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:

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

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


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

Optimized brute force

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public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        int row = A.length, column = B[0].length, colA = A[0].length;
        int[][] res = new int[row][column];
        
        for (int i = 0; i < row; i++){
            for (int j = 0; j < colA; j++){
                if (A[i][j] != 0){
                    for (int k = 0; k < column; k++){
                        if (B[j][k] != 0){
                            res[i][k] += A[i][j] * B[j][k];
                        }
                    }
                }
            }
        }
        return res;
    }
}

However, this solution still checks matrix B multiple times.

One hash table that build index for non-zero values in each row of Matrix B

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public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || A[0] == null || B == null || B[0] == null) return null;
        int m = A.length, n = A[0].length, l = B[0].length;
        int[][] C = new int[m][l];
        Map<Integer, HashMap<Integer, Integer>> tableB = new HashMap<>(); // 

        for(int k = 0; k < n; k++) {
            tableB.put(k, new HashMap<Integer, Integer>());
            for(int j = 0; j < l; j++) {
                if (B[k][j] != 0){
                    tableB.get(k).put(j, B[k][j]);
                }
            }
        }

        for(int i = 0; i < m; i++) {
            for(int k = 0; k < n; k++) {
                if (A[i][k] != 0){
                    for (Integer j: tableB.get(k).keySet()) {
                        C[i][j] += A[i][k] * tableB.get(k).get(j);
                    }
                }
            }
        }
        return C;   
    }
}

Two hash tables for both matrix A and B

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public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || B == null || A.length == 0 || B.length == 0 || A[0].length == 0 || B[0].length == 0)
            return null;
        int m = A.length, n = A[0].length, l = B[0].length;
        if (n != l) 
            throw new IllegalArgumentException("A's column number must be equal to B's row number.");
        int[][] res = new int[m][l];
        
        HashMap<Integer, HashMap<Integer, Integer>> matrixA = convertToMatrix(A);
        HashMap<Integer, HashMap<Integer, Integer>> matrixB = convertToMatrix(B);
        
        for (Integer i : matrixA.keySet()) {
            for (Integer k : matrixA.get(i).keySet()) {
                if (matrixB.containsKey(k)) {
                    for (Integer j : matrixB.get(k).keySet()) {
                        res[i][j] += matrixA.get(i).get(k) * matrixB.get(k).get(j);
                    }
                }
            }
        }
        return res;
    }
    
    public HashMap<Integer, HashMap<Integer, Integer>> convertToMatrix(int[][] A) {
        HashMap<Integer, HashMap<Integer, Integer>> map = new HashMap<>();
        for (int i = 0; i < A.length; i++) {
            map.put(i, new HashMap<Integer, Integer>());
            for (int j = 0; j < A[i].length; j++) {
                if (A[i][j] != 0) {
                    map.get(i).put(j, A[i][j]);
                }
            }
        }
        return map;
    }
}