Given two sparse matrices A and B, return the result of AB. You may assume that A’s column number is equal to B’s row number.

Example:
Input:

1 2 3 4

A = [   [ 1, 0, 0],   [-1, 0, 3] ]

A = [
  [ 1, 0, 0],
  [-1, 0, 3]
]

1 2 3 4 5

B = [   [ 7, 0, 0 ],   [ 0, 0, 0 ],   [ 0, 0, 1 ] ]

B = [
  [ 7, 0, 0 ],
  [ 0, 0, 0 ],
  [ 0, 0, 1 ]
]

Output:

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |

The Matrix Multiplication Algorithm in C++

Given two matrices, if either one of them is empty, the multiplication result should be empty as well. The result matrix dimension is the [rowA, colB] and each element in the matrix should be the sum of the dot products for each row in A and each column in B i.e. r[i][j] = sum(A[i][k] * B[k][j]) where k is within [0, colA).

The straightforward/naive implementation of two matrix muplication using std::vector is as follows:

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class Solution {
public:
    vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;        
        int rowA = A.size();
        if (rowA == 0) return r;
        int colA = A[0].size();
        if (colA == 0) return r;
        int rowB = B.size();
        if (rowB == 0) return r;
        int colB = B[0].size();
        if (colB == 0) return r;
        r.resize(rowA);
        for (int i = 0; i < rowA; ++ i) {
            r[i].resize(colB);
        }
        for (int i = 0; i < rowA; ++ i) {
            for (int j = 0; j < colB; ++ j) {
                for (int k = 0; k < colA; ++ k) {
                    r[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        return r;
    }
};

class Solution {
public:
    vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;        
        int rowA = A.size();
        if (rowA == 0) return r;
        int colA = A[0].size();
        if (colA == 0) return r;
        int rowB = B.size();
        if (rowB == 0) return r;
        int colB = B[0].size();
        if (colB == 0) return r;
        r.resize(rowA);
        for (int i = 0; i < rowA; ++ i) {
            r[i].resize(colB);
        }
        for (int i = 0; i < rowA; ++ i) {
            for (int j = 0; j < colB; ++ j) {
                for (int k = 0; k < colA; ++ k) {
                    r[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        return r;
    }
};

The time complexity is O(rowA * colB * colA).

–EOF (The Ultimate Computing & Technology Blog) —