Leetcode: Sparse Matrix Multiplication

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:

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 |

Code (Java):

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || A.length == 0 ||
            B == null || B.length == 0) {
            return new int[0][0];    
        }
        
        int m = A.length;
        int n = A[0].length;
        int l = B[0].length;
        
        int[][] C = new int[m][l];
        
        // Step 1: convert the sparse A to dense format
        Map<Integer, Map<Integer, Integer>> denseA = new HashMap<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (A[i][j] != 0) {
                    if (!denseA.containsKey(i)) {
                        denseA.put(i, new HashMap<>());
                    }
                    denseA.get(i).put(j, A[i][j]);
                }
            }
        }
        
        // Step 2: convert the sparse B to dense format
        Map<Integer, Map<Integer, Integer>> denseB = new HashMap<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < l; j++) {
                if (B[i][j] != 0) {
                    if (!denseB.containsKey(i)) {
                        denseB.put(i, new HashMap<>());
                    }
                    denseB.get(i).put(j, B[i][j]);
                }
            }
        }
        
        // Step3: calculate the denseA * denseB
        for (int i : denseA.keySet()) {
            for (int j : denseA.get(i).keySet()) {
                if (!denseB.containsKey(j)) {
                    continue;
                }
                
                for (int k : denseB.get(j).keySet()) {
                    C[i][k] += denseA.get(i).get(j) * denseB.get(j).get(k);
                }
            }
        }
        
        return C;
    }
} 

Another solution using one table:
https://leetcode.com/discuss/71912/easiest-java-solution

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        int m = A.length, n = A[0].length, nB = B[0].length;
        int[][] C = new int[m][nB];

        for(int i = 0; i < m; i++) {
            for(int k = 0; k < n; k++) {
                if (A[i][k] != 0) {
                    for (int j = 0; j < nB; j++) {
                        if (B[k][j] != 0) C[i][j] += A[i][k] * B[k][j];
                    }
                }
            }
        }
        return C;   
    }
}

---

The followings is the original 75ms solution:

The idea is derived from a CMU lecture.

A sparse matrix can be represented as a sequence of rows, each of which is a sequence of (column-number, value) pairs of the nonzero values in the row.

So let's create a non-zero array for A, and do multiplication on B.

Hope it helps!

---

public int[][] multiply(int[][] A, int[][] B) {
    int m = A.length, n = A[0].length, nB = B[0].length;
    int[][] result = new int[m][nB];

    List[] indexA = new List[m];
    for(int i = 0; i < m; i++) {
        List<Integer> numsA = new ArrayList<>();
        for(int j = 0; j < n; j++) {
            if(A[i][j] != 0){
                numsA.add(j); 
                numsA.add(A[i][j]);
            }
        }
        indexA[i] = numsA;
    }

    for(int i = 0; i < m; i++) {
        List<Integer> numsA = indexA[i];
        for(int p = 0; p < numsA.size() - 1; p += 2) {
            int colA = numsA.get(p);
            int valA = numsA.get(p + 1);
            for(int j = 0; j < nB; j ++) {
                int valB = B[colA][j];
                result[i][j] += valA * valB;
            }
        }
    }

    return result;   
}