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View PersistTheList3.kt
fun main(args: Array<String>) {
val mutableList: MutableList<Int> = mutableListOf(4, 5, 6)
val listNotToBeAmended: List<Int> = mutableList
println("listNotToBeAmended before add: $listNotToBeAmended")
mutableList.add(3)
println("listNotToBeAmended after add to underlying list: $listNotToBeAmended")
}
View PersistTheList2.kt
val mutableList: MutableList<Int> = mutableListOf(4, 5, 6)
View PersistTheList1.kt
val readOnlyList: List<Int> = listOf(1, 2, 3, 4)
View Anagram.fs
module Anagram
//Hackerrank problem
//Anagram
// Sid is obsessed with reading short stories. Being a CS student, he is doing some interesting frequency
// analysis with the books. He chooses strings and in such a way that .
// Your task is to help him find the minimum number of characters of the first string he needs to change to
// enable him to make it an anagram of the second string.
// Note: A word x is an anagram of another word y if we can produce y by rearranging the letters of x.
// Input Format
View PalindromeIndex.fs
module PalindromeIndex
//Hackertank problem
//Palindrome Index
// Given a string, , of lowercase letters, determine the index of the character whose removal will make a
// palindrome. If is already a palindrome or no such character exists, then print . There will always be a
// valid solution, and any correct answer is acceptable. For example, if "bcbc", we can either remove 'b'
// at index or 'c' at index .
// Input Format
// The first line contains an integer, , denoting the number of test cases.
// Each line of the subsequent lines (where ) describes a test case in the form of a single string,
View BeginningHaskellTalkExamples.hs
module BeginningHaskellTalkExamples where
isPalindromeWithRecursion :: String -> Bool
isPalindromeWithRecursion str =
loop strEndIndex where
strEndIndex = length str - 1
loop i =
if (i <= (div strEndIndex 2)) then True
else if (str !! i) /= str !! (strEndIndex - i) then False