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CornerRectangles.java
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/* (C) 2024 YourCompanyName */
package dynamic_programming;
/**
* Created by gouthamvidyapradhan on 26/12/2017. Given a grid where each entry is only 0 or 1, find
* the number of corner rectangles.
*
* <p>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.
*
* <p>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[1][2], grid[1][4],
* grid[3][2], grid[3][4]. 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: The number of rows and columns of grid will each be in the range [1, 200]. Each
* grid[i][j] will be either 0 or 1. The number of 1s in the grid will be at most 6000.
*
* <p>Solution O(n + m ^ 2): For every row, consider each pair of 1s (every column pairs) and sum up
* the previous occurrence of 1s for the same column.
*/
public class CornerRectangles {
/**
* Main method
*
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception {
int[][] A = {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}};
System.out.println(new CornerRectangles().countCornerRectangles(A));
}
public int countCornerRectangles(int[][] grid) {
int[][] count = new int[grid[0].length][grid[0].length];
int result = 0;
for (int[] row : grid) {
for (int i = 0; i < row.length; i++) {
if (row[i] == 1) {
for (int j = i + 1; j < row.length; j++) {
if (row[j] == 1) {
if (count[i][j] > 0) {
result += count[i][j];
}
count[i][j]++;
}
}
}
}
}
return result;
}
}