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Solution: Number of Submatrices That Sum to Target

seanpgallivan profile image seanpgallivan ・4 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #1074 (Hard): Number of Submatrices That Sum to Target


Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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'.


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 1 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 <= 100
  • 1 <= 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:


(Jump to: Problem Description || Solution Idea)

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
};
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Python Code:


(Jump to: Problem Description || Solution Idea)

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
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Java Code:


(Jump to: Problem Description || Solution Idea)

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;
    }
}
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C++ Code:


(Jump to: Problem Description || Solution Idea)

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;
    }
};
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Discussion (1)

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rohithv07 profile image
Rohith V

My Code on this problem :

class Solution {
    public int numSubmatrixSumTarget(int[][] matrix, int target) {
        int count = 0;
        int line = matrix.length;
        int column = matrix[0].length + 1;
        int[][] sum = new int[line][column];

        for (int i = 0; i < sum.length; i++){
            for (int j = 1; j < sum[0].length; j++){
                sum[i][j] = sum[i][j - 1] + matrix[i][j - 1];
            }
        }


        for (int start = 0; start < column; start++){
            for (int end = start + 1; end < column; end++ ){

                int sumOfSubMatrix = 0;
                Map<Integer, Integer> map = new HashMap<Integer, Integer>();
                map.put(0, 1);
                for(int l = 0; l < line; l++){
                    sumOfSubMatrix += sum[l][end] - sum[l][start];
                    if (map.containsKey(sumOfSubMatrix - target))
                        count += map.get(sumOfSubMatrix - target);
                    map.put(sumOfSubMatrix, map.getOrDefault(sumOfSubMatrix, 0) + 1);

                }
            }
        }

        return count;

    }
}
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Repo : github.com/Rohithv07/LeetCodeTopIn...