claytonjwong

#!/bin/bash

git config --global user.name "claytonjwong"
git config --global user.email claytonjwong@gmail.com
git config --global alias.co checkout
git config --global alias.br branch
git config --global alias.ci commit
git config --global alias.st status
git config --global alias.au 'add . -u'
git config --global alias.di diff

claytonjwong

# https://en.wikipedia.org/wiki/Monty_Hall_problem

import random

N = int(1e5)

door = [1, 2, 3]

same = 0
diff = 0

claytonjwong

Create a map m to count the frequency of each implicit character c of the string s.

• Note: The character c is explicit in Rust and C++

---

Kotlin

var m = s.groupingBy{ it }.eachCount()

claytonjwong

Kotlin

var transpose = { A: Array<IntArray> -> A[0].mapIndexed{ j, _ -> A.mapIndexed{ i, _ -> A[i][j] }.toIntArray() }.toTypedArray() }

Javascript

let transpose = A => A[0].map((_, j) => A.map((_, i) => A[i][j]));

claytonjwong

//
// Recursively change each node of the linked list such that next
// is the last node seen during a linear scan of the linked list.
//
class Solution {
    fun reverseList(head: ListNode?): ListNode? {
        fun go(node: ListNode? = head, last: ListNode? = null): ListNode? {
            var next = node?.next
            node?.next = last
            if (next == null)

claytonjwong

//
// Recursively change each node of the linked list such that next
// is the last node seen during a linear scan of the linked list.
//
let reverseList = head => {
    let go = (node = head, last = null) => {
        let next = node.next;
        node.next = last;
        if (!next)
            return node;

claytonjwong

#
# Recursively change each node of the linked list such that next
# is the last node seen during a linear scan of the linked list.
#
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        def go(node = head, last = None):
            next = node.next
            node.next = last
            if not next:

claytonjwong

//
// Recursively change each node of the linked list such that next
// is the last node seen during a linear scan of the linked list.
//
class Solution {
public:
    using fun = function<ListNode*(ListNode*, ListNode*)>;
    ListNode* reverseList(ListNode* head) {
        fun go = [&](auto node, auto last) {
            auto next = node->next;

claytonjwong

fun permutations(A: IntArray): Array<IntArray> {
    var N = A.size
    var perms = mutableListOf<IntArray>()
    fun go(i: Int = 0) {
        if (i == N) {
            perms.add(A.copyOf())
            return
        }
        for (k in i until N) {
            A[i] = A[k].also{ A[k] = A[i] }

claytonjwong

let permutations = A => {
    let N = A.length;
    let perms = [];
    let go = (i = 0) => {
        if (i == N) {
            perms.push([...A]);
            return;
        }
        for (let k = i; k < N; ++k) {
            [A[i], A[k]] = [A[k], A[i]];