/rwc

## Rendezvous with Cassidoo Solutions

Solutions to the interview question of the week from the [rendezvous with cassidoo newsletter](https://buttondown.email/cassidoo)

Solutions in the following languages:
- TypeScript
- Elixir

Inspired and forked from [jda0](https://gist.github.com/jda0/01070831825ed63efcd1f626653a16a3)

/**
 *
 * Sort an array of strings based on the number of distinct characters that
 * occur in the word (followed by the length of the word).
 *
 * $ charNumSort([“Bananas”, “do”, “not”, “grow”, “in”, “Mississippi”])
 * $ do in not Mississippi Bananas grow
 *
 *
 */

const charNumSort = (input: Array<string>) =>
  input
    .sort((first, second) => {
      // use Set() to find unique char count
      const compare = [...new Set(first)].length - [...new Set(second)].length;

      return compare ? compare : second.length - first.length;
    })
    .join(" ");

console.log(charNumSort(["Bananas", "do", "not", "grow", "in", "Mississippi"]));

// https://codesandbox.io/s/modest-sanderson-wxs33?file=/src/index.ts

/**
 *
 * Given an array of random integers, move all the zeros in the array to the end
 * of the array. Try to keep this in O(n) time (or better)! Example:
 * $ moveZeros([1, 2, 0, 1, 0, 0, 3, 6])
 * $ [1, 2, 1, 3, 6, 0, 0, 0]
 *
 */

const moveZeros = (arr: Array<number>) => {
  let numberOfZeroes = 0;
  const result = arr.reduce((agg, curr) => {
    curr ? agg.push(curr) : numberOfZeroes++;
    return agg;
  }, [] as number[]);

  return [...result, ...Array(numberOfZeroes).fill(0)];
};

console.log(moveZeros([1, 2, 0, 1, 0, 0, 3, 6]));

// https://codesandbox.io/s/misty-forest-pxdog?file=/src/index.ts

/**
 *
 * Given a string s and a character c, return the number of occurrences of c in
 * s. Example:
 *
 * $ numChars(‘oh heavens’, ‘h’)
 * $ 2
 *
 */

const numChars = (input: string, char: string) =>
  /**
   *
   * yay regular expressions! global flag since we wanna count *all* instances
   * return 0 in case of no matches
   *
   */
  (input.match(RegExp(char, "g")) || []).length;

console.log(numChars("oh heavens", "h"));

// https://codesandbox.io/s/mystifying-swirles-8h5it?file=/src/index.ts

/**
 *
 * Given an array of numbers that represent stock prices (where each number is
 * the price for a certain day), find 2 days when you should buy and sell your
 * stock for the highest profit. Example:
 *
 * $ stockBuySell([110, 180, 260, 40, 310, 535, 695])
 * $ “buy on day 4, sell on day 7”
 *
 */

const stockBuySell = (prices: Array<number>) =>
  `buy on day ${prices.indexOf(Math.min(...prices)) + 1}\
, sell on day ${prices.indexOf(Math.max(...prices)) + 1}`;

// A much faster way.
// The earlier one is ~88% slower: https://jsbench.me/54keiiwuvy/1
const stockBuySellReduce = (prices: Array<number>) => {
  const minMax = prices.reduce(
    (agg, curr, index, arr) => {
      if (curr < arr[agg.min]) agg.min = index;
      else if (curr > arr[agg.max]) agg.max = index;
      return agg;
    },
    { min: 0, max: 0 },
  );
  return `buy on day ${minMax.min + 1}, sell on day ${minMax.max + 1}`;
};

console.log(stockBuySell([110, 180, 260, 40, 310, 535, 695]));
console.log(stockBuySellReduce([110, 180, 260, 40, 310, 535, 695]));

// https://codesandbox.io/s/sharp-jepsen-8npl1?file=/src/index.ts

/**
 *
 * Given an array of people objects (where each person has a name and a number
 * of pizza slices they’re hungry for) and a number for the number of slices
 * that the pizza can be sliced into, return the number of pizzas you need to
 * buy.
 *
 * $ arr = [
 * $   { name: Joe, num: 9 },
 * $   { name: Cami, num: 3 },
 * $   { name: Cassidy, num: 4 },
 * $ ]
 * $ gimmePizza(arr, 8)
 * $ 2
 * $ // 16 slices needed, pizzas can be sliced into 8 pieces,
 * $ // so 2 pizzas should be ordered
 *
 */

type TPeopleSlicesMap = {
  name: string;
  num: number;
};
const gimmePizza = (arr: Array<TPeopleSlicesMap>, maxSlices: number) => {
  const totalSlices = arr.reduce((total, { num }) => {
    return (total += num);
  }, 0);
  return Math.ceil(totalSlices / maxSlices);
};

const inputMap = [
  { name: "Joe", num: 9 },
  { name: "Cami", num: 3 },
  { name: "Cassidy", num: 4 },
];
console.log(gimmePizza(inputMap, 8));

// https://codesandbox.io/s/young-darkness-61nq3?file=/src/index.ts

/**
 *
 * Given a positive integer n, write a function that returns true if it is a
 * perfect square and false otherwise. Don’t use any built-in math functions
 * like sqrt. Hint: Use binary search!
 *
 * Examples:
 * $ perfectSquare(25)
 * $ true
 *
 * $ perfectSquare(10)
 * $ false
 *
 */

const isPerfectSquare = (input: number) => {
  const binarySearchPerfectSquare = (
    input: number,
    start: number,
    end: number,
  ): number | boolean => {
    if (start > end) return false;

    // mid value, parsed to an int using bitwise operator
    const mid = ((start + end) / 2) >> 0;
    if (mid * mid < input)
      return binarySearchPerfectSquare(input, mid + 1, end);
    if (mid * mid > input)
      return binarySearchPerfectSquare(input, start, mid - 1);
    // mid*mid === input, perfect square!
    return true;
  };
  return binarySearchPerfectSquare(input, 1, input);
};

console.log(isPerfectSquare(25));
console.log(isPerfectSquare(10));

