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.

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(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

 

Approach #1: Array. [Java]

class Solution {
    public int largestOverlap(int[][] A, int[][] B) {
        int N = A.length;
        List<Integer> LA = new ArrayList<>();
        List<Integer> LB = new ArrayList<>();
        HashMap<Integer, Integer> count = new HashMap<>();
        for (int i = 0; i < N * N; ++i) if (A[i/N][i%N] == 1)
            LA.add(i / N * 100 + i % N);
        for (int i = 0; i < N * N; ++i) if (B[i/N][i%N] == 1)
            LB.add(i / N * 100 + i % N);
        for (int i : LA) for (int j : LB)
            count.put(i-j, count.getOrDefault(i-j, 0) + 1);
        int res = 0;
        for (int i : count.values()) res = Math.max(res, i);
        return res;
    }
}

　　

Analysis:

Assume index in A and B is [0, N*N-1]

 

Loop on A, if value == 1, save a coordinates i / N * 100 + i % N to LA.

Loop on B, if value == 1, save a coordinates i / N * 100 + i % N to LB.

Loop on combination (i, j) of LA and LB, increase count[i-j] by 1.

If we slid to make A[i] orverlap B[j], we can get 1 point.

Loop on count and return max values.

 

Time Complexity:

O(N^2) for preparing, and O(AB) for loop.

O(AB + N^2)

 

Reference:

https://leetcode.com/problems/image-overlap/discuss/130623/C%2B%2BJavaPython-Straight-Forward

 

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