diff --git a/DiagonalTraversal.java b/DiagonalTraversal.java
new file mode 100644
index 00000000..5461af62
--- /dev/null
+++ b/DiagonalTraversal.java
@@ -0,0 +1,49 @@
+// Time Complexity : O(mn)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+
+// Your code here along with comments explaining your approach
+// 1: We maintain a boolean flag that holds the value of whether we are traversing upwards or downwards
+// 2: For upwards conditions, we increment/decrement the row and column accordingly, while keeping in mind the condition of reaching the last column
+// 3: For downwards conditions, we increment/decrement the row and column accordingly, while keeping in mind the condition of reaching the last row
+class Solution {
+    public int[] findDiagonalOrder(int[][] mat) {
+        int m = mat.length; // number of columns
+        int n = mat[0].length; // number of rows;
+
+        int[] result = new int[m * n];
+        int row = 0, col = 0; // pointers to move through matrix array
+
+        boolean flag = true; // starting in upwards direction
+
+        for (int idx = 0; idx < m * n; idx++) {
+            result[idx] = mat[row][col];
+            if (flag) { // upwards
+                if (row == 0 && col != n - 1) {
+                    col++;
+                    flag = false;
+                } else if (col == n - 1) {
+                    flag = false;
+                    row++;
+                } else {
+                    row--;
+                    col++;
+                }
+            } else { //downwards
+                if (col == 0 && row != m - 1) {
+                    row++;
+                    flag = true;
+                } else if (row == m - 1) {
+                    flag = true;
+                    col++;
+                } else {
+                    row++;
+                    col--;
+                }
+
+            }
+        }
+        return result;
+
+    }
+}
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diff --git a/ProductExceptSelf.java b/ProductExceptSelf.java
new file mode 100644
index 00000000..7e83ea56
--- /dev/null
+++ b/ProductExceptSelf.java
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+// Time Complexity : O(n) average
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+
+// Your code here along with comments explaining your approach
+// 1: We go through two passes for each element in the array, calculating the left and right product for each element
+// 2: For each pass, we maintain a running product that contains either the left or right product that we have come across thus far
+// 3: After calculating the left array, we use the same array to find the runningProduct and thus the rightProduct and can return the same array
+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        int rp = 1;
+        int n = nums.length;
+        int[] result = new int[n];
+        // left pass,
+        // we add 1 to the beginning of the array
+        result[0] = 1;
+
+        for (int i = 1; i < n; i++) {
+            rp = rp * nums[i - 1];
+            result[i] = rp;
+        }
+
+        // right pass
+        rp = 1;
+        for (int i = n - 2; i >= 0; i--) {
+            rp = rp * nums[i + 1];
+            result[i] = result[i] * rp;
+
+        }
+
+        return result;
+
+    }
+}
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diff --git a/SpiralMatrix.java b/SpiralMatrix.java
new file mode 100644
index 00000000..bb62e85f
--- /dev/null
+++ b/SpiralMatrix.java
@@ -0,0 +1,51 @@
+// Time Complexity : O(mn)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+
+// Your code here along with comments explaining your approach
+// 1: We have 4 variables to maintain our position in the matrix array
+// 2: For each type of traversal, we have a for loop that uses the necessary conditions to determine which elements to add in the output ArrayList
+// 3: Since we are using a while loop, we have additional checks with if statements to ensure that our pointers are still valid in the matrix
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> li = new ArrayList<>();
+        int m = matrix.length;
+        int n = matrix[0].length;
+
+        int top = 0;
+        int bottom = m - 1;
+        int left = 0;
+        int right = n - 1;
+
+        while (top <= bottom && left <= right) {
+            // top row
+            for (int j = left; j <= right; j++) {
+                li.add(matrix[top][j]);
+            }
+            top++;
+
+            // right column
+            for (int i = top; i <= bottom; i++) {
+                li.add(matrix[i][right]);
+            }
+            right--;
+
+            // bottom row
+            if (top <= bottom) {
+                for (int j = right; j >= left; j--) {
+                    li.add(matrix[bottom][j]);
+                }
+            }
+            bottom--;
+
+            // left column
+            if (left <= right) {
+                for (int j = bottom; j >= top; j--) {
+                    li.add(matrix[j][left]);
+                }
+            }
+            left++;
+        }
+        return li;
+    }
+}
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