairs using the frequency map
    for count in frequencyMap.values {
        anagramCount += (count * (count - 1)) / 2
    }

    return anagramCount
}