// https://codesandbox.io/s/relaxed-pine-yofdz?file=/src/index.ts

/**
 *
 * Given an array of integers and a target value, return the number of pairs of
 * array elements that have a difference equal to a target value.
 *
 * Example:
 * $ arrayDiff([1, 2, 3, 4], 1)
 * $ 3 // 2 - 1 = 1, 3 - 2 = 1, and 4 - 3 = 1
 *
 */

const arrayDiff = (arr: Array<number>, target: number) => {
  let count = 0,
    flag = false;

  arr
    .sort()
    .reverse()
    .forEach((baseNum, i, reverseSortedArr) => {
      if (i === arr.length) return;
      flag = false;
      reverseSortedArr.slice(i + 1).forEach((num, j) => {
        // arr is sorted, so we won't get the target diff again
        if (flag) return;
        if (j === reverseSortedArr.length) return;
        if (baseNum - num === target) {
          count += 1;
          flag = true;
        }
      });
    });
  return count;
};

console.log(arrayDiff([1, 2, 3, 4], 1));

// https://codesandbox.io/s/cool-satoshi-84ku4?file=/src/index.ts

/**
 *
 * You’re given a string of characters that are only 2s and 0s. Return the index
 * of the first occurrence of “2020” without using the indexOf (or similar)
 * function, and -1 if it’s not found in the string.
 *
 * Example:
 * $ find2020(‘2220000202220020200’)
 * $ 14
 *
 */

const find2020 = (input: string) => {
  let index = -1;
  input.split("").reduce(
    (_shouldBe2020, _curr, i) => {
      const next4 = input.slice(i, i + 4);
      if (next4 === "2020") index = i;
      return next4;
    },
    input.slice(0, 4),
  );
  return index;
};

// https://codesandbox.io/s/sharp-mountain-wddiy?file=/src/index.ts

/**
 *
 * Given a rowIndex, return an array of the values in that row of Pascal’s
 * Triangle.
 *
 */

const getCurrentRow = (previousRow: Array<number>): Array<number> => {
  return Array.from(Array(previousRow.length + 1)).map((_, i) => {
    if (i === 0) return 1;
    else if (i === previousRow.length) return 1;
    else return previousRow[i - 1] + previousRow[i];
  });
};

const getPascalRow = (index: number): Array<number> => {
  if (index === 0) return [1];
  return getCurrentRow(getPascalRow(index - 1));
};

console.log(getPascalRow(0)); // [1]
console.log(getPascalRow(1)); // [1, 1]
console.log(getPascalRow(2)); // [1, 2, 1]
console.log(getPascalRow(3)); // [1, 3, 3, 1]
console.log(getPascalRow(4)); // [1, 4, 6, 4, 1]

// https://codesandbox.io/s/heuristic-wood-qzzbd?file=/src/index.ts

/**
 *
 * Given an integer n, return true if n^3 and n have the same set of digits.
 * Example:
 *
 * $ sameDigits(1) // true
 * $ sameDigits(10) // true
 * $ sameDigits(251894) // true
 * $ sameDigits(251895) // false
 *
 */

const INPUTS = [1, 10, 251894, 251895];

const getUniqueDigitsAsString = (input: number) =>
  [...new Set(input.toString().split("").sort())].join();

const sameDigits = (input: number) => {
  const inputDigits = getUniqueDigitsAsString(input);
  const cubeDigits = getUniqueDigitsAsString(Math.pow(input, 3));
  return inputDigits === cubeDigits;
};

INPUTS.forEach((input) => console.log(sameDigits(input)));
// true, true, true, false

// https://codesandbox.io/s/cold-cdn-85qcb?file=/src/index.ts

/**
 *
 * Given a list, return a list of all its prefixes in ascending order of their
 * length. You’re essentially implementing the inits function in Haskell!
 *
 * Example:
 * $ inits([4, 3, 2, 1])
 * $ [[], [4], [4,3], [4,3,2], [4,3,2,1]]
 *
 * $ inits([144])
 * $ [[], [144]]
 *
 */

const INPUT = [4, 3, 2, 1];

const inits = (input: Array<number>) =>
  input.reduce(
    (arr, curr, index) => {
      arr.push([...arr[index], curr]);
      return arr;
    },
    [[]] as number[][],
  );

console.log(inits(INPUT));

// https://codesandbox.io/s/determined-swanson-qn27b?file=/src/index.ts

/**
 *
 * Given a direction and a number of columns, write a function that outputs an
 * arrow of asterisks (see the pattern in the examples below)!
 *
 * Example:
 *
 * $ printArrow('right', 3)
 * Output:
 *
 *
 *
 *
 *
 *
 *
 * $ printArrow('left', 5)
 * Output:
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 */

const printArrow = (direction: "right" | "left", len: number) => {
  Array.from(Array(2 * (len - 1) + 1)).forEach((_, index) => {
    let numSpaces;
    if (direction === "right") {
      numSpaces = index > len - 1 ? 2 * len - 2 - index : index;
    } else {
      numSpaces = len - (index > len - 1 ? 2 * len - 1 - index : index + 1);
    }
    console.log(`${" ".repeat(numSpaces)}*`);
  });
};

// https://codesandbox.io/s/gifted-perlman-0yvo0?file=/src/index.ts

const LINE = "█".repeat(10);
const PRIDE_COLORS = [
  "#e40303",
  "#ff8c00",
  "#ffed00",
  "#008026",
  "#004dff",
  "#750787",
];

const PrideFlag = () => {
  PRIDE_COLORS.forEach((color) => {
    console.log(`%c${LINE}`, `color: ${color}`);
  });
};

