diff --git a/DiagonalTraverse.java b/DiagonalTraverse.java
new file mode 100644
index 00000000..4f30df50
--- /dev/null
+++ b/DiagonalTraverse.java
@@ -0,0 +1,57 @@
+// Time Complexity : O(n*m) where n is the number of rows and m is the number of columns in the matrix
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode :Yes
+// Three line explanation of solution in plain english 
+// 1. We traverse the matrix in a diagonal manner, starting from the top-left corner.
+// 2. We use a boolean variable to keep track of the direction of traversal (up-right or down-left).
+// 3. We update the row and column indices based on the current direction 
+// and the boundaries of the matrix. We store the elements in the result array as we traverse.
+// The auxiliary space is O(1) as we are using only a few variables to keep track of the current position and direction,
+//  and the result array is not considered auxiliary space as it is required for the output.
+
+// Your code here along with comments explaining your approac
+
+class Solution {
+    public int[] findDiagonalOrder(int[][] mat) {
+
+        boolean direction = true; // true = up-right, false = down-left
+
+        int row = mat.length;
+        int col = mat[0].length;
+
+        int[] result = new int[row * col];
+
+        int r = 0, c = 0;
+
+        for (int i = 0; i < row * col; i++) {
+
+            result[i] = mat[r][c];
+
+            if (direction) {
+                if (c == col - 1) {
+                    r++;
+                    direction = false;
+                } else if (r == 0) {
+                    c++;
+                    direction = false;
+                } else {
+                    r--;
+                    c++;
+                }
+            } else {
+                if (r == row - 1) {
+                    c++;
+                    direction = true;
+                } else if (c == 0) {
+                    r++;
+                    direction = true;
+                } else {
+                    r++;
+                    c--;
+                }
+            }
+        }
+
+        return result;
+    }
+}
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diff --git a/ProductofArray.java b/ProductofArray.java
new file mode 100644
index 00000000..0d1cccb1
--- /dev/null
+++ b/ProductofArray.java
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+
+// Time Complexity : O(N)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode :Yes
+// Three line explanation of solution in plain english 
+// 1. Calculate the product of all elements to the left of each element
+// 2. Calculate the product of all elements to the right of each element
+// 3. Multiply the left and right products for each element
+// Product of all no's of element to left * element
+// Product of all no's of element to right * element
+// Auxuliary space is not used as we are storing the left product in the result array and using a variable to store the right product.
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        
+        int[] result = new int[nums.length];
+        
+        int rightProduct = 1;
+        result[0] = 1;
+
+        // Left products
+        for (int i = 1; i < nums.length; i++) {
+            rightProduct = rightProduct * nums[i - 1];
+            result[i] = rightProduct;
+        }
+
+        // Right products
+        rightProduct = 1;
+        for (int i = nums.length - 2; i >= 0; i--) {
+            rightProduct = rightProduct * nums[i + 1];
+            result[i] = result[i] * rightProduct;
+        }
+
+        return result;
+    }
+}
\ No newline at end of file
diff --git a/SprialMatrix.java b/SprialMatrix.java
new file mode 100644
index 00000000..75873aab
--- /dev/null
+++ b/SprialMatrix.java
@@ -0,0 +1,57 @@
+// Time Complexity : O(n*m) where n is the number of rows and m is the number of columns in the matrix
+// Did this code successfully run on Leetcode :Yes
+// Three line explanation of solution in plain english 
+// 1. We maintain four pointers to keep track of the current boundaries of the matrix (rowBegin, rowEnd, colBegin, colEnd).
+// 2. We traverse the matrix in a spiral manner, moving right, down, left, and up,
+//  while updating the boundaries after each traversal.
+// 3. We continue this process until the pointers cross each other, indicating that we have traversed the entire matrix. 
+// We store the elements in the result list as we traverse.
+// The auxiliary space is O(1) as we are using only a few variables to keep track of the current position and boundaries, and the result list is not considered auxiliary space as it is required for the output.
+
+// Your code here along with comments explaining your approac
+import java.util.*;
+
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+
+        List<Integer> res = new ArrayList<>();
+
+        int rowBegin = 0;
+        int colBegin = 0;
+        int rowEnd = matrix.length - 1;
+        int colEnd = matrix[0].length - 1;
+
+        while (rowBegin <= rowEnd && colBegin <= colEnd) {
+
+            // move right
+            for (int j = colBegin; j <= colEnd; j++) {
+                res.add(matrix[rowBegin][j]);
+            }
+            rowBegin++;
+
+            // move down
+            for (int j = rowBegin; j <= rowEnd; j++) {
+                res.add(matrix[j][colEnd]);
+            }
+            colEnd--;
+
+            // move left
+            if (rowBegin <= rowEnd) {
+                for (int j = colEnd; j >= colBegin; j--) {
+                    res.add(matrix[rowEnd][j]);
+                }
+                rowEnd--;
+            }
+
+            // move up
+            if (colBegin <= colEnd) {
+                for (int j = rowEnd; j >= rowBegin; j--) {
+                    res.add(matrix[j][colBegin]);
+                }
+                colBegin++;
+            }
+        }
+
+        return res;
+    }
+}
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