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leetcode_java/src/main/java/LeetCodeJava/Array/SparseMatrixMultiplication.java
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| package LeetCodeJava.Array; | ||
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| // https://leetcode.com/problems/sparse-matrix-multiplication/description/ | ||
| // https://leetcode.ca/all/311.html | ||
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| import java.util.ArrayList; | ||
| import java.util.Arrays; | ||
| import java.util.List; | ||
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| /** | ||
| * 311. 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: | ||
| * | ||
| * Input: | ||
| * | ||
| * A = [ | ||
| * [ 1, 0, 0], | ||
| * [-1, 0, 3] | ||
| * ] | ||
| * | ||
| * B = [ | ||
| * [ 7, 0, 0 ], | ||
| * [ 0, 0, 0 ], | ||
| * [ 0, 0, 1 ] | ||
| * ] | ||
| * | ||
| * Output: | ||
| * | ||
| * | 1 0 0 | | 7 0 0 | | 7 0 0 | | ||
| * AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 | | ||
| * | 0 0 1 | | ||
| * Difficulty: | ||
| * Medium | ||
| * Lock: | ||
| * Prime | ||
| * Company: | ||
| * Amazon Apple Bloomberg Facebook Goldman Sachs Google LinkedIn Microsoft Snapchat | ||
| * | ||
| */ | ||
| public class SparseMatrixMultiplication { | ||
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| // V0 | ||
| // IDEA : ARRAY OP (fix by gpt) | ||
| /** | ||
| * Why there is 3 loop ? | ||
| * | ||
| * -> Matrix Multiplication: Ci,j = Sigma(Aik * Bkj) | ||
| * | ||
| * -> so we have 3 layer loop as below: | ||
| * - i : Iterates over the rows of A (outer loop). | ||
| * - j : Iterates over the columns of B (second loop). | ||
| * - k : Iterates over the “shared dimension” (columns of A or rows of B ) to compute the dot product (inner loop). | ||
| * | ||
| * | ||
| * -> | ||
| * | ||
| * The Role of the Loops | ||
| * 1. Outer loop ( i ): Iterates over the rows of mat1 to calculate each row of the result matrix. | ||
| * 2. Middle loop ( j ): Iterates over the columns of mat2 to compute each element in a row of the result matrix. | ||
| * 3. Inner loop ( k ): Iterates over the “shared dimension” to compute the dot product of the i^{th} row of mat1 and the j^{th} column of mat2. | ||
| * | ||
| * | ||
| * -> Why the Inner Loop ( k ) Exists ? | ||
| * | ||
| * -> The inner loop is necessary | ||
| * because each element of the result matrix | ||
| * is computed as the dot product of a | ||
| * row from mat1 and a column from mat2. | ||
| * Without this loop, the computation of the dot product would be incomplete. | ||
| */ | ||
| public static int[][] multiply(int[][] mat1, int[][] mat2) { | ||
| // Edge case: Single element matrices | ||
| if (mat1.length == 1 && mat1[0].length == 1 && mat2.length == 1 && mat2[0].length == 1) { | ||
| return new int[][]{{mat1[0][0] * mat2[0][0]}}; | ||
| } | ||
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| int l_1 = mat1.length; // Number of rows in mat1 | ||
| int w_1 = mat1[0].length; // Number of columns in mat1 (and rows in mat2) | ||
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| int w_2 = mat2[0].length; // Number of columns in mat2 | ||
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| // Initialize the result matrix | ||
| int[][] res = new int[l_1][w_2]; | ||
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| // Perform matrix multiplication | ||
| for (int i = 0; i < l_1; i++) { | ||
| for (int j = 0; j < w_2; j++) { | ||
| int sum = 0; // Sum for res[i][j] | ||
| for (int k = 0; k < w_1; k++) { | ||
| sum += mat1[i][k] * mat2[k][j]; | ||
| } | ||
| res[i][j] = sum; | ||
| } | ||
| } | ||
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| return res; | ||
| } | ||
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| // V1 | ||
| // IDEA : ARRAY OP (gpt) | ||
| public int[][] multiply_1(int[][] A, int[][] B) { | ||
| int m = A.length; // Number of rows in A | ||
| int n = A[0].length; // Number of columns in A | ||
| int p = B[0].length; // Number of columns in B | ||
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| int[][] result = new int[m][p]; | ||
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| // Iterate through rows of A | ||
| for (int i = 0; i < m; i++) { | ||
| for (int k = 0; k < n; k++) { | ||
| if (A[i][k] != 0) { // Skip zero entries in A | ||
| for (int j = 0; j < p; j++) { | ||
| if (B[k][j] != 0) { // Skip zero entries in B | ||
| result[i][j] += A[i][k] * B[k][j]; | ||
| } | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| return result; | ||
| } | ||
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| // V2 | ||
| // https://leetcode.ca/2016-10-06-311-Sparse-Matrix-Multiplication/ | ||
| public int[][] multiply_2(int[][] mat1, int[][] mat2) { | ||
| int m = mat1.length, n = mat2[0].length; | ||
| int[][] ans = new int[m][n]; | ||
| List<int[]>[] g1 = f(mat1); | ||
| List<int[]>[] g2 = f(mat2); | ||
| for (int i = 0; i < m; ++i) { | ||
| for (int[] p : g1[i]) { | ||
| int k = p[0], x = p[1]; | ||
| for (int[] q : g2[k]) { | ||
| int j = q[0], y = q[1]; | ||
| ans[i][j] += x * y; | ||
| } | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
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| private List<int[]>[] f(int[][] mat) { | ||
| int m = mat.length, n = mat[0].length; | ||
| List<int[]>[] g = new List[m]; | ||
| Arrays.setAll(g, i -> new ArrayList<>()); | ||
| for (int i = 0; i < m; ++i) { | ||
| for (int j = 0; j < n; ++j) { | ||
| if (mat[i][j] != 0) { | ||
| g[i].add(new int[] {j, mat[i][j]}); | ||
| } | ||
| } | ||
| } | ||
| return g; | ||
| } | ||
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| // V3 | ||
| } |
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