You are given two images img1 and img2 both of size n x n, 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: img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]]
Output: 3
Explanation: We slide 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
n == img1[i].length
n == img2.length 
n == img2[i].length
1 <= n <= 30
img1[i][j] is 0 or 1.
img2[i][j] is 0 or 1.

Difficulty: MEDIUM
Tags: Matrices

Data structures
Skill types: Data structures

Starter code:
class Solution:
    def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
        

Solutions:
class Solution:

	def largestOverlap(self, A, B) -> int:
		leng = len(A[0])
		a = 0
		b = 0
		for i in range(0, leng * leng):
			row = int(i % leng)
			col = int(i / leng)
			a = (a << 1) + A[col][row]
			b = (b << 1) + B[col][row]
		maxsum = 0
		for i in range(-leng + 1, leng):
			if i < 0:
				mask = ('0' * abs(i) + '1' * (leng - abs(i))) * leng
				bp = (b & int(mask, 2)) << abs(i)
			elif i > 0:
				mask = ('1' * (leng - abs(i)) + '0' * abs(i)) * leng
				bp = (b & int(mask, 2)) >> abs(i)
			else:
				bp = b
			for j in range(-leng + 1, leng):
				if j < 0:
					bpp = bp >> leng * abs(j)
				elif j > 0:
					bpp = bp << leng * abs(j) & 2 ** (leng * leng) - 1
				else:
					bpp = bp
				maxsum = max(maxsum, bin(a & bpp).count('1'))
		return maxsum


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		N = len(A)
		ct = Counter()
		for (i, row) in enumerate(A):
			for (j, col) in enumerate(row):
				if col:
					for (i2, row2) in enumerate(B):
						for (j2, col2) in enumerate(row2):
							if col2:
								ct[i - i2, j - j2] += 1
		return max(ct.values() or [0])


class Solution(object):

	def largestOverlap(self, A, B):
		count = [0] * (2 * len(A) - 1) ** 2
		for (i, row) in enumerate(A):
			for (j, v) in enumerate(row):
				if not v:
					continue
				for (i2, row2) in enumerate(B):
					for (j2, v2) in enumerate(row2):
						if not v2:
							continue
						count[(len(A) - 1 + i - i2) * (2 * len(A) - 1) + len(A) - 1 + j - j2] += 1
		return max(count)


import collections

class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		count = collections.Counter()
		for (i, rowa) in enumerate(A):
			for j in range(len(rowa)):
				if rowa[j]:
					for (m, rowb) in enumerate(B):
						for n in range(len(rowb)):
							if rowb[n]:
								count[i - m, j - n] += 1
		return max(list(count.values()) or [0])


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		dic = {}
		for i in range(len(A)):
			for j in range(len(A[i])):
				pos = A[i][j]
				if pos != 1:
					continue
				for x in range(len(B)):
					for y in range(len(B[x])):
						pos = B[x][y]
						if pos != 1:
							continue
						(xdiff, ydiff) = (y - j, x - i)
						if (xdiff, ydiff) not in dic:
							dic[xdiff, ydiff] = 1
						else:
							dic[xdiff, ydiff] += 1
		if len(dic) == 0:
			return 0
		return max(dic.values())


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:

		def nonzero_cells(matrix):
			ones = []
			for r in range(len(matrix)):
				for c in range(len(matrix[r])):
					if matrix[r][c] == 1:
						ones.append((r, c))
			return ones
		a = nonzero_cells(A)
		b = nonzero_cells(B)
		overlap_vectors = defaultdict(int)
		max_overlap = 0
		for (x_a, y_a) in a:
			for (x_b, y_b) in b:
				overlap_vector = (x_b - x_a, y_b - y_a)
				overlap_vectors[overlap_vector] += 1
				max_overlap = max(max_overlap, overlap_vectors[overlap_vector])
		return max_overlap