PrideFlag();

// https://codesandbox.io/s/blissful-leavitt-86k1s?file=/src/index.ts

/**
 *
 * Write a function to find the longest common prefix string in an array of
 * strings.
 *
 * Example:
 *
 * $ longestPrefix(["cranberry","crawfish","crap"])
 * $ "cra"
 *
 * $ longestPrefix(["parrot", "poodle", "fish"])
 * $ ""
 *
 */

function longestPrefix(input: Array<string>) {
  let result = "";

  input[0].split("").reduce((prefix, currentLetter) => {
    const isPrefixValid = input.every((word) => word.startsWith(prefix));
    if (isPrefixValid) {
      result = prefix;
      return `${prefix}${currentLetter}`;
    }
    return result;
  }, "");
  return result;
}

// https://codesandbox.io/s/heuristic-dan-pl0nn?file=/src/index.ts

/**
 *
 * An “odious number” is a non-negative number that has an odd number of 1s in
 * its binary expansion. Write a function that returns true if a given number is
 * odious.
 *
 * Example:
 *
 * $ isOdious(14)
 * $ true
 *
 * $ isOdious(5)
 * $ false
 *
 */

const isOdious = (input: number) =>
  (input.toString(2).match(/1/g) ?? []).length % 2 === 1;

// https://codesandbox.io/s/admiring-poitras-8l6sd?file=/src/index.ts

/**
 *
 * Given an array of objects A, and an array of indexes B, reorder the objects
 * in array A with the given indexes in array B.
 *
 * Example:
 *
 * let a = [C, D, E, F, G, H]; let b = [3, 0, 4, 1, 2, 5];
 *
 * $ reorder(a, b) // a is now [D, F, G, C, E, H]
 *
 */

const reorder = (
  a: Array<Record<string, unknown> | string>,
  b: Array<number>,
) =>
  [...Array(a.length)].reduce((res, _, index) => {
    res[b[index]] = a[index];
    return res;
  }, []);

// https://codesandbox.io/s/agitated-bash-tmu25?file=/src/index.ts

defmodule RWC_259 do
  @doc """
  Given an integer n, count the total number of 1 digits appearing in all
  non-negative integers less than or equal to n.

  Example:

  > numberOfOnes(14)
  > 7 // 1, 10, 11, 12, 13, 14
  """
  defp count_one_recursively(input, count) when input == 1, do: count + 1

  defp count_one_recursively(input, count) do
    input
    |> Integer.to_string()
    |> String.split("")
    |> Enum.count(&(&1 == "1"))
    |> then(fn x ->
      case x > 0 do
        true ->
          count_one_recursively(input - 1, count + x)

        _ ->
          count_one_recursively(input - 1, count)
      end
    end)
  end

  def number_of_ones(input) do
    input
    |> count_one_recursively(0)
    |> IO.puts()
  end
end

RWC_259.number_of_ones(14)

defmodule RWC_270 do
  @doc """
  Let’s say you have n doors that start out as closed. With the first pass
  across the doors, you toggle every door open. With the second pass, you toggle
  every second door. With the third, every third door, and so on. Write a
  function that takes in an integer numberOfPasses, and returns how many doors
  are open after the number of passes.

  Example:

  let n = 7 let numberOfPasses = 3

  > passDoors(n, numberOfPasses)
  > 4

  // Explanation: // 0 means open, 1 means closed // Initial: 1 1 1 1 1 1 1 //
  Pass 1:  0 0 0 0 0 0 0 // Pass 2:  0 1 0 1 0 1 0 // Pass 3:  0 1 1 1 0 0 0
  """
  @spec pass_doors(n :: non_neg_integer, number_of_passes :: non_neg_integer) ::
          non_neg_integer
  def pass_doors(n, number_of_passes) do
    1..number_of_passes
    # duplicate with n+1 because map_every/3 starts from first elem indexed 0
    # open => true, closed => false
    |> Enum.reduce(List.duplicate(false, n + 1), fn nth, doors ->
      Enum.map_every(doors, nth, &(not &1))
    end)
    # ignore the extra door we added at the start
    |> Enum.drop(1)
    |> Enum.count(& &1)
  end
end

RWC_270.pass_doors(7, 3)

defmodule RWC_281 do
  @doc """
  Given an array of integers arr and an integer n, return a subarray of arr of
  length n where the sum is the largest.

  Make sure you maintain the order of the original array, and if n is greater
  than arr.length, you can choose what you want to return.

  > maxSubarray([-4,2,-5,1,2,3,6,-5,1], 4)
  > [1,2,3,6]

  > maxSubarray([1,2,0,5], 2)
  > [0,5]
  """

  def maxSubarray(input, size) do
    input
    |> Enum.chunk_every(size, 1, :discard)
    |> Enum.max_by(&Enum.sum(&1))
  end
end

IO.inspect(RWC_281.maxSubarray([-4, 2, -5, 1, 2, 3, 6, -5, 1], 4))
IO.inspect(RWC_281.maxSubarray([1, 2, 0, 5], 2))

defmodule RWC_282 do
  @doc """
  Given a number, sum every second digit in that number.

  Example:

  > sumEveryOther(548915381)
  > 26 // 4+9+5+8

  > sumEveryOther(10)
  > 0

  > sumEveryOther(1010.11)
  > 1 // 0+0+1
  """

  def sumEveryOther(input) do
    input
    |> to_string()
    |> String.graphemes()
    |> List.delete_at(0)
    |> Enum.filter(&(&1 !== "."))
    |> Enum.take_every(2)
    |> Enum.map(&String.to_integer/1)
    |> Enum.sum()
  end
end

defmodule RWC_284 do
  @doc """
  You are given a list of positive integers which represents some range of
  integers which has been truncated. Find the missing bits, insert ellipses to
  show that that part has been truncated, and print it. If the consecutive
  values differ by exactly two, then insert the missing value.

