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Today's code

Q1 First dictionary type Pete, the baker [5kyu]

Function of how many cakes can be made Make def cakes (recipe, available):

Example 1

Input: Recipe and ingredients

cakes({flour: 500, sugar: 200, eggs: 1}, {flour: 1200, sugar: 1200, eggs: 5, milk: 200})

Output: Number that can be made

2

Example 2

Input: Recipe and ingredients

{apples: 3, flour: 300, sugar: 150, milk: 100, oil: 100}, {sugar: 500, flour: 2000, milk: 2000})

Output: Number that can be made

0

MyAnswer

    list = []
    for key,value in recipe.items():
        if key in available:
            list.append(available[key]//value)
        else:
            list.append(0)
    return min(list)

It seems that you can get the contents with items () if you gg with the first dictionary type, so get it with for Save the ingredients // recipes in the list for each ingredient (key) If there is no material, save 0 Change the minimum value of list

Straightforward code

BestAnswers

def cakes(recipe, available):
  return min(available.get(k, 0)/recipe[k] for k in recipe)

It seems that you can get the key with for k in dict Get value with dict.get (k, 0), get second argument if no key

Almost the same with min (available.get (k, 0) // v for k, v in recipe.items ()) I have to use dict.get return min ([available [i] // recipe [i] if i in available else 0 for i in recipe]) and if can be included

Q2 OX game Tic-Tac-Toe Checker [5kyu]

If you think that a tic tac game is something, it seems to be an OX game If you pass the board and classify the results, it's OK -1: Incomplete 1: X wins 2: O wins 0: Draw (it's a cat's game) It looks difficult ...

Example 1

Input: Board

[[0, 0, 1], [0, 1, 2], [2, 1, 0]]

• 0: Blank 1: X 2: O

Output: Result

-1

MyAnswer

def ox(board):
    lines = []
    for i in range(3):
        lines.append(board[i])
        lines.append([board[j][i] for j in range(3)])
    lines.append([board[0][0],board[1][1],board[2][2]])
    lines.append([board[0][2],board[1][1],board[2][0]])

    yetfin = 0
    for k in range(8):
        if lines[k] == [1,1,1] or lines[k] == [2,2,2]:
            return lines[k][0] 

        if set(lines[k]) == {0,1} or set(lines[k]) == {0,2} :
            yetfin = 1
            
    if yetfin:
        return -1
    else:
        return 0

It is a program that saves the numerical values of each of the 8 vertical and horizontal columns in lines and then searches for which of the following Win: [111] [222] Incomplete: {0,1} {0,2} Draw: Other than that

BestAnswers

def ox2(board):
  for i in range(0,3):
    if board[i][0] == board[i][1] == board[i][2] != 0:
      return board[i][0]
    elif board[0][i] == board[1][i] == board[2][i] != 0:
      return board[0][i]
      
  if board[0][0] == board[1][1] == board[2][2] != 0:
    return board[0][0]
  elif board[0][2] == board[1][1] == board[2][0] != 0:
    return board[0][0]

  elif 0 not in board[0] and 0 not in board[1] and 0 not in board[2]:
    return 0
  else:
    return -1

What you are doing is almost the same except that you have not made lines until the winning decision (This is smart because it saves memory) But I'm just looking for 0 in the draw decision ... In other words XXO OOX XO_ The behavior has changed with a stacking game with 0 like

board = [[1,1,2],
          [2,2,1], 
          [1,2,0]]
print('MyAnswer:',ox(board))
print('BestPractice:',ox2(board))

MyAnswer: 0
BestPractice: -1

It will be.

that? If you look closely at the problem statement

-1 if the board is not yet finished (there are empty spots),

In other words, if there is even one empty, it is incomplete, so the latter seems to be the correct Best Practice.

Afterword

I don't feel like leaving the house before noon on Saturday, so I can't eat anything

Template for yourself

Today's code

Q

Example 1

input:

output:

Example 2

input:

output:

MyAnswer

BestAnswers