1074. Number of Submatrices That Sum to Target

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. 1 <= matrix.length <= 300
2. 1 <= matrix[0].length <= 300
3. -1000 <= matrix[i] <= 1000
4. -10^8 <= target <= 10^8

---

 

思路：一开始的想法是：求一个二维数组的running sum，然后维护一个set数组，一边遍历一边算“以当前点结尾的情况下贡献了多少结果”，具体如下

class Solution(object):
    def numSubmatrixSumTarget(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: int
        """
        n,m=len(matrix),len(matrix[0])
        run=[[matrix[i][j] for j in range(m)] for i in range(n)]
        for i in range(n):
            for j in range(m):
                if i>0: run[i][j]+=run[i-1][j]
                if j>0: run[i][j]+=run[i][j-1]
                if i>0 and j>0: run[i][j]-=run[i-1][j-1]
        
#        print(run)
        res=0
        ds=[{} for _ in range(m)]
        for i in range(n):
            for j in range(m):
                s=run[i][j]
                ds[j][s]=ds[j].get(s,0)+1
            
            for j in range(m):
                if run[i][j]==0: res+=1
                for k in range(j):
                    res+=ds[k].get(run[i][j],0)
                if ds[j][run[i][j]]>1: res+=ds[j][run[i][j]]-1
        
        if matrix[-1][-1]==0: res+=1
        
        return res
    
s=Solution()
#print(s.numSubmatrixSumTarget(matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0))
#print(s.numSubmatrixSumTarget(matrix = [[1,-1],[-1,1]], target = 0))
print(s.numSubmatrixSumTarget(matrix = [[0,0,0,1,1],[1,1,1,0,1],[1,1,1,1,0],[0,0,0,1,0],[0,0,0,1,1]], target = 0))

但是会WA，因为Sum[ii,jj]-Sum[i,j]并不是[i~ii,j~jj]这个小矩形的sum，就像在求running sum的时候要用容斥原理来求一样

究其WA的原因就是“Sum计算结果错误”，那有没有一种方式能准确计算Sum，然后套用一下上面呢？直观想法是怎么修改Sum的定义，但是如果还是像上面那样“遍历（限定）一个维度（比如遍历每行，分别求出每行作为小矩形底端情况下的结果，然后累加）的话”，是做不到的

所以只能多“套一层循环（多加一些限定）”，即“限制小矩形的起始行和结束行”

class Solution(object):
    def numSubmatrixSumTarget(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: int
        """
        n,m=len(matrix),len(matrix[0])
        run=[[matrix[i][j] for j in range(m)] for i in range(n)]
        for i in range(1,n):
            for j in range(m):
                run[i][j]+=run[i-1][j]
        run=[[0]*m]+run
#        print(run)

        def helper(a):
            d={0:1}
            s=res=0
            for i in range(len(a)):
                s+=a[i]
                res+=d.get(s-target,0)
                d[s]=d.get(s,0)+1
            return res
        
        res=0
        for i in range(n+1):
            for j in range(i+1,n+1):
                a=[k2-k1 for k1,k2 in zip(run[i],run[j])]
#                print(a)
                res+=helper(a)
        return res
    
s=Solution()
print(s.numSubmatrixSumTarget(matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0))
print(s.numSubmatrixSumTarget(matrix = [[1,-1],[-1,1]], target = 0))
print(s.numSubmatrixSumTarget(matrix = [[0,0,0,1,1],[1,1,1,0,1],[1,1,1,1,0],[0,0,0,1,0],[0,0,0,1,1]], target = 0))