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Leetcode Problem #1074 (Hard): Number of Submatrices That Sum to Target
Description:
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Given a
matrixand atarget, return the number of non-empty submatrices that sum totarget.A submatrix
x1, y1, x2, y2is the set of all cellsmatrix[x][y]withx1 <= x <= x2andy1 <= 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, ifx1 != x1'.
Examples:
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. Visual:
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
Idea:
(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)
This problem is essentially a 2-dimensional version of #560. Subarray Sum Equals K (S.S.E.K). By using a prefix sum on each row or each column, we can compress this problem down to either N^2 iterations of the O(M) SSEK, or M^2 iterations of the O(N) SSEK.
In the SSEK solution, we can find the number of subarrays with the target sum by utilizing a result map (res) to store the different values found as we iterate through the array while keeping a running sum (csum). Just as in the case with a prefix sum array, the sum of a subarray between i and j is equal to the sum of the subarray from 0 to j minus the sum of the subarray from 0 to i-1.
Rather than iteratively checking if sum[0,j] - sum[0,i-1] = T for every pair of i, j values, we can flip it around to sum[0,j] - T = sum[0,i-1] and since every earlier sum value has been stored in res, we can simply perform a lookup on sum[0,j] - T to see if there are any matches.
When extrapolating this solution to our 2-dimensional matrix (M), we will need to first prefix sum the rows or columns, (which we can do in-place to avoid extra space, as we will not need the original values again). Then we should iterate through M again in the opposite order of rows/columns where the prefix sums will allow us to treat a group of columns or rows as if it were a 1-dimensional array and apply the SSEK algorithm.
Implementation:
There are only minor differences in the code of all four languages.
Javascript Code:
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var numSubmatrixSumTarget = function(M, T) { let xlen = M[0].length, ylen = M.length, ans = 0, res = new Map() for (let i = 0, r = M[0]; i < ylen; r = M[++i]) for (let j = 1; j < xlen; j++) r[j] += r[j-1] for (let j = 0; j < xlen; j++) for (let k = j; k < xlen; k++) { res.clear(), res.set(0,1), csum = 0 for (let i = 0; i < ylen; i++) { csum += M[i][k] - (j ? M[i][j-1] : 0) ans += (res.get(csum - T) || 0) res.set(csum, (res.get(csum) || 0) + 1) } } return ans};
Python Code:
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class Solution: def numSubmatrixSumTarget(self, M: List[List[int]], T: int) -> int: xlen, ylen, ans, res = len(M[0]), len(M), 0, defaultdict(int) for r in M: for j in range(1, xlen): r[j] += r[j-1] for j in range(xlen): for k in range(j, xlen): res.clear() res[0], csum = 1, 0 for i in range(ylen): csum += M[i][k] - (M[i][j-1] if j else 0) ans += res[csum - T] res[csum] += 1 return ans
Java Code:
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class Solution { public int numSubmatrixSumTarget(int[][] M, int T) { int xlen = M[0].length, ylen = M.length, ans = 0; Map<Integer, Integer> res = new HashMap<>(); for (int[] r : M) for (int j = 1; j < xlen; j++) r[j] += r[j-1]; for (int j = 0; j < xlen; j++) for (int k = j; k < xlen; k++) { res.clear(); res.put(0,1); int csum = 0; for (int i = 0; i < ylen; i++) { csum += M[i][k] - (j > 0 ? M[i][j-1] : 0); ans += res.getOrDefault(csum - T, 0); res.put(csum, res.getOrDefault(csum, 0) + 1); } } return ans; }}
C++ Code:
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class Solution {public: int numSubmatrixSumTarget(vector<vector<int>>& M, int T) { int xlen = M[0].size(), ylen = M.size(), ans = 0; unordered_map<int, int> res; for (int i = 0; i < ylen; i++) for (int j = 1; j < xlen; j++) M[i][j] += M[i][j-1]; for (int j = 0; j < xlen; j++) for (int k = j; k < xlen; k++) { res.clear(); res[0] = 1; int csum = 0; for (int i = 0; i < ylen; i++) { csum += M[i][k] - (j ? M[i][j-1] : 0); ans += res.find(csum - T) != res.end() ? res[csum - T] : 0; res[csum]++; } } return ans; }};