  Examples:

  > missingBits([1,2,3,4,20,21,22,23])
  > "[1,2,3,4,...,20,21,22,23]"

  > missingBits([1,2,3,5,6])
  > "[1,2,3,4,5,6]"

  > missingBits([1,3,20,27])
  > "[1,2,3,...,20,...,27]"
  """

  def missingBits(input) do
    input
    |> Enum.with_index()
    |> then(fn input_with_index ->
      input_with_index
      |> Enum.reduce([], fn {num, index}, acc ->
        {next_num, _} = Enum.at(input_with_index, index + 1, {num, nil})

        case next_num - num do
          # Last index
          0 -> [num | acc]
          1 -> [num | acc]
          # When step is exactly two
          2 -> [num + 1, num | acc]
          _ -> ["...", num | acc]
        end
      end)
      |> Enum.reverse()
      |> Enum.join(",")
      |> then(&"[#{&1}]")
    end)
  end
end

defmodule RWC_285 do
  @doc """
  Given a positive integer, generate an array in which every element is an array
  that goes from 1 to the index of that array.

  Example:

  > generateArrays(4)
  > [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]]

  > generateArrays(1)
  > [[1]]
  """

  def generateArrays(input) do
    1..input
    |> Enum.to_list()
    |> Enum.map(&Enum.to_list(1..&1))
  end
end

defmodule RWC_286 do
  @doc """
  Spreadsheet programs often use the A1 Reference Style to refer to columns.
  Given a column name in this style, return its column number.

  Examples of column names to their numbers:

  A -> 1 B -> 2 C -> 3 // etc Z -> 26 AA -> 27 AB -> 28 // etc AAA -> 703
  """

  def get_weight(char, power) do
    char
    |> String.to_charlist()
    |> hd()
    # A => 65, B => 66, so on...
    |> then(&(&1 - 64))
    |> then(&(&1 * 26 ** power))
  end

  def recusively_get_weight([{char, index}]), do: get_weight(char, index)

  def recusively_get_weight([left | rest]),
    do: recusively_get_weight([left]) + recusively_get_weight(rest)

  def get_column_number(input) do
    input
    |> String.split("", trim: true)
    |> Enum.reverse()
    |> Enum.with_index()
    |> recusively_get_weight()
  end
end

defmodule RWC_287 do
  @doc """
  Print the digits 0 through 100 without using the characters 1, 2, 3, 4, 5, 6,
  7, 8, or 9 in your code. Get creative!
  """

  def print_digits() do
    hundred =
      "d"
      |> to_charlist()
      |> hd()

    zero =
      [hundred - hundred]
      |> hd()

    zero..hundred
    |> Enum.join(", ")
  end
end

defmodule RWC_288 do
  @doc """
  Given a string of parenthesis, return the number of parenthesis you need to
  add to the string in order for it to be balanced.

  Examples:

  > numBalanced(`()`)
  > 0

  > numBalanced(`(()`)
  > 1

  > numBalanced(`))()))))()`)
  > 6

  > numBalanced(`)))))`)
  > 5
  """

  def num_balanced(input) do
    input
    |> String.split("", trim: true)
    |> Enum.frequencies()
    |> Map.values()
    |> then(
      &case &1 do
        [only_one_kind] -> only_one_kind
        [one, other] -> abs(one - other)
      end
    )
  end
end

defmodule RWC_289 do
  @doc """
  Given a list of numbers, return all groups of repeating consecutive numbers.

  Examples:

  > repeatedGroups([1, 2, 2, 4, 5])
  [[2, 2]]

  > repeatedGroups([1, 1, 0, 0, 8, 4, 4, 4, 3, 2, 1, 9, 9])
  [[1, 1], [0, 0], [4, 4, 4], [9, 9]]
  """

  def chunk_repeats([head | rest]), do: chunk_repeats(rest, [[head]])
  def chunk_repeats([], done), do: done

  def chunk_repeats([head | rest], [
        [head | head_repeated] | previously_repeated
      ]),
      do:
        chunk_repeats(rest, [
          [head | [head | head_repeated]] | previously_repeated
        ])

  def chunk_repeats([head | rest], previously_repeated),
    do: chunk_repeats(rest, [[head] | previously_repeated])

  def repeated_groups(input) do
    input
    |> chunk_repeats()
    |> Enum.reverse()
    |> Enum.filter(&(length(&1) > 1))
  end
end

defmodule RWC_300 do
  @doc """
  Write a function to find out whether the binary representation of a number is
  palindrome or not.

  Example:

  > binaryPal(5) > true

  > binaryPal(10) > false
  """

  def binary_pal(input) do
    input
    |> Integer.to_charlist(2)
    |> List.to_string()
    |> then(fn digits ->
      digits
      |> String.reverse()
      |> then(
        &case &1 do
          ^digits -> true
          _ -> false
        end
      )
    end)
  end
end

defmodule RWC_301 do
  @doc """
  Given a string, calculate the score that it would get in a game of Scrabble.
  For extra credit, try verifying if the string is a valid word, or take into
  account premium squares!

  Scoring and example:

  1 point: E, A, I, O, N, R, T, L, S, U 2 points: D, G 3 points: B, C, M, P 4
  points: F, H, V, W, Y 5 points: K 8 points: J, X 10 points: Q, Z

  > scrabbleScore('FIZZBUZZ')
  > 49
  """

  @score_map %{
    "EAIONRTLSU" => 1,
    "DG" => 2,
    "BCMP" => 3,
    "FHVWY" => 4,
    "K" => 5,
    "JX" => 8,
    "QZ" => 10
  }

  def scrabble_score(input) do
    input
    |> String.split("", trim: true)
    |> Enum.reduce(0, fn letter, score ->
      @score_map
      |> Map.keys()
      |> Enum.find(&String.contains?(&1, letter))
      |> then(&Map.get(@score_map, &1))
      |> Kernel.+(score)
    end)
  end
end

defmodule RWC_303 do
  @doc """
  Given an array of people objects (where each person has a name and a number of
  pie pieces they’re hungry for) and a number for the number of pieces that the
  pie can be cut into, return the number of pies you need to buy.

