1074-number-of-submatrices-that-sum-to-target¶
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Description¶
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.
Example 3:
Input: matrix = [[904]], target = 0 Output: 0
Constraints:
1 <= matrix.length <= 1001 <= matrix[0].length <= 100-1000 <= matrix[i] <= 1000-10^8 <= target <= 10^8
Solution(Python)¶
class Solution:
def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
return self.optimal(matrix, target)
# Time Complexity: O((m*n)^3)
# Space Complexity: O(1)
def bruteforce(self, matrix: List[List[int]], target: int) -> int:
res = 0
m = len(matrix)
n = len(matrix[0])
def submatrixSum(rowStart, rowSize, colStart, colSize):
subMatrixSum = 0
for i in range(rowStart, rowstart + rowSize):
for j in range(colStart, colStart + colSize):
subMatrixSum += matrix[i][j]
return subMatrixSum
for rowStart in range(m):
for rowSize in range(1, m + 1):
for colStart in range(n):
for colSize in range(1, n + 1):
if submatrixSum(rowStart, rowSize, colStart, colSize) == target:
res += 1
return res
# Time Complexity: O((m*n)^2)
# Space Complexity: O(m*n)
def better(self, matrix: List[List[int]], target: int) -> int:
res = 0
m = len(matrix)
n = len(matrix[0])
for row in range(m):
for col in range(1, n):
matrix[row][col] += matrix[row][col - 1]
for colstart in range(n):
for colend in range(colstart, n):
for rowstart in range(m):
sub_sum = 0
for rowend in range(rowstart, m):
sub_sum += (
matrix[rowend][colend] -
matrix[rowend][colstart - 1]
if colstart
else 0
)
if sub_sum == target:
res += 1
return res
# Time Complexity: O((m^2*n))
# Space Complexity: O(m*n)
def optimal(self, matrix: List[List[int]], target: int) -> int:
res = 0
m = len(matrix)
n = len(matrix[0])
for row in range(m):
for col in range(1, n):
matrix[row][col] += matrix[row][col - 1]
for colstart in range(n):
for colend in range(colstart, n):
cur_sum = 0
hash_map = {0: 1}
for rowstart in range(m):
cur_sum += matrix[rowstart][colend] - (
matrix[rowstart][colstart - 1] if colstart > 0 else 0
)
res += hash_map.get(cur_sum - target, 0)
hash_map[cur_sum] = hash_map.get(cur_sum, 0) + 1
return res