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You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

The first, the third and the last rows have an edge at 1, thus you can draw a line that crosses three bricks less than the height of the wall. You need to find the edge that appears the most.

In this case it would be four it appears in rows 2,3,5,6.

After transforming the wall to a wall of edges, a SelectMany into a GroupBy and Max will do the trick.

You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

The first, the third and the last row have an edge at 1, thus you can draw a line that crosses three bricks less than the height of the wall. You need to find the edge that appears the most.

In this case it would be four it appears in rows 2,3,5,6.

After transforming the wall to a wall of edges, a SelectMany into a GroupBy and Max will do the trick.

You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

The first, the third and the last rows have an edge at 1, thus you can draw a line that crosses three bricks less than the height of the wall. You need to find the edge that appearances the most.

In this case it would be four it appearance in rows 2,3,5,6.

After transforming the wall to a wall of edges, a SelectMany into a GroupBy and Max will do the trick.

You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

The first, the third and the last rows have an edge at 1, thus you can draw a line that crosses three bricks less than the height of the wall. You need to find the edge that appears the most.

In this case it would be four it appears in rows 2,3,5,6.

After transforming the wall to a wall of edges, a SelectMany into a GroupBy and Max will do the trick.

You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

if you look at the middle column you are crossing the 4th edge, that is the edge after a total width of 4 four times more than any other.

height minus maximum edges that can be crossed in a column is your answer.

You could look at it in a different way, transform the wall of bricks to a wall of edges, in your example it would be

1, 3, 5
3, 4
1, 4
2
3, 4
1, 4, 5

The first, the third and the last rows have an edge at 1, thus you can draw a line that crosses three bricks less than the height of the wall. You need to find the edge that appearances the most.

In this case it would be four it appearance in rows 2,3,5,6.

After transforming the wall to a wall of edges, a SelectMany into a GroupBy and Max will do the trick.