  Example:

  iex>
  ...>  arr = [
  ...>    %{ name: Joe, num: 9 },
  ...>    %{ name: Cami, num: 3 },
  ...>    %{ name: Cassidy, num: 4 }
  ...>  ]
  iex> RWC_303.num_pie(arr, 8)
  2 # 16 pieces needed, pies can be cut into 8 pieces, so 2 pies should be
  bought
  """

  def num_pie(input, count) do
    input
    |> Enum.reduce(0, &(&2 + &1[:num]))
    |> div(count)
  end
end

defmodule RWC_304 do
  @doc """
  Given an array arr and integers n and m, remove n elements from the front of
  the array, and m elements from the back. Assume that n + m <= arr.length.

  Example:

  iex> RWC_304.trim_array([1, 2, 3, 4, 5, 6], 2, 1)
  [3, 4, 5]

  iex> RWC_304.trim_array([6, 2, 4, 3, 7, 1, 3], 5, 0)
  [1, 3]

  iex> RWC_304.trim_array([1, 7], 0, 0)
  [1, 7]
  """

  def trim_array(input, n, m) do
    input
    |> Enum.slice(n, length(input) - n - m)
  end
end

defmodule RWC_305 do
  @doc """
  Given some JSON data, calculate the maximum depth reached. Both arrays and
  dictionaries increase the depth! If the input is invalid data, the response
  should be undefined (you decide how you want that to return).

  iex> RWC_305.depth_json([])
  1

  iex> RWC_305.depth_json([1, 2, 3, 4, 5])
  1

  iex> RWC_305.depth_json([%{a: []}, ["abc"]])
  3
  """
  def depth_json(input), do: compute_depth(input)

  defp compute_depth(input) when is_list(input) or is_map(input) do
    cond do
      is_list(input) and input == [] ->
        1

      true ->
        input
        |> flatten_if_map()
        |> Enum.map(&compute_depth/1)
        |> Enum.max()
        |> Kernel.+(1)
    end
  end

  defp compute_depth(_), do: 0

  defp flatten_if_map(input) when is_map(input), do: Map.values(input)
  defp flatten_if_map(input), do: input
end

defmodule RWC_306 do
  @doc """
  Write a function that takes an array of consecutive, increasing letters as
  input, and returns any missing letters in the array between the first and last
  letter.

  Example:

  iex> RWC_306.missing_letters(["a","b","c","d","f"])
  ["e"]


  iex> RWC_306.missing_letters([
  ...>   "a", "b", "c", "d", "e", "h", "i", "j", "k", "l", "m", "n", "o",
  ...>   "p", "q", "r", "s", "t", "u", "w", "x", "y", "z"
  ...> ])
  ["f", "g", "v"]
  """

  def missing_letters(input) do
    input_complete_mapset =
      input
      |> then(fn list ->
        [head | _] = list
        tail = List.last(list)

        hd(to_charlist(head))..hd(to_charlist(tail))
        |> MapSet.new()
      end)

    input_mapset =
      input
      |> MapSet.new(&hd(to_charlist(&1)))

    input_complete_mapset
    |> MapSet.difference(input_mapset)
    |> MapSet.to_list()
    |> List.to_string()
    |> String.graphemes()
  end
end

defmodule RWC_307 do
  @doc """
  Given an integer n, return true if it's a perfect square AND when reversed, is
  still a perfect square.


  iex> RWC_307.reversed_squares(9)
  true

  iex> RWC_307.reversed_squares(441)
  true

  iex> RWC_307.reversed_squares(25)
  false
  """

  def reversed_squares(input) do
    input
    |> Integer.digits()
    |> Enum.reverse()
    |> Integer.undigits()
    |> then(
      &cond do
        is_perfect_square(&1) and is_perfect_square(input) -> true
        true -> false
      end
    )
  end

  defp is_perfect_square(input) do
    input
    |> :math.sqrt()
    |> then(&if Kernel.trunc(&1) == &1, do: true, else: false)
  end
end

defmodule RWC_308 do
  @doc """
  Given an array of strings and a max width, format the text such that each line
  has exactly maxWidth characters and is fully justified. You can choose how you
  want to wrap a word.

  Example:

  iex> RWC_308.justify_text(
  ...>   ["This", "is", "an", "example", "of", "text", "justification."],
  ...>   16
  ...> )
  [
    "This    is    an", "example  of text", "justification.  "
  ]
  """

  def justify_text(input, max_length) do
    input
    |> Enum.reduce([[]], fn word, phrase_list ->
      phrase_list
      |> List.last([])
      |> then(fn last_phrase ->
        current_phrase_length = last_phrase |> Enum.join(".") |> String.length()

        cond do
          current_phrase_length + String.length(word) < max_length ->
            # update last element
            last_list =
              phrase_list
              |> Enum.at(-1)
              |> Enum.concat([word])

            phrase_list
            |> Enum.drop(-1)
            |> Enum.concat([last_list])

          true ->
            # move to new phrase
            phrase_list
            |> Enum.concat([[word]])
        end
      end)
    end)
    |> Enum.map(&justify_phrase(&1, max_length))
  end

  defp justify_phrase(input, max_length) do
    number_of_spaces_to_add =
      max_length - (input |> Enum.join("") |> String.length())

    number_of_breaks = if length(input) == 1, do: 1, else: length(input) - 1
    each_space_block_length = div(number_of_spaces_to_add, number_of_breaks)
    remainder = rem(number_of_spaces_to_add, number_of_breaks)

    spaces_list =
      " "
      |> list_from(each_space_block_length)
      |> list_from(number_of_breaks)
      |> then(fn [first_break_spaces | rest] ->
        case remainder do
          0 ->
            [first_break_spaces | rest]

