StephenClarkApps

[
{
"title": "Lemon cheesecake",
"desc": "A cheesecake made of lemon",
"image": "https://s3-eu-west-1.amazonaws.com/s3.mediafileserver.co.uk/carnation/WebFiles/RecipeImages/lemoncheesecake_lg.jpg"
},
{
"title": "victoria sponge",
"desc": "sponge with jam",
"image": "https://www.bbcgoodfood.com/sites/bbcgoodfood.com/files/recipe_images/recipe-image-legacy-id--1001468_10.jpg"

StephenClarkApps

package com.somepackage;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

public class Main {
  public static void main (String[] args) {
    List<Integer> numbers = Arrays.asList(1, 4, 10, 15);
    List<Integer> result2 = numbers.stream() // convert list to stream
      .filter(num -> (num%2 ==0))            // only results divisible by 0

StephenClarkApps

// square all elements of an array using map
let values = [2.0, 3.0, 5.0, 8.0]
let squares = values.map {$0 * $0}
print(squares) // [4.0, 9.0, 25.0, 64.0]

StephenClarkApps

let fahrenheit = []

let celsius = fahrenheit.map({
  (value: Int) -> Float in
  return ((Float(5)/9) * Float(Float(value - 32))
})
// [37.77778, 48.88889, 32.22223]

StephenClarkApps

/* Example of using the golden ratio, which 
is basically binet's formula to find a fib number */

import Foundation

let sqrt_of_5 = sqrt(Double(5))
let Φ: Double = ((1 + sqrt(5.0)) / 2.0)

func fibonacciBinet(_ index: Int) -> Int {
  

StephenClarkApps

#!/bin/python3
from functools import lru_cache

"""good method to find fib for n using python 3 lru_cache"""
@lru_cache(maxsize = 1000)
def fibonacciLRU(n):
  if n == 1:
    return 1
  elif n == 2:
    return 1

StephenClarkApps

// Memoized recursive implementation in Swift
var fibCache: [Int: Int] = [:] // This is a Swift Dictionary

func fibonacciAtIndex(_ index: Int) -> Int {
  if let fibonacci = fibCache[index] {
    return fibonacci
  }
  guard index >= 2 else {
    fibCache[index] = index
    return index

StephenClarkApps

#!/bin/python3
from functools import lru_cache

"""method to find fib for n using python lru cache"""
@lru_cache(maxsize = 1000)
def fibonacciLRU(n):
  if n == 1:
    return 1
  elif n == 2:
    return 1

StephenClarkApps

fibCache = {}
"""return the nth term of the sequence using explicit memoization (start with 1)"""
def fibonacciWithCache(n):
  # If we have cached the value then we return it
  if n in fibCache:
    return fibCache[n]
  # Compute the Nth term
  if n == 1:
    value = 1
  if n == 2:

StephenClarkApps

"""return the nth term of a fib sequence using a naive method (start with 0)"""
def fibonacci_modern (n):
    if n == 1:
        return 0
    elif n == 2:
        return 1
    elif n > 2:
        return fibonacci_modern(n - 1) + fibonacci_modern(n - 2)

"""return the nth term of a fib sequence using a naive method (start with 1)"""