> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/hashmap-and-hashset-and-sliding-window/image-overlap.md).

# Image Overlap

Two images `A` and `B` are given, represented as binary, square matrices of the same size.  (A binary matrix has only 0s and 1s as values.)

We translate one image however we choose (sliding it left, right, up, or down any number of units), and place it on top of the other image.  After, the *overlap* of this translation is the number of positions that have a 1 in both images.

(Note also that a translation does **not** include any kind of rotation.)

What is the largest possible overlap?

**Example 1:**

```
Input: A = [[1,1,0],
            [0,1,0],
            [0,1,0]]
       B = [[0,0,0],
            [0,1,1],
            [0,0,1]]
Output: 3
Explanation: We slide A to right by 1 unit and down by 1 unit.
```

**Notes:** 

1. `1 <= A.length = A[0].length = B.length = B[0].length <= 30`
2. `0 <= A[i][j], B[i][j] <= 1`

```java
class Solution {
    public int largestOverlap(int[][] A, int[][] B) {
        int rows = A.length, cols = A[0].length;
        List<int[]> la = new ArrayList<>(), lb = new ArrayList<>(); // two lists to save pixel coordinates
        for (int r = 0; r < rows; r++)
            for (int c = 0; c < cols; c++) {
                // save the pixel coordinates
                if (A[r][c] == 1)
                    la.add(new int[] { r, c });
                if (B[r][c] == 1)
                    lb.add(new int[] { r, c });
            }
        Map<String, Integer> map = new HashMap<>();
        // map to map the vector (from a pixel in A to a pixel in B) to its count
        for (int[] pixelA : la)
            for (int[] pixelB : lb) {
                String s = (pixelA[0] - pixelB[0]) + " " + (pixelA[1] - pixelB[1]); 
                // get the vector from a pixel in A to a pixel in B
                map.put(s, map.getOrDefault(s, 0) + 1); // count the number of same vectors
            }
        int max = 0;
        for (int count : map.values())
            max = Math.max(max, count);
        return max;
    }
}
```


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