          _ ->
            first_break_spaces
            |> Enum.concat(list_from(" ", remainder))
            |> then(&[&1 | rest])
        end
      end)
      |> then(
        &cond do
          length(input) == 1 -> &1
          true -> Enum.concat(&1, [[""]])
        end
      )

    input
    |> Enum.zip(spaces_list)
    |> Enum.reduce("", fn {word, spaces}, line ->
      "#{line}#{word}#{Enum.join(spaces)}"
    end)
  end

  defp list_from(el, count), do: for(_ <- 1..count, do: el)
end

defmodule RWC_309 do
  @doc """
  Given a string, separate it into groups of non-space equal characters, sorted.

  Example:

  iex> RWC_309.explode_string("Ahh, abracadabra!")
  ["!",",","A","aaaaa","bb","c","d","hh","rr"]
  """

  def explode_string(input) do
    input
    |> String.replace(" ", "")
    |> String.split("", trim: true)
    |> Enum.frequencies()
    |> Map.to_list()
    |> Enum.map(fn {char, count} ->
      list_from(char, count)
      |> Enum.join()
    end)
  end

  defp list_from(el, count), do: for(_ <- 1..count, do: el)
end

defmodule RWC_310 do
  @doc """
  Given an array where each element is the price of a given stock on that
  index's day, choose a single day to buy a stock and a different day (in the
  future/later in the array) to sell the stock to maximize your profit. Return
  the maximum profit that you can get from a given input. If you can't profit,
  return 0.

  Example:

  iex> RWC_310.maximum_profit([7, 1, 5, 3, 6, 4])
  5    # Buy on day 2, and sell on day 5, your profit = 6 - 1 = 5.
  """

  def maximum_profit(input) do
    input
    |> Enum.with_index()
    |> Enum.reduce(0, fn {price, index}, max_profit ->
      input
      |> Enum.slice((index + 1)..-1)
      |> Enum.max(&>=/2, fn -> 0 end)
      |> Kernel.-(price)
      |> then(
        &cond do
          &1 > max_profit -> &1
          true -> max_profit
        end
      )
    end)
  end
end

defmodule RWC_311 do
  @doc """
  Given two strings s and t, return true if t is an anagram of s, and false
  otherwise. Try this in a language you're not comfortable with!

  Example:

  iex> RWC_311.is_anagram("barbie", "oppenheimer")
  false

  iex> RWC_311.is_anagram("race", "care")
  true
  """

  def is_anagram(a, b), do: get_chars(a) === get_chars(b)

  defp get_chars(input),
    do: input |> String.split("", trim: true) |> Enum.frequencies()
end

defmodule RWC_312 do
  @doc """
  Implement the Luhn algorithm to validate a credit card number. Bonus points if
  you can identify what brand of credit card the user inputted!

  iex> RWC_312.luhn_check(123456789)
  false

  iex> RWC_312.luhn_check(5555555555554444)
  true
  """

  def luhn_check(input) do
    input
    |> Integer.digits()
    |> Enum.with_index(1)
    |> Enum.reduce(0, fn {digit, index}, sum ->
      cond do
        rem(index, 2) == 0 -> digit * 2
        true -> digit
      end
      |> Integer.digits()
      |> Enum.sum()
      |> Kernel.+(sum)
    end)
    |> then(&(rem(&1, 10) == 0))
  end
end

defmodule RWC_313 do
  @doc """
  You have a faulty keyboard. Whenever you type a vowel on it (a,e,i,o,u,y), it
  reverses the string that you have written, instead of typing the character.
  Typing other characters works as expected. Given a string, return what will be
  on the screen after typing with your faulty keyboard.

  Example:

  iex> RWC_313.faulty_keeb("string")
  "rtsng"

  iex> RWC_313.faulty_keeb("hello world!")
  "w hllrld!"
  """

  @vowels ["a", "e", "i", "o", "u", "y"]

  def faulty_keeb(input) do
    input
    |> String.graphemes()
    |> Enum.reduce([], fn char, word ->
      cond do
        Enum.member?(@vowels, char) -> Enum.reverse(word)
        true -> Enum.concat(word, [char])
      end
    end)
    |> Enum.join()
  end
end

defmodule RWC_314 do
  @doc """
  Make a "guessing game" where there is a target number, and as the user makes
  guesses, the output returns higher or lower until the user is correct.

  Example usage:

  Guess the number!
  > 10
  higher
  > 20
  higher
  > 30
  lower
  > 25
  higher
  > 27
  Correct! You won in 5 guesses!
  """

  def solve(guess) do
    IO.puts("Guess the number!")

    input()
    |> solve(guess)
  end

  def solve(guess, answer, guess_count \\ 1)

  def solve(guess, answer, 1) when guess === answer,
    do: IO.puts("Correct! You won in 1 guess!") |> exit()

  def solve(guess, answer, guess_count) when guess === answer,
    do: IO.puts("Correct! You won in #{guess_count} guesses!") |> exit()

  def solve(guess, answer, guess_count) do
    cond do
      guess > answer ->
        IO.puts("lower")

      guess < answer ->
        IO.puts("higher")
    end

    input()
    |> solve(answer, guess_count + 1)
  end

  defp input() do
    IO.gets("> ")
    |> Integer.parse()
    |> then(&elem(&1, 0))
  end
end

RWC_314.solve(:rand.uniform(100))

defmodule RWC_315 do
  @moduledoc """
  Given a sequence of numbers, generate a "count and say" string.

