LeetCode – Sparse Matrix Multiplication (Java)

Category: Algorithms >> Interview October 26, 2014

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.

1. Naive Method

We can implement Sum(A_ik * B_kj) -> C_ij as a naive solution.

public int[][] multiply(int[][] A, int[][] B) {
    //validity check
 
    int[][] C = new int[A.length][B[0].length];
 
    for(int i=0; i<C.length; i++){
        for(int j=0; j<C[0].length; j++){
            int sum=0;
            for(int k=0; k<A[0].length; k++){
                sum += A[i][k]*B[k][j];
            }
            C[i][j] = sum;
        }
    }
 
    return C;
}

Time complexity is O(n^3).

2. Optimized Method

From the formula: Sum(A_ik * B_kj) -> C_ij

We can see that when A_ik is 0, there is no need to compute B_kj. So we switch the inner two loops and add a 0-checking condition.

public int[][] multiply(int[][] A, int[][] B) {
    //validity check
 
    int[][] C = new int[A.length][B[0].length];
 
    for(int i=0; i<C.length; i++){
        for(int k=0; k<A[0].length; k++){
            if(A[i][k]!=0){
                for(int j=0; j<C[0].length; j++){
                    C[i][j] += A[i][k]*B[k][j];
                }
            }
        }
    }
 
    return C;
}

Since the matrix is sparse, time complexity is ~O(n^2) which is much faster than O(n^3).

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Donne Numme

I am unable to understand how the optimized solution works. If A[0][0] = 0, then C[0][0] = A[0][0] * B[0][0] + A[0][1] * B[1][0]. In that case, we still need to iterate each element. Can some one explain how this works?
ryanlr

It’s not compressed. Just take shortcut to skip computation when an element is 0.
nik vamv

Is this compressed sparse row multiplication ?

LeetCode – Sparse Matrix Multiplication (Java)

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