zahash

#@time_it
def dict_method_v2(arr):
    ''' In this method we will remove some of the redundancies
        of the dict_method
        
        please note that this method takes a bit more time
        because we are making more comparisions for the max_value
        than before
        
        Complexity : O(n)

zahash

def time_it(func):
    def wrapper(*args, **kwargs):
        start = time.time()
        func(*args, **kwargs)
        end = time.time()
        return (end - start)*1000
    return wrapper

zahash

#@time_it
def dict_method(arr):
    ''' In this method we will use dictionary to store the 
        count of each element and then search the dictionary 
        (or) hash map for the element with maximum count
        
        this is one of the quickest methods
        
        Complexity : O(n)
    '''

zahash

#@time_it
def dict_method(st):
    '''
    This function will return the non-repeating
    characters in a string 
    
    Args:
        st: the input string
    
    Return:

zahash

#@time_it
def set_method(st):
    s = set()
    repeat = set()
    for letter in st:
        if letter in s:
            repeat.add(letter)
        else:
            s.add(letter)
    return s-repeat

zahash

def cartesian_product(x,y):
    '''
    This function will take two iterables and returns 
    the cartesian product of them as a list
    '''
    return [a+b for a in x for b in y]

zahash

def display(values):
    """
    Display the values as a 2-D grid.
    Input: The sudoku in dictionary form
    Output: None
    """
    print('')
    
    rows = 'ABCDEFGHI'
    cols = '123456789'

zahash

start = '53..7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79'
    
rows = 'ABCDEFGHI'
cols = '123456789'
boxes = cartesian_product(rows, cols)

row_units = [cartesian_product(r, cols) for r in rows]
col_units = [cartesian_product(rows, c) for c in cols]
box_units = [cartesian_product(r,c) 
                    for r in ['ABC', 'DEF', 'GHI'] 

zahash

def eliminate(grid_dict):
    for k,v in grid_dict.items():
        if len(v) != 1: # if the box needs elimination
            peers = peer_dict[k]    # get all the peers
            peer_values = set([grid_dict[p] for p in peers if len(grid_dict[p]) == 1])
            grid_dict[k] = ''.join(set(grid_dict[k]) - peer_values)
    return grid_dict

zahash

def only_choice(grid_dict):
    for unit in unit_list:
        for num in '123456789':
            num_places = [box for box in unit if num in grid_dict[box]]
            if len(num_places) == 1:
                grid_dict[num_places[0]] = num
    return grid_dict