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		(counts, best) = ({}, 0)
		size = len(A[0])
		for x1 in range(size):
			for y1 in range(size):
				if A[y1][x1] == 1:
					for x2 in range(size):
						for y2 in range(size):
							if B[y2][x2] == 1:
								diff = (x2 - x1, y2 - y1)
								counts[diff] = counts.get(diff, 0) + 1
								best = counts[diff] if counts[diff] > best else best
		return best


from functools import lru_cache
from copy import deepcopy

@lru_cache(None)
def countBits(n):
	return 0 if n == 0 else (n & 1) + countBits(n >> 1)

class Solution:

	def stateCompress(self, img):
		(n, m) = (len(img), len(img[0]))
		rows = [0 for _ in range(n)]
		for (i, row) in enumerate(img):
			for e in row:
				rows[i] = (rows[i] << 1) + e
		return (n, m, rows)

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		(n, m, rA) = self.stateCompress(A)
		(_, _, rB) = self.stateCompress(B)

		def shift(arr1, arr2, l):
			return sum((countBits(e1 & e2) for (e1, e2) in zip(arr1[:l], arr2[n - l:n])))
		maxOverLap = 0
		tA = deepcopy(rA)
		tB = deepcopy(rB)
		for dc in range(m):
			for dr in range(1, n + 1):
				maxOverLap = max(maxOverLap, shift(tA, rB, dr))
				maxOverLap = max(maxOverLap, shift(tB, rA, dr))
			tA = [e >> 1 for e in tA]
			tB = [e >> 1 for e in tB]
		return maxOverLap


class Solution:

	def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
		n = len(img1)
		s1 = set()
		s2 = set()
		for (j, row) in enumerate(img1):
			for (i, v) in enumerate(row):
				if v == 1:
					s1.add((j, i))
		for (j, row) in enumerate(img2):
			for (i, v) in enumerate(row):
				if v == 1:
					s2.add((j, i))
		d = {}
		for x in s1:
			for y in s2:
				v0 = x[0] - y[0]
				v1 = x[1] - y[1]
				vv = (v0, v1)
				if vv in d:
					d[vv] += 1
				else:
					d[vv] = 1
		dl = sorted(list(d.items()), key=lambda x: x[1])
		if not len(dl):
			return 0
		return dl[-1][1]


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:

		def get_movement(a, b):
			x = a[0] - b[0]
			y = a[1] - b[1]
			return (x, y)
		A_elements = []
		B_elements = []
		dimension = None
		for (i, row) in enumerate(A):
			for (j, val) in enumerate(row):
				if val == 1:
					A_elements.append((i, j))
				if B[i][j] == 1:
					B_elements.append((i, j))
		dimension = i + 1
		maximo = 0
		A_len = len(A_elements)
		B_len = len(B_elements)
		maximo_posible = min(A_len, B_len)
		moved = {}
		for elem_A in A_elements:
			for elem_B in B_elements:
				move = get_movement(elem_A, elem_B)
				if move not in list(moved.keys()):
					moved[move] = 1
				else:
					moved[move] += 1
				if moved[move] > maximo:
					maximo = moved[move]
					if maximo >= maximo_posible:
						break
		return maximo


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		offset_map = {}
		max_overlap = 0
		for row in range(0, len(A)):
			offset_map[row] = {}
			offset_map[row * -1] = {}
			for col in range(0, len(A[row])):
				r_row = row
				r_col = col
				if not offset_map[r_row] or not r_col in offset_map[r_row] or (not offset_map[r_row][r_col]):
					overlap = self.shift_and_count(A, B, r_row, r_col)
					overlap = max(overlap, self.shift_and_count(B, A, r_row, r_col))
					max_overlap = max(overlap, max_overlap)
					offset_map[r_row][r_col] = overlap
		return max_overlap

	def shift_and_count(self, A: List[List[int]], B: List[List[int]], row_shift: int, col_shift: int) -> int:
		result = 0
		for row in range(len(A)):
			new_row = row - row_shift
			if new_row >= 0 and new_row < len(A):
				for col in range(len(A[row])):
					new_col = col - col_shift
					if new_col >= 0 and new_col < len(A[row]):
						if A[new_row][new_col] == 1 and B[row][col] == 1:
							result += 1
		return result

