@@ -23,6 +23,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Find the Celebrity](problems/src/array/FindTheCelebrity.java) (Medium)
 - [Meeting Rooms](problems/src/array/MeetingRooms.java) (Easy)
 - [Longest Continuous Increasing Subsequence](problems/src/array/LongestIncreasingSubsequence.java) (Easy)
+- [Sparse Matrix Multiplication](problems/src/array/SparseMatrixMultiplication.java) (Medium)
 
 #### [Backtracking](problems/src/backtracking)
 
 
@@ -0,0 +1,80 @@
+package array;
+
+/**
+ * Created by gouthamvidyapradhan on 03/12/2017.
+ *
+ * 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:
+
+ A = [
+ [ 1, 0, 0],
+ [-1, 0, 3]
+ ]
+
+ B = [
+ [ 7, 0, 0 ],
+ [ 0, 0, 0 ],
+ [ 0, 0, 1 ]
+ ]
+
+
+ |  1 0 0 |   | 7 0 0 |   |  7 0 0 |
+ AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
+ | 0 0 1 |
+
+ Solution: Skip the rows and columns which you already know would result in zero after multiplication.
+ */
+public class SparseMatrixMultiplication {
+
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        int[][] A = {{1,0,0,1}};
+        int[][] B = {{7,0,0}, {0,0,0}, {0,0,1}, {0,0,1}};
+        int[][] C = new SparseMatrixMultiplication().multiply(A, B);
+    }
+
+    public int[][] multiply(int[][] A, int[][] B) {
+        boolean[] AH = new boolean[A.length]; // Metadata for matrix A
+        boolean[] BH = new boolean[B[0].length]; // Metadata for matrix A
+        for(int i = 0; i < A.length; i ++){
+            for(int j = 0; j < A[0].length; j ++){
+                if(A[i][j] != 0){
+                    AH[i] = true;
+                    break;
+                }
+            }
+        }
+
+        for(int i = 0; i < B[0].length; i ++){
+            for(int j = 0; j < B.length; j ++){
+                if(B[j][i] != 0){
+                    BH[i] = true;
+                    break;
+                }
+            }
+        }
+
+        int[][] C = new int[A.length][B[0].length];
+        for(int i = 0; i < C.length; i ++){
+            if(AH[i]){
+                for(int j = 0; j < C[0].length; j++){
+                    if(BH[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;
+    }
+}