835. Image Overlap

最新推荐文章于 2021-09-08 20:41:00 发布

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最新推荐文章于 2021-09-08 20:41:00 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10669690.html

本文探讨了两个相同大小的二进制矩阵图像A和B的最大重叠问题。通过滑动其中一个图像并将其放置在另一个图像之上，寻找位置重叠最多的方案。文章提供了详细的Java代码实现及算法分析。

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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 <= A.length = A[0].length = B.length = B[0].length <= 30
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)