## 750. Number Of Corner Rectangles

Given a grid where each entry is only 0 or 1, find the number of corner rectangles.

A corner rectangle is 4 distinct 1s on the grid that form an axis-aligned rectangle. Note that only the corners need to have the value 1. Also, all four 1s used must be distinct.

Example 1:

Input: grid =
[[1, 0, 0, 1, 0],
[0, 0, 1, 0, 1],
[0, 0, 0, 1, 0],
[1, 0, 1, 0, 1]]
Output: 1
Explanation: There is only one corner rectangle, with corners grid, grid, grid, grid.


Example 2:

Input: grid =
[[1, 1, 1],
[1, 1, 1],
[1, 1, 1]]
Output: 9
Explanation: There are four 2x2 rectangles, four 2x3 and 3x2 rectangles, and one 3x3 rectangle.


Example 3:

Input: grid =
[[1, 1, 1, 1]]
Output: 0
Explanation: Rectangles must have four distinct corners.


Note:

1. The number of rows and columns of grid will each be in the range [1, 200].
2. Each grid[i][j] will be either 0 or 1.
3. The number of 1s in the grid will be at most 6000.

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#### Approach #1: Count Corners [Accepted]

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Intuition

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We ask the question: for each additional row, how many more rectangles are added?

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For each pair of 1s in the new row (say at new_row[i] and new_row[j]), we could create more rectangles where that pair forms the base. The number of new rectangles is the number of times some previous row had row[i] = row[j] = 1.

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Algorithm

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Let\'s maintain a count count[i, j], the number of times we saw row[i] = row[j] = 1. When we process a new row, for every pair new_row[i] = new_row[j] = 1, we add count[i, j] to the answer, then we increment count[i, j].

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Complexity Analysis

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Time Complexity: where is the number of rows and columns.

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Space Complexity: in additional space.

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#### Approach #2: Heavy and Light Rows [Accepted]

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Intuition and Algorithm

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Can we improve on the ideas in Approach #1? When a row is filled with 1s, we do work to enumerate every pair of 1s. This is okay when is small, but expensive when is big.

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Say the entire top row is filled with 1s. When looking at the next row with say, f 1s that match the top row, the number of rectangles created is just the number of pairs of 1s, which is f * (f-1) / 2. We could find each f quickly using a Set and a simple linear scan of each row.

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Let\'s call a row to be heavy if it has more than points. The above algorithm changes the complexity of counting a heavy row from to , and there are at most heavy rows.

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Complexity Analysis

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Time Complexity: where is the number of ones in the grid.

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Space Complexity: in additional space, for rows, target, and count.

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Analysis written by: @awice.

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