  Example:

  iex> RWC_315.count_and_say(112222555)
  "two 1s, then four 2s, then three 5s"

  iex> RWC_315.count_and_say(3333333333)
  "ten 3s"
  """

  @number_say_map %{
    1 => "one",
    2 => "two",
    3 => "three",
    4 => "four",
    5 => "five",
    6 => "six",
    7 => "seven",
    8 => "eight",
    9 => "nine",
    10 => "ten"
  }

  def count_and_say(input) do
    input
    |> Integer.digits()
    # count
    |> Enum.reduce([], fn number, list ->
      current_list = List.first(list, [nil])

      current_list
      |> List.first()
      |> then(fn current_digit ->
        case current_digit do
          # same digit as current
          ^number ->
            list
            |> List.delete_at(0)
            |> then(&[[number | current_list] | &1])

          # new digit
          _ ->
            [[number] | list]
        end
      end)
    end)
    |> Enum.reverse()
    # say
    |> Enum.map_join(", then ", fn list ->
      len = Map.get(@number_say_map, length(list))
      "#{len} #{List.first(list)}s"
    end)
  end
end

defmodule RWC_316 do
  @moduledoc """
  Given an array of integers and a number k (where k is guaranteed to be less
  than the array's length), return a subarray of length k with the minimum
  possible sum. Maintain the order of the original array!

  Example:

  iex> RWC_316.min_subs([1, 3, 20, 4, 8, 9, 11], 3)
  [4, 8, 9]

  iex> RWC_316.min_subs([4, 4, 4, 4, 8], 2)
  [4, 4]
  """

  def min_subs(input, len) do
    input
    |> Enum.chunk_every(len, 1, :discard)
    |> Enum.min_by(&Enum.sum/1)
  end
end

defmodule RWC_317 do
  @moduledoc """
  Given an array of integers, sort them into two separate sorted arrays of even
  and odd numbers. If you see a zero, skip it.

  Example:

  iex> RWC_317.separate_and_sort([4, 3, 2, 1, 5, 7, 8, 9])
  [[2, 4, 8], [1, 3, 5, 7, 9]]

  iex> RWC_317.separate_and_sort([1,1,1,1])
  [[], [1,1,1,1]]
  """

  def separate_and_sort(input) do
    input
    |> Enum.sort()
    |> Enum.split_with(&(rem(&1, 2) == 0))
    |> Tuple.to_list()
  end
end

defmodule RWC_318 do
  @moduledoc """
  You have n equal-sized blocks and you want to build a staircase with them.
  Return the number of steps you can fully build.

  iex> RWC_318.build_staircase(6)
  3

  #=> #
  #=> ##
  #=> ###

  iex> RWC_318.build_staircase(9)
  3 #=> it takes 10 blocks to make 4 steps
  """

  def build_staircase(input) do
    1..input
    |> Enum.reduce_while(0, fn step, total ->
      cond do
        total + step > input -> {:halt, step - 1}
        true -> {:cont, total + step}
      end
    end)
  end
end

defmodule RWC_322 do
  @moduledoc """
  Given two strings s and t, determine if they are isomorphic. Two strings are
  isomorphic if there is a one-to-one mapping possible for every character of
  the first string to every character of the second string.

  iex> RWC_322.is_isomorphic("abb", "cdd")
  true # "a" maps to "c" and "b" maps to "d"

  iex> RWC_322.is_isomorphic("cassidy", "1234567")
  false # "s" cannot have a mapping to both "3" and "4"

  iex> RWC_322.is_isomorphic("cass", "1233")
  true
  """

  def is_isomorphic(first, second) do
    [first, second]
    |> Enum.map(fn string ->
      string
      |> String.split("", trim: true)
      |> Enum.frequencies()
      |> Map.values()
    end)
    |> then(fn [f1, f2] -> f1 == f2 end)
  end
end

defmodule RWC_323 do
  @moduledoc """
  Given a string s, you are allowed to delete at most k characters. Find if the
  string can be a palindrome after deleting at most k characters.

  iex> RWC_323.k_pal("abcweca", 2)
  true

  iex> RWC_323.k_pal("abcweca", 1)
  false

  iex> RWC_323.k_pal("acwca", 2)
  true

  iex> RWC_323.k_pal("acxcb", 1)
  false
  """

  def k_pal(string, count) do
    string
    |> String.graphemes()
    |> check_first_and_last_characters(count)
  end

  defp check_first_and_last_characters([_char], _count), do: true
  defp check_first_and_last_characters([char | [char]], _count), do: true

  defp check_first_and_last_characters(string, count) do
    [first | with_last] = string
    [last | remaining_reversed] = Enum.reverse(with_last)
    remaining = Enum.reverse(remaining_reversed)

    cond do
      first == last ->
        check_first_and_last_characters(remaining, count)

      first != last and count == 0 ->
        false

      first != last and count > 0 ->
        check_first_and_last_characters([first | remaining], count - 1) or
          check_first_and_last_characters(with_last, count - 1)

      true ->
        false
    end
  end
end

defmodule RWC_325 do
  @moduledoc """
  Given a list of words and a dictionary of letter scores, find the word with
  the highest score according to the rules:

  score = word_length * (sum of letter scores in the word)

  If there are multiple words with the same highest score, return the
  lexicographically smallest one.

  iex> word_list = ["apple", "banana", "cherry", "date", "fig"];
  ...> RWC_325.score_word_game(word_list)
  "cherry"
  """

  def score_word_game(word_list, _letter_scores \\ nil) do
    word_list
    |> Enum.map(&get_word_score/1)
    |> Enum.max_by(fn {_word, score} -> score end)
    |> then(&elem(&1, 0))
  end

  defp get_word_score(word) do
    word
    |> String.downcase()
    |> String.graphemes()
    |> Enum.reduce(0, fn letter, total ->
      score = (to_charlist(letter) |> hd()) - 96

      total + score
    end)
    |> then(&{word, &1})
  end
end

defmodule RWC_326 do
  @moduledoc """
  Given a list of tasks, where each task has a duration, and a limited amount of
  available time to work, write a function to return the tasks that can be
  completed within the given time, without re-ordering the original list of
  tasks. Try to maximize the number of tasks that can be completed!

