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Maarten Fabré
Maarten Fabré
  • 9.4k
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  • 16
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Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    if len(edges) == 1: # only one brick wide
        return len(edgeswall)
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    if len(edges) == 1: # only one brick wide
        return len(edges)
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    if len(edges) == 1: # only one brick wide
        return len(wall)
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges
added 82 characters in body
Source Link
Maarten Fabré
Maarten Fabré
  • 9.4k
  • 1
  • 16
  • 27

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    if len(edges) == 1: # only one brick wide
        return len(edges)
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    if len(edges) == 1: # only one brick wide
        return len(edges)
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges
Source Link
Maarten Fabré
Maarten Fabré
  • 9.4k
  • 1
  • 16
  • 27

Counter

Great use of Counter. One thing that can be improved, is to use most_common instead of max.

If edges is the Counter of all the edges in the wall, _, max_edges = edges.most_common(1) gives you the index with the most edges, and how many edges it has. Since this is the right side of the wall, you need the 2nd most common element: edges.most_common(2)[1]

aggregating the counter

Python is batteries included. Whenever you want to do advanced iteration, it can pay to look at the itertools module. For this, you can use accumulate (as @sedsarq notes in his answer) and chain.

edges = collections.Counter(
    chain.from_iterable(accumulate(row) for row in wall)
)

def least_bricks2(wall):
    edges = collections.Counter(
        chain.from_iterable(accumulate(row) for row in wall)
    )
    _, max_edges = edges.most_common(2)[1]  #skipping the right edge of the wall
    return len(wall) - max_edges