README_EN.md

311. Sparse Matrix Multiplication

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Description

Given two sparse matrices mat1 of size m x k and mat2 of size k x n, return the result of mat1 x mat2. You may assume that multiplication is always possible.

 

Example 1:

Input: mat1 = [[1,0,0],[-1,0,3]], mat2 = [[7,0,0],[0,0,0],[0,0,1]]
Output: [[7,0,0],[-7,0,3]]

Example 2:

Input: mat1 = [[0]], mat2 = [[0]]
Output: [[0]]

 

Constraints:

• m == mat1.length
• k == mat1[i].length == mat2.length
• n == mat2[i].length
• 1 <= m, n, k <= 100
• -100 <= mat1[i][j], mat2[i][j] <= 100

Solutions

Python3

class Solution:
    def multiply(self, mat1: List[List[int]], mat2: List[List[int]]) -> List[List[int]]:
        r1, c1, c2 = len(mat1), len(mat1[0]), len(mat2[0])
        res = [[0] * c2 for _ in range(r1)]
        for i in range(r1):
            for j in range(c2):
                for k in range(c1):
                    res[i][j] += mat1[i][k] * mat2[k][j]
        return res

class Solution:
    def multiply(self, mat1: List[List[int]], mat2: List[List[int]]) -> List[List[int]]:
        r1, c1, c2 = len(mat1), len(mat1[0]), len(mat2[0])
        res = [[0] * c2 for _ in range(r1)]
        mp = defaultdict(list)
        for i in range(r1):
            for j in range(c1):
                if mat1[i][j] != 0:
                    mp[i].append(j)
        for i in range(r1):
            for j in range(c2):
                for k in mp[i]:
                    res[i][j] += mat1[i][k] * mat2[k][j]
        return res

Java

class Solution {
    public int[][] multiply(int[][] mat1, int[][] mat2) {
        int r1 = mat1.length, c1 = mat1[0].length, c2 = mat2[0].length;
        int[][] res = new int[r1][c2];
        Map<Integer, List<Integer>> mp = new HashMap<>();
        for (int i = 0; i < r1; ++i) {
            for (int j = 0; j < c1; ++j) {
                if (mat1[i][j] != 0) {
                    mp.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
                }
            }
        }
        for (int i = 0; i < r1; ++i) {
            for (int j = 0; j < c2; ++j) {
                if (mp.containsKey(i)) {
                    for (int k : mp.get(i)) {
                        res[i][j] += mat1[i][k] * mat2[k][j];
                    }
                }
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    vector<vector<int>> multiply(vector<vector<int>>& mat1, vector<vector<int>>& mat2) {
        int r1 = mat1.size(), c1 = mat1[0].size(), c2 = mat2[0].size();
        vector<vector<int>> res(r1, vector<int>(c2));
        unordered_map<int, vector<int>> mp;
        for (int i = 0; i < r1; ++i) {
            for (int j = 0; j < c1; ++j) {
                if (mat1[i][j] != 0) mp[i].push_back(j);
            }
        }
        for (int i = 0; i < r1; ++i) {
            for (int j = 0; j < c2; ++j) {
                for (int k : mp[i]) res[i][j] += mat1[i][k] * mat2[k][j];
            }
        }
        return res;
    }
};

Go

func multiply(mat1 [][]int, mat2 [][]int) [][]int {
	r1, c1, c2 := len(mat1), len(mat1[0]), len(mat2[0])
	res := make([][]int, r1)
	for i := range res {
		res[i] = make([]int, c2)
	}
	mp := make(map[int][]int)
	for i := 0; i < r1; i++ {
		for j := 0; j < c1; j++ {
			if mat1[i][j] != 0 {
				mp[i] = append(mp[i], j)
			}
		}
	}
	for i := 0; i < r1; i++ {
		for j := 0; j < c2; j++ {
			for _, k := range mp[i] {
				res[i][j] += mat1[i][k] * mat2[k][j]
			}
		}
	}
	return res
}

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