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.
Note:
1 <= matrix.length <= 300
1 <= matrix[0].length <= 300
-1000 <= matrix[i] <= 1000
-10^8 <= target <= 10^8
publicclassSolution {publicinthelper(int[] sums,int target) {int sum =0, ans =0;HashMap<Integer,Integer> map =newHashMap<>();map.put(0,1);for (int x : sums) { sum += x; ans +=map.getOrDefault(sum - target,0);map.put(sum,map.getOrDefault(sum,0) +1); }return ans; }publicintnumSubmatrixSumTarget(int[][] A,int target) {if (A.length==0)return0;int ans =0;// Take 2 pointersfor (int i =0; i <A[0].length; i++) {int sum[] =newint[A.length];for (int j = i; j <A[0].length; j++) {// Find the prefix sum array of columns between these 2 pointersfor (int k =0; k <A.length; k++) sum[k] +=A[k][j];// Find subarays with sum == target ans +=helper(sum, target); } }return ans; }}