LeetCode 솔루션 분류
[10/27] 835. Image Overlap
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Medium
987358Add to ListShareYou 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].lengthn == img2.length == img2[i].length1 <= n <= 30img1[i][j]is either0or1.img2[i][j]is either0or1.
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Runtime: 1250 ms, faster than 46.20% of Python3 online submissions for Image Overlap.
Memory Usage: 14.7 MB, less than 43.86% of Python3 online submissions for Image Overlap.
Memory Usage: 14.7 MB, less than 43.86% of Python3 online submissions for Image Overlap.
from collections import defaultdict
class Solution:
def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
a = []
b = []
N = len(img1)
for row in range(N):
for col in range(N):
if img1[row][col] == 1: a.append((row,col))
if img2[row][col] == 1: b.append((row,col))
diff_dic = defaultdict(int)
for point1 in a:
for point2 in b:
diff_dic[(point1[0]-point2[0], point1[1]-point2[1])] += 1
return max(diff_dic.values()) if diff_dic.values() else 0
