LeetCode 311: Sparse Matrix Multiplication
LeetCode 311 Solution Explanation
Explanation:
To perform matrix multiplication for sparse matrices, we can optimize the process by only computing and storing the non-zero elements. The algorithm involves iterating through the non-zero elements of the first matrix and multiplying them with the corresponding elements in the second matrix to compute the result matrix.
- Iterate through the non-zero elements of the first matrix, and for each non-zero element A[i][j], iterate through the corresponding row in the second matrix B[j][k].
- Multiply A[i][j] with B[j][k] and add the result to the corresponding element in the result matrix C[i][k].
- Repeat this process for all non-zero elements in the first matrix.
- The resulting matrix C will be the product of the two sparse matrices.
Time complexity:
- Let n be the number of rows in the first matrix, m be the number of columns in the first matrix, and p be the number of columns in the second matrix.
- The time complexity of this algorithm is O(n * m * p), as we iterate through the non-zero elements of the matrices.
Space complexity:
- We only store the non-zero elements and their positions in the result matrix, so the space complexity is O(n * p).
:
LeetCode 311 Solutions in Java, C++, Python
class Solution {
public int[][] multiply(int[][] A, int[][] B) {
int n = A.length;
int m = A[0].length;
int p = B[0].length;
int[][] C = new int[n][p];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (A[i][j] != 0) {
for (int k = 0; k < p; k++) {
if (B[j][k] != 0) {
C[i][k] += A[i][j] * B[j][k];
}
}
}
}
}
return C;
}
}Interactive Code Editor for LeetCode 311
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