# Sparse Matrix Multiplication in C++

C++Server Side ProgrammingProgramming

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Suppose we have two matrices A and B, we have to find the result of AB. We may assume that A's column number is equal to B's row number.

So, if the input is like [[1,0,0],[-1,0,3]] [[7,0,0],[0,0,0],[0,0,1]],

 1 0 0 -1 0 3

 7 0 0 0 0 0 0 0 1

then the output will be [[7,0,0],[-7,0,3]]

 7 0 0 -7 0 3

To solve this, we will follow these steps −

• r1 := size of A, r2 := size of B

• c1 := size of A[0], c2 := size of B[0]

• Define one 2D array ret of order r1 x c2

• Define an array sparseA[r1] of pairs

• for initialize i := 0, when i < r1, update (increase i by 1), do −

• for initialize j := 0, when j < c1, update (increase j by 1), do −

• if A[i, j] is not equal to 0, then −

• insert { j, A[i, j] } at the end of sparseA[i]

• for initialize i := 0, when i < r1, update (increase i by 1), do −

• for initialize j := 0, when j < size of sparseA[i], update (increase j by 1), do −

• for initialize k := 0, when k < c2, update (increase k by 1), do −

• x := first element of sparseA[i, j]

• if B[x, k] is not equal to 0, then −

• ret[i, k] := ret[i, k] + second element of sparseA[i, j] * B[x, k]

• return ret

## Example

Let us see the following implementation to get better understanding −

class Solution {
public:
vector<vector<int<> multiply(vector<vector<int<>& A, vector<vector<int<>& B) {
int r1 = A.size();
int r2 = B.size();
int c1 = A[0].size();
int c2 = B[0].size();
vector < vector <int< > ret(r1, vector <int< (c2));
vector < pair <int, int> > sparseA[r1];
for(int i = 0; i < r1; i++){
for(int j = 0; j < c1; j++){
if(A[i][j] != 0)sparseA[i].push_back({j, A[i][j]});
}
}
for(int i = 0; i < r1; i++){
for(int j = 0; j < sparseA[i].size(); j++){
for(int k = 0; k < c2; k++){
int x = sparseA[i][j].first;
if(B[x][k] != 0){
ret[i][k] += sparseA[i][j].second * B[x][k];
}
}
}
}
return ret;
}
};

## Input

{{1,0,0},{-1,0,3}},{{7,0,0},{0,0,0},{0,0,1}}

## Output

[[7, 0, 0, ],[-7, 0, 3, ],]
Updated on 18-Nov-2020 12:03:42