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835. Image Overlap

Description

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:

  • n == img1.length == img1[i].length
  • n == img2.length == img2[i].length
  • 1 <= n <= 30
  • img1[i][j] is either 0 or 1.
  • img2[i][j] is either 0 or 1.

Solutions

  • class Solution {
    public int largestOverlap(int[][] img1, int[][] img2) {
        int n = img1.length;
        Map<List<Integer>, Integer> cnt = new HashMap<>();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (img1[i][j] == 1) {
                    for (int h = 0; h < n; ++h) {
                        for (int k = 0; k < n; ++k) {
                            if (img2[h][k] == 1) {
                                List<Integer> t = List.of(i - h, j - k);
                                ans = Math.max(ans, cnt.merge(t, 1, Integer::sum));
                            }
                        }
                    }
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int largestOverlap(vector<vector<int>>& img1, vector<vector<int>>& img2) {
        int n = img1.size();
        map<pair<int, int>, int> cnt;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (img1[i][j]) {
                    for (int h = 0; h < n; ++h) {
                        for (int k = 0; k < n; ++k) {
                            if (img2[h][k]) {
                                ans = max(ans, ++cnt[{i - h, j - k}]);
                            }
                        }
                    }
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
        n = len(img1)
        cnt = Counter()
        for i in range(n):
            for j in range(n):
                if img1[i][j]:
                    for h in range(n):
                        for k in range(n):
                            if img2[h][k]:
                                cnt[(i - h, j - k)] += 1
        return max(cnt.values()) if cnt else 0


  • func largestOverlap(img1 [][]int, img2 [][]int) (ans int) {
	type pair struct{ x, y int }
	cnt := map[pair]int{}
	for i, row1 := range img1 {
		for j, x1 := range row1 {
			if x1 == 1 {
				for h, row2 := range img2 {
					for k, x2 := range row2 {
						if x2 == 1 {
							t := pair{i - h, j - k}
							cnt[t]++
							ans = max(ans, cnt[t])
						}
					}
				}
			}
		}
	}
	return
}

  • function largestOverlap(img1: number[][], img2: number[][]): number {
    const n = img1.length;
    const cnt: Map<number, number> = new Map();
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (img1[i][j] === 1) {
                for (let h = 0; h < n; ++h) {
                    for (let k = 0; k < n; ++k) {
                        if (img2[h][k] === 1) {
                            const t = (i - h) * 200 + (j - k);
                            cnt.set(t, (cnt.get(t) ?? 0) + 1);
                            ans = Math.max(ans, cnt.get(t)!);
                        }
                    }
                }
            }
        }
    }
    return ans;
}

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