Number of submatrices that sum to target¶
Time: O(M^2xN); Space: O(N); hard
Given a matrix, and a target, return the number of non-empty submatrices that sum to target.
A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2.
Two submatrices (x1, y1, x2, y2) and (x1’, y1’, x2’, y2’) are different if they have some coordinate that is different: for example, if x1 != x1’.
Example 1:
Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
Output: 4
Explanation:
The four 1x1 submatrices that only contain 0.
Example 2:
Input: matrix = [[1,-1],[-1,1]], target = 0
Output: 5
Explanation:
The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
Notes:
1 <= len(matrix) <= 300
1 <= len(matrix[0]) <= 300
-1000 <= matrix[i] <= 1000
-10^8 <= target <= 10^8
[1]:
import collections
class Solution1(object):
"""
Time: O(m^2*n), m is min(r, c), n is max(r, c)
Space: O(n), which doesn't include transposed space
"""
def numSubmatrixSumTarget(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: int
"""
if len(matrix) > len(matrix[0]):
return self.numSubmatrixSumTarget(map(list, zip(*matrix)), target)
for i in range(len(matrix)):
for j in range(len(matrix[i])-1):
matrix[i][j+1] += matrix[i][j]
result = 0
for i in range(len(matrix)):
prefix_sum = [0] * len(matrix[i])
for j in range(i, len(matrix)):
lookup = collections.defaultdict(int)
lookup[0] = 1
for k in range(len(matrix[j])):
prefix_sum[k] += matrix[j][k]
if prefix_sum[k] - target in lookup:
result += lookup[prefix_sum[k] - target]
lookup[prefix_sum[k]] += 1
return result
[2]:
s = Solution1()
matrix = [[0,1,0], [1,1,1], [0,1,0]]
target = 0
assert s.numSubmatrixSumTarget(matrix, target) == 4
matrix = [[1,-1], [-1,1]]
target = 0
assert s.numSubmatrixSumTarget(matrix, target) == 5