	def overlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		overlap = 0
		if len(A) != len(B):
			raise ValueError
		for row in range(0, len(A)):
			for col in range(0, len(A[row])):
				if len(A[row]) != len(B[row]):
					raise ValueError
				if A[row][col] == 1 and B[row][col] == 1:
					overlap += 1
		return overlap


def mat_to_set(A):
	return {(i, j) for (i, row) in enumerate(A) for (j, bit) in enumerate(row) if bit}

class Solution:

	def largestOverlap(self, A, B) -> int:
		N = len(A)
		(A, B) = (mat_to_set(A), mat_to_set(B))
		if len(A) > len(B):
			(A, B) = (B, A)
		max_overlap = 0
		for di in range(1 - N, N):
			for dj in range(1 - N, N):
				overlap = 0
				for (i, j) in A:
					if (i + di, j + dj) in B:
						overlap += 1
				max_overlap = max(max_overlap, overlap)
		return max_overlap


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		max_count = 0
		a_coords = self.coordinates(A)
		b_coords = self.coordinates(B)
		for v in range(-1 * (len(A) - 1), len(A)):
			for h in range(-1 * (len(A[0]) - 1), len(A[0])):
				overlap_count = 0
				for (r, c) in a_coords:
					if (r + v, c + h) in b_coords:
						overlap_count += 1
				max_count = max(max_count, overlap_count)
		return max_count

	def coordinates(self, image):
		coords = set()
		for (r, row) in enumerate(image):
			for (c, value) in enumerate(row):
				if value == 1:
					coords.add((r, c))
		return coords


from functools import lru_cache

@lru_cache
def countBits(n):
	if n == 0:
		return 0
	return (n & 1) + countBits(n >> 1)

class Solution:

	def stateCompress(self, img):
		(n, m) = (len(img), len(img[0]))
		rows = [0 for _ in range(n)]
		for (i, row) in enumerate(img):
			for (j, e) in enumerate(row):
				rows[i] = (rows[i] << 1) + e
		return (n, m, rows)

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		(n, m, rA) = self.stateCompress(A)
		(_, _, rB) = self.stateCompress(B)

		@lru_cache
		def shift(xAs, xAe, xBs, xBe, dy):
			if dy > 0:
				ret = list((countBits(e1 << dy & e2) for (e1, e2) in zip(rA[xAs:xAe], rB[xBs:xBe])))
				return ret
			else:
				dy = -dy
				ret = list((countBits(e1 >> dy & e2) for (e1, e2) in zip(rA[xAs:xAe], rB[xBs:xBe])))
				return ret
		print((rA, rB))
		maxOverLap = 0
		for dr in range(n):
			for dc in range(m):
				maxOverLap = max(maxOverLap, sum(shift(dr, n, 0, n - dr, dc)))
				maxOverLap = max(maxOverLap, sum(shift(0, dr, n - dr, n, dc)))
				maxOverLap = max(maxOverLap, sum(shift(dr, n, 0, n - dr, -dc)))
				maxOverLap = max(maxOverLap, sum(shift(0, dr, n - dr, n, -dc)))
		return maxOverLap


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		size = len(A)
		shifts = [ii for ii in range(size)]
		max_overlap = 0
		for x in shifts:
			for y in shifts:
				max_overlap = max(self.shift_and_count(x_shift=x, y_shift=y, array_reference=A, array_to_be_shifted=B), max_overlap)
				max_overlap = max(self.shift_and_count(x_shift=x, y_shift=y, array_reference=B, array_to_be_shifted=A), max_overlap)
		return max_overlap

	def shift_and_count(self, x_shift: int, y_shift: int, array_reference: List[List[int]], array_to_be_shifted: List[List[int]]) -> int:
		if x_shift > 0:
			array_to_be_shifted = self.shift_left(array=array_to_be_shifted, shifts=x_shift)
		if y_shift > 0:
			array_to_be_shifted = self.shift_up(array=array_to_be_shifted, shifts=y_shift)
		return self.overlap(array_a=array_reference, array_b=array_to_be_shifted)

