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  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and Anagrams
  5. Discussions

Sherlock and Anagrams

  • saberelharti
     + 0 comments

    Here's the approch that I implement to solve this problem with Kotlin: 

    fun sherlockAndAnagrams(s: String): Int {
    var windowedSize = s.length - 1 // define possible windowed steps for that string
    var listOfWindows : List<List<Char>> = listOf()
    var anagramsCounter = 0
    
    // itterate based on possible windowed steps
    while(windowedSize > 0){
        // get all windowed sets
        listOfWindows = s.toList().windowed(size = windowedSize, step = 1) { it ->
            it.sortedBy { it }
        }
        // count all anagrams
        listOfWindows.forEachIndexed{ index, window ->
            for(j in (index + 1) .. listOfWindows.lastIndex) {
                if(window == listOfWindows[j])  
                    anagramsCounter++
            }
        }
        windowedSize-- // reduce the windowed size
    }
    return anagramsCounter
}