<p>Given a <code>matrix</code>&nbsp;and a <code>target</code>, return the number of non-empty submatrices that sum to <font face="monospace">target</font>.</p>

<p>A submatrix <code>x1, y1, x2, y2</code> is the set of all cells <code>matrix[x][y]</code> with <code>x1 &lt;= x &lt;= x2</code> and <code>y1 &lt;= y &lt;= y2</code>.</p>

<p>Two submatrices <code>(x1, y1, x2, y2)</code> and <code>(x1&#39;, y1&#39;, x2&#39;, y2&#39;)</code> are different if they have some coordinate&nbsp;that is different: for example, if <code>x1 != x1&#39;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
<strong>Output:</strong> 4
<strong>Explanation:</strong> The four 1x1 submatrices that only contain 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[1,-1],[-1,1]], target = 0
<strong>Output:</strong> 5
<strong>Explanation:</strong> The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[904]], target = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= matrix.length &lt;= 100</code></li>
	<li><code>1 &lt;= matrix[0].length &lt;= 100</code></li>
	<li><code>-1000 &lt;= matrix[i][j] &lt;= 1000</code></li>
	<li><code>-10^8 &lt;= target &lt;= 10^8</code></li>
</ul>

class Solution:
    def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
        count = 0
        m, n = len(matrix), len(matrix[0])
        prefixSums = [[0] * n for _ in range(m)]
        # calculate row prefix sums (O(n^2))
        for i in range(m):
            cur = 0
            for j in range(n):
                cur += matrix[i][j]
                prefixSums[i][j] = cur
        # compress column into one prefix sum col (O(n^3))
        for c1 in range(n):
            for c2 in range(c1, n):
                # prefixSum -> count
                hash_map = defaultdict(int)
                hash_map[0] = 1
                prefixSum = 0
                for i in range(m):
                    prefixSum += prefixSums[i][c2] - (prefixSums[i][c1 - 1] if c1 > 0 else 0)
                    count += hash_map[prefixSum - target]
                    hash_map[prefixSum] += 1
        return count

    # def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
    #     count = 0
    #     m, n = len(matrix), len(matrix[0])
    #     prefixSums = [[0] * n for _ in range(m)]
    #     # calculate row prefix sums (O(n^2))
    #     for i in range(m):
    #         cur = 0
    #         for j in range(n):
    #             cur += matrix[i][j]
    #             prefixSums[i][j] = cur
    #     col = [0] * m
    #     # compress column into one prefix sum col (O(n^3))
    #     for c1 in range(n):
    #         for c2 in range(c1, n):
    #             for i in range(m):
    #                 col[i] = prefixSums[i][c2] - (prefixSums[i][c1 - 1] if c1 > 0 else 0)
    #             # prefixSum -> count
    #             hash_map = defaultdict(int)
    #             hash_map[0] = 1
    #             prefixSum = 0
    #             for i in range(m):
    #                 prefixSum += col[i]
    #                 count += hash_map[prefixSum - target]
    #                 hash_map[prefixSum] += 1
    #     return count


17500