Given the left view and front view of the skyline of a block as two arrays, find the maximum total height of buildings in the block.

For both of the matrices below,

Left view of the skyline is [4, 2, 3].

Front View of the skyline is [3, 2, 3, 4]

Based on the left and front view, the volume we'd be able to build on the block could be

[3, 2, 3, 4] [3, 2, 3, 4] [2, 2, 2, 2] or [2, 2, 2, 1] [3, 2, 3, 3] [1, 2, 3, 3]

The total height of building in the first matrix is the largest possible.

Therefore the maximum total volume for the left and front views given is sum of the first matrix 31.

The solution should take no more than O(NlogN + MlogM) time and O(1) space.

Input gives two arrays.

left view [3, 5]

front view [2, 5]

Output 12.

Because the largest block volume to build is

[2, 3] [2, 5]

Return sum 2 + 3 + 2 + 5 = 12

Analyse the test cases and we'll find a pattern of the maximum height of a building at (i,j) is

min(left_view[i], front_view[j])

Therefore the naive solution is iterate through every cell in the MxN matrix and figure out the max height of each building, which takes O(MN) time.

A more optimal solution is to sort the input arrays and find the sum of each row based on the sum of previous row.

Assume after sorting, left_view [2, 3, 4], front_view is [2, 3, 3, 4].

Assume after sorting, left_view [2, 3, 4], front_view is [2, 3, 3, 4].

When i = 0, left_view[0] is 2. Every building in front_view is no less than 2.

Therefore the tallest building possible for the row 0 is 2. The total volume for row 0 is 2 * len(front_view) = 8.

When i = 1, left_view[1] is 3.

One value in front_view is 2, which is less than 3. Therefore the max volume for row 1 is [2, 3, 3, 3].

When i = 2, left_view[2] is 3.

One value in front_view is 2, which is less than 3. Therefore the max volume for row 2 is [2, 3, 3, 3].

When i = 3, left_view[3] is 4.

First three values in front_view is is less than 4. Therefore the max volume for row 2 is [2, 3, 3, 4].

The total for each row is sum(front_view elements < left_view[i]) + remaining element size * left_view[i].

We just need to find the split point index j in front_view[j - 1] < left_view[i] and front_view[j] >= left_view[i].

The total for each row will be sum(front_view[0:j]) + left_view[i] * (len(front_view) - j).

Find the split index j for every row and sum up the row totals. Getting the total takes O(M + N) if arrays are sorted. Sorting two arrays takes O(MlogM + NlogN).

Getting the total takes O(M + N) if arrays are sorted. Sorting two arrays takes O(MlogM + NlogN).

```
def max_building_volume(left, front):
left = sorted(left)
front = sorted(front)
i = 0
sum_to_ith = 0 # prefix sum of 0 to ith building in current row up
total = 0 #total volume
for h in left:
while i < len(front) and h >= front[i]:
sum_to_ith += front[i]
i += 1
total += (sum_to_ith + (len(front) - i) * h)
return total
```

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