1074. Number of Submatrices That Sum to Target

Hard

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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 <= 100

• 1 <= matrix[0].length <= 100

• -1000 <= matrix[i] <= 1000

• -10^8 <= target <= 10^8

Same concept as https://leetcode.com/problems/subarray-sum-equals-k/

class Solution:
    def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
        #.
        prefix = matrix
        for i in range(len(matrix)):
            for j in range(1, len(matrix[0])):
                prefix[i][j] += prefix[i][j-1]
        counter = collections.Counter()
        count = 0
        # We calculate the prefix sum by row thus to get submatrix sum
        # we use startColumn and endColumn window to sub matrix sum.
        for i in range(len(matrix[0])):
            for j in range(i, len(matrix[0])):
                counter.clear()
                counter[0] = 1 # Simple denotion that sum == k
                s = 0
                for k in range(len(matrix)):
                    # Consider [j] as endColumn and [i-1] as startColumn
                    s += prefix[k][j] - (prefix[k][i-1] if i > 0 else 0)
                    count += counter[s - target]
                    counter[s] += 1
        return count