soru.js

/**
 * The first, reverseArray, takes an array as argument and
 * produces a new array that has the same elements in
 * the inverse order
 * **/
// input
// [1, 3, 4, 5, 6, 7, 9]

// output 1
// [9, 7, 6, 5, 4, 3, 1]

// output
// [9, 3, 4, 5, 6, 7, 1]
// [9, 7, 4, 5, 6, 3, 1]
// [9, 7, 6, 5, 4, 3, 1]

// input
// [1, 3, 4, 5, 6, 7]
// [7, 3, 4, 5, 6, 1]  i=0 arr[0] <==> arr[5]
// [7, 6, 4, 5, 3, 1]  i=1 arr[1] <==> arr[4]
// [7, 6, 5, 4, 3, 1]  i=2

// Big O notation
// N sayili dizi icin O(N)
const reverseArray = (arr) => {
  const newArray = [];
  for (let i = arr.length - 1; i > 0; i--) {
    newArray.push(arr[i]);
  }
  return newArray;
};

// Big O notation
// N sayili dizi icin; O(N/2)
const reverArrayInPlace = (arr) => {
  const halfTheArrayCount = arr.length / 2;
  for (let i = 0; i < halfTheArrayCount; i++) {
    const temp = arr[i];
    arr[i] = arr[arr.length - i];
    arr[arr.length - i] = temp;
  }
  return arr;
};

/**
 * Write a range function that takes two arguments, start and end,
 * and returns an array
 * containing all the numbers from start up to
 * (and including) end.
 */

const range = (start, end) => {
  const array = [];
  for (let i = start; i <= end; i++) {
    array.push(i);
  }
  return array;
};

/**
 * Next, write a sum function that takes an array of numbers and
 * returns the sum of these numbers.
 * Run the example program and see whether it does indeed return 55.
 */

const sum = (array) => {
  let totalSum = 0;
  for (let i = 0; i < array.length; i++) {
    totalSum = totalSum + array[i];
  }
  return totalSum;
};

console.log(sum(range(1, 10)));