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835. Image Overlap

MediumOpen on LeetCodeProblem statement

Problem Statement

835. Image Overlap

Medium

You are given two images, img1 and img2, represented as binary, square matrices of size n x n. A binary matrix has only 0s and 1s as values.

We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the overlap by counting the number of positions that have a 1 in both images.

Note also that a translation does not include any kind of rotation. Any 1 bits that are translated outside of the matrix borders are erased.

Return the largest possible overlap.

 

Example 1:

Input: img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]]
Output: 3
Explanation: We translate img1 to right by 1 unit and down by 1 unit.

The number of positions that have a 1 in both images is 3 (shown in red).

Example 2:

Input: img1 = [[1]], img2 = [[1]]
Output: 1

Example 3:

Input: img1 = [[0]], img2 = [[0]]
Output: 0

 

Constraints:

Java

Source file
class Solution {
    public int largestOverlap(int[][] img1, int[][] img2) {
        int n = img1.length, maxOverlap = 0;
        int[][] imgBig = new int[n * 3][n * 3];
        for (int i = n; i < 2 * n; i++)
            for (int j = n; j < 2 * n; j++)
                imgBig[i][j] = img2[i - n][j - n];
        for (int x = 0; x < 2 * n; x++) {
            for (int y = 0; y < 2 * n; y++) {
                int overlap = 0;
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < n; j++) {
                        overlap += (img1[i][j] & imgBig[i + x][j + y]);
                    }
                }
                maxOverlap = Math.max(overlap, maxOverlap);
            }
        }
        return maxOverlap;
    }
}