  Example:

  iex> tasks = [
  ...>    %{ name: "Task 1", duration: 4 },
  ...>    %{ name: "Task 2", duration: 2 },
  ...>    %{ name: "Task 3", duration: 7 },
  ...>    %{ name: "Task 4", duration: 5 },
  ...>    %{ name: "Task 5", duration: 1 },
  ...>    %{ name: "Task 6", duration: 3 }
  ...>  ];
  ...> time_to_work = 6;
  ...> RWC_326.do_tasks(tasks, time_to_work)
  ["Task 2", "Task 5", "Task 6"]
  """

  def do_tasks(task_list, time_to_work) do
    task_list
    |> Enum.with_index()
    |> Enum.sort_by(fn {%{duration: duration}, _index} -> duration end)
    |> Enum.reduce_while({0, []}, fn {%{duration: duration}, index},
                                     {time_sum, indices} ->
      cond do
        time_sum + duration > time_to_work -> {:halt, indices}
        true -> {:cont, {time_sum + duration, [index | indices]}}
      end
    end)
    |> Enum.sort()
    |> Enum.map(
      &(task_list
        |> Enum.at(&1)
        |> then(fn %{name: name} -> name end))
    )
  end
end

defmodule RWC_327 do
  @moduledoc """
  Given a start number a, an ending number b, and a string str that can be
  "odd", "even", or "prime", return all of the numbers that are odd, even, or
  prime between a and b. a will not always necessarily be less than b!


  iex> RWC_327.between_nums(3, 11, "even")
  [4,6,8,10]

  iex> RWC_327.between_nums(15, 1, "prime")
  [2,3,5,7,11,13]
  """

  defp is_prime(num) do
    cond do
      num <= 1 ->
        false

      num == 2 ->
        true

      rem(num, 2) == 0 ->
        false

      true ->
        3..(num - 1)//2
        |> Enum.all?(&(rem(num, &1) != 0))
    end
  end

  def between_nums(start, finish, option) when start > finish,
    do: between_nums(finish, start, option)

  def between_nums(start, finish, option) do
    start..finish
    |> Enum.filter(fn num ->
      case option do
        "prime" ->
          is_prime(num)

        "even" ->
          rem(num, 2) == 0

        _ ->
          rem(num, 2) != 0
      end
    end)
  end
end

defmodule RWC_328 do
  @moduledoc """
  Given two arrays calories and prices, where calories[i] and prices[i]
  represent the calorie content and price of the ith food item, and a daily
  calorie goal, find the minimum cost to achieve or exceed the daily calorie
  goal. If it's impossible to meet the goal, return -1.

  Example:

  iex> calories = [200, 400, 600, 800]
  ...> prices = [50, 60, 80, 100]
  ...> daily_goal = 1200
  ...> RWC_328.min_cost_for_calories(calories, prices, daily_goal)

  160 # the 2nd and 4th items add up to 1200 calories for the minimum cost
  """

  def min_cost_for_calories(calories, prices, daily_goal) do
    calories
    |> Enum.with_index()
    |> Enum.flat_map(fn {calorie, index} ->
      calories
      |> Enum.with_index()
      |> Enum.slice(index..-1)
      |> Enum.filter(&(elem(&1, 0) != calorie))
      |> Enum.map(&{elem(&1, 0) + calorie, {index, elem(&1, 1)}})
    end)
    |> Enum.filter(&(elem(&1, 0) >= daily_goal))
    |> Enum.reduce(:infinity, fn {_calories, {index_a, index_b}}, total_price ->
      current_combo_price = Enum.at(prices, index_a) + Enum.at(prices, index_b)

      cond do
        current_combo_price < total_price -> current_combo_price
        true -> total_price
      end
    end)
  end
end

defmodule RWC_329 do
  @moduledoc """
  There is a sorted integer array that has been rotated an unknown number of
  times. Given that rotated array, return how many times it has been rotated. It
  may contain duplicate numbers!

  iex> RWC_329.rotated_num([4, 0, 1, 2, 3])
  1

  iex> RWC_329.rotated_num([7, 9, 20])
  0

  iex> RWC_329.rotated_num([7, 7, 314, 1337, 7])
  4
  """

  def rotated_num(nums) do
    cond do
      Enum.sort(nums) == nums -> 0
      true -> 1 + rotated_num(Enum.slide(nums, 0, -1))
    end
  end
end

defmodule RWC_330 do
  @moduledoc """
  Given an array of integers, return the majority element. If there is no
  majority element, return if the array is majority even or odd numbers, and if
  there is none, say so.

  iex> RWC_330.majority([3,1,4,1])
  "1"

  iex> RWC_330.majority([33,44,55,66,77])
  "Majority odds"

  iex> RWC_330.majority([1,2,3,4])
  "No majority"
  """

  def majority(nums) do
    nums
    |> Enum.frequencies()
    |> Map.to_list()
    |> Enum.sort_by(fn {_, count} -> count end, :desc)
    |> then(fn [{possible_majority, first} | [{_, second} | _]] ->
      cond do
        first == second ->
          nums
          |> Enum.group_by(&(rem(&1, 2) == 0))
          |> then(fn %{false: odds, true: evens} ->
            cond do
              length(odds) > length(evens) -> "Majority odds"
              length(odds) > length(evens) -> "Majority evens"
              true -> "No majority"
            end
          end)

        true ->
          Integer.to_string(possible_majority)
      end
    end)
  end
end