	def overlap(self, array_a: List[List[int]], array_b: List[List[int]], value: int=1) -> int:
		common_values = 0
		for (row_a, row_b) in zip(array_a, array_b):
			for (a, b) in zip(row_a, row_b):
				if a == b == value:
					common_values += 1
		return common_values

	def shift_up(self, array: List[List[int]], shifts: int, fill_value: int=0) -> List[List[int]]:
		a = array[shifts:]
		for _ in range(shifts):
			a.append([fill_value] * len(array[0]))
		return a

	def shift_left(self, array: List[List[int]], shifts: int, fill_value: int=0) -> List[List[int]]:
		a = []
		for sublist in array:
			last_elements = [fill_value] * shifts
			sublist = sublist[shifts:] + last_elements
			a.append(sublist)
		return a


def matrix_to_set(A):
	return {(i, j) for (i, row) in enumerate(A) for (j, pixel) in enumerate(row) if pixel}

class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		N = len(A)
		(set_a, set_b) = (matrix_to_set(A), matrix_to_set(B))
		if len(set_b) < len(set_a):
			(set_b, set_a) = (set_a, set_b)
		return max((sum(((i + di, j + dj) in set_b for (i, j) in set_a)) for di in range(-N + 1, N) for dj in range(-N + 1, N)))


class Solution:

	def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
		N = len(img1)
		if N == 1:
			return 1 if img1[0][0] == 1 and img2[0][0] == 1 else 0
		max_slide = N
		sliders = set()
		for i in range(N):
			for j in range(N):
				sliders.add((i, j))
				sliders.add((-i, j))
				sliders.add((i, -j))
				sliders.add((-i, -j))

		def slide_and_count(slide):
			res = 0
			(left, down) = slide
			for i in range(min(N, N - left)):
				i2 = i + left
				for j in range(min(N, N - down)):
					j2 = j + down
					if i2 < 0 or j2 < 0:
						continue
					try:
						if img1[i][j] == 1 and img2[i2][j2] == 1:
							res += 1
					except:
						pass
			print(slide, res)
			return res
		return max((slide_and_count(slide) for slide in sliders))


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		return max(self.largestOverlapAtoB(A, B), self.largestOverlapAtoB(B, A))

	def largestOverlapAtoB(self, A, B):
		shift_row = 0
		shift_col = 0
		overlap = 0
		for shift_row in range(len(A[0])):
			for shift_col in range(len(A[0])):
				shifted_A = self.shiftMatrix(A, shift_row, shift_col)
				overlap = max(overlap, self.findOverlap(B, shifted_A))
		return overlap

	def findOverlap(self, A, B):
		lenght = len(A[0])
		overlap = 0
		for ii in range(lenght):
			for jj in range(lenght):
				if A[ii][jj] == B[ii][jj] and A[ii][jj] == 1:
					overlap += 1
		return overlap

	def shiftMatrix(self, M, shift_row, shift_col):
		lenght = len(M[0])
		out = []
		for _ in range(shift_row):
			out.append(lenght * [0])
		for ix_r in range(shift_row, lenght):
			out_row = []
			for _ in range(shift_col):
				out_row.append(0)
			for ix_c in range(shift_col, lenght):
				out_row.append(M[ix_r - shift_row][ix_c - shift_col])
			out.append(out_row)
		return out


def check_overlap(side, down, A, B):
	overlap = 0
	for i in range(len(A)):
		for j in range(len(A)):
			try:
				overlap += A[i + side][j + down] & B[i][j]
			except:
				pass
	return overlap

class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		max_overlap = 0
		for i in range(len(A)):
			for j in range(len(A)):
				max_overlap = max(max_overlap, check_overlap(i, j, A, B))
				max_overlap = max(max_overlap, check_overlap(i, j, B, A))
		return max_overlap


