Formatted question description: https://leetcode.ca/all/750.html

# 750. Number Of Corner Rectangles (Medium)

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.

Companies:

Related Topics:
Dynamic Programming

## Solution 1.

// OJ: https://leetcode.com/problems/number-of-corner-rectangles/

// Time: O(MN)
// Space: O(C^2) where C is the count of 1s.
class Solution {
public:
int countCornerRectangles(vector<vector<int>>& grid) {
int M = grid.size(), N = grid.size(), ans = 0;
map<int, set<int>> m;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (!grid[i][j]) continue;
m[i].insert(j);
}
}
for (auto i = m.begin(); i != m.end(); ++i) {
for (auto j = next(i); j != m.end(); ++j) {
int cnt = 0;
for (int y : i->second) {
if (j->second.find(y) == j->second.end()) continue;
++cnt;
}
ans += cnt * (cnt - 1) / 2;
}
}
return ans;
}
};


Java

class Solution {
public int countCornerRectangles(int[][] grid) {
Map<String, Integer> map = new HashMap<String, Integer>();
int rows = grid.length, columns = grid.length;
for (int i = 0; i < rows; i++) {
List<Integer> onesColumns = new ArrayList<Integer>();
for (int j = 0; j < columns; j++) {
if (grid[i][j] == 1)
}
if (onesColumns.size() > 1) {
int size = onesColumns.size();
for (int j = 0; j < size; j++) {
for (int k = j + 1; k < size; k++) {
int[] array = {onesColumns.get(j), onesColumns.get(k)};
String arrayStr = Arrays.toString(array);
int count = map.getOrDefault(arrayStr, 0);
count++;
map.put(arrayStr, count);
}
}
}
}
int totalCount = 0;
Set<String> set = map.keySet();
for (String key : set) {
int count = map.getOrDefault(key, 0);
totalCount += count * (count - 1) / 2;
}