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):

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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;   
}