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:

		def match(m1, m2):
			prod = 0
			for (r1, r2) in zip(m1, m2):
				for (v1, v2) in zip(r1, r2):
					prod += v1 * v2
			return prod
		pad = [0] * (3 * (len(A) - 1))
		largeA = [pad] * (len(A) - 1)
		for i in range(len(A)):
			largeA.append([0] * (len(A) - 1) + A[i] + [0] * (len(A) - 1))
		largeA += [pad] * (len(A) - 1)
		maxProd = 0
		for i in range(len(largeA) - len(A) + 1):
			for j in range(len(largeA) - len(A) + 1):
				subA = []
				for r in largeA[i:len(A) + i]:
					subA.append(r[j:len(A) + j])
				prod = match(subA, B)
				if prod > maxProd:
					maxProd = prod
		return maxProd


from typing import List

class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:

		def check_overlap(side_x, down_x, A, B):
			overlap = 0
			for i in range(len(A)):
				for j in range(len(A)):
					try:
						overlap += A[i + side_x][j + down_x] & B[i][j]
					except:
						pass
			return overlap
		max_overlap = 0
		for i in range(len(A)):
			for j in range(len(A)):
				max_overlap = max(max_overlap, check_overlap(i, j, A, B))
				max_overlap = max(max_overlap, check_overlap(i, j, B, A))
		return max_overlap


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		a_1_locations = set()
		b_1_locations = set()
		for (grid, locations) in [(A, a_1_locations), (B, b_1_locations)]:
			for (i, col) in enumerate(grid):
				for (j, val) in enumerate(col):
					if val == 1:
						locations.add((i, j))
		max_overlap = 0
		max_abs_x1_offset = len(A)
		max_abs_x2_offset = len(A[0])
		a_offsets = itertools.product(range(-1 * max_abs_x1_offset, max_abs_x1_offset), range(-1 * max_abs_x2_offset, max_abs_x2_offset))
		for (a_offset_x, a_offset_y) in a_offsets:
			a_1_locations_offset = set(map(lambda location: (location[0] + a_offset_x, location[1] + a_offset_y), a_1_locations))
			overlap = len(a_1_locations_offset & b_1_locations)
			if overlap > max_overlap:
				max_overlap = overlap
		return max_overlap


class Solution:

	def largestOverlap(self, img1: List[List[int]], img2: List[List[int]]) -> int:
		A = [(i, j) for (i, row) in enumerate(img1) for (j, item) in enumerate(row) if item]
		B = [(i, j) for (i, row) in enumerate(img2) for (j, item) in enumerate(row) if item]
		count = collections.Counter(((ax - bx, ay - by) for (ax, ay) in A for (bx, by) in B))
		return max(list(count.values()) or [0])


class Solution:

	def getCoordinate(self, A: List[List[int]]):
		list_a = []
		for i in range(len(A)):
			for j in range(len(A[i])):
				if A[i][j] == 1:
					list_a.append(i * 100 + j)
		return list_a

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		list_a = self.getCoordinate(A)
		list_b = self.getCoordinate(B)
		diff = [a - b for a in list_a for b in list_b]
		counter = {}
		for v in diff:
			if counter.get(v, None) == None:
				counter[v] = 1
			else:
				counter[v] += 1
		max_value = 0
		for key in list(counter.keys()):
			if counter[key] > max_value:
				max_value = counter[key]
		return max_value


class Solution:

	def largestOverlap(self, A: List[List[int]], B: List[List[int]]) -> int:
		index_A = [(i, j) for (i, row) in enumerate(A) for (j, item) in enumerate(row) if item]
		index_B = [(i, j) for (i, row) in enumerate(B) for (j, item) in enumerate(row) if item]
		cnt = collections.Counter([(ax - bx, ay - by) for (ax, ay) in index_A for (bx, by) in index_B])
		if len(cnt) == 0:
			return 0
		return max(cnt.values())


class Solution:

	def largestOverlap(self, A, B):
		A = [(i, j) for (i, row) in enumerate(A) for (j, item) in enumerate(row) if item]
		B = [(i, j) for (i, row) in enumerate(B) for (j, item) in enumerate(row) if item]
		count = {}
		for (ax, ay) in A:
			for (bx, by) in B:
				delta = (ax - bx, ay - by)
				if not delta in count:
					count[delta] = 0
				count[delta] += 1
		return max(count.values()) if count else 0