diff --git a/DiagonalTraverse.java b/DiagonalTraverse.java
new file mode 100644
index 00000000..00404d0b
--- /dev/null
+++ b/DiagonalTraverse.java
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+// Time Complexity : O(m*n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode :Yes
+// Three line explanation of solution in plain english
+// we start with r=c=0, with upward direction
+//for up we decrese row, increse col, when out of boundary adjust row and column and make down
+//for down increase row, decrese col, when out of boundary adjust row and column and make down
+//have counetr to exit when all traversed
+// Your code here along with comments explaining your approach
+class DiagonalTraverse {
+    public int[] findDiagonalOrder(int[][] mat) {
+        int[] ans = new int[mat.length*mat[0].length];
+        int counter = 0;
+        //initial row, column
+        int r = 0;
+        int c = 0;
+        //intial up
+        boolean up = true;
+        //boundary
+        int rLen = mat.length-1;
+        int cLen = mat[0].length-1;
+        while(true)
+        {
+            if(up)
+            {
+                //when up add to ans, decrese row, increase col
+                ans[counter] = mat[r][c];
+                counter++;
+                r--;
+                c++;
+
+                //boundary check, adjust r and col, make down
+                //cental diagonal amy be up or down
+                if(r<0 && c>cLen)
+                {
+                    c = cLen;
+                    r = 1;
+                    up = false;
+                }
+                else if(r<0)
+                {
+                    r = 0;
+                    c = c;
+                    up = false;
+                }
+                else if(c > cLen)
+                {
+                    r = r + 2;
+                    c = cLen;
+                    up = false;
+                }
+            }
+            else
+            {
+                //down, incrase row, decrease col
+                ans[counter] = mat[r][c];
+                counter++;
+                r++;
+                c--;
+
+                //boundary check, adjust
+                //cental diagonal amy be up or down
+                if(c<0 && r>rLen)
+                {
+                    r = rLen;
+                    c = 1;
+                    up = true;
+                }
+                else if(c<0)
+                {
+                    c = 0;
+                    r = r;
+                    up = true;
+                }
+                else if(r>rLen)
+                {
+                    r = rLen;
+                    c = c + 2;
+                    up = true;
+                }
+
+            }
+            //if counter==mat len return
+            if(counter==ans.length)
+            {
+                break;
+            }
+        }
+        return ans;
+    }
+}
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diff --git a/Matrix.java b/Matrix.java
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+++ b/Matrix.java
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+// Time Complexity : O (m*n)
+// Space Complexity : o(1)
+// Did this code successfully run on Leetcode :Yes
+// Three line explanation of solution in plain english
+//traverse left to right,top to bottome, right to left, bottom to top
+//update boundaries and do boundary check everytime
+// Your code here along with comments explaining your approach
+class Matrix {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> list = new LinkedList<>();
+        int lB = 0;
+        int rB = matrix[0].length-1;
+        int uB = 0;
+        int loB = matrix.length-1;
+        int counter = 0;
+        int totalLen = matrix.length * matrix[0].length;
+
+        while(counter < totalLen)
+        {
+            //left to right
+            if(uB <= loB) {
+                for(int i = lB; i <= rB; i++)
+                {
+                    list.add(matrix[uB][i]);
+                    counter++;
+                }
+                uB++;
+            }
+
+            //top to bottom
+            if(lB <= rB) {
+                for(int i = uB; i <= loB; i++)
+                {
+                    list.add(matrix[i][rB]);
+                    counter++;
+                }
+                rB--;
+            }
+
+            //right to left
+            if(uB <= loB) {
+                for(int i = rB; i >= lB; i--)
+                {
+                    list.add(matrix[loB][i]);
+                    counter++;
+                }
+                loB--;
+            }
+
+            //bottom to top
+            if(lB <= rB) {
+                for(int i = loB; i >= uB; i--)
+                {
+                    list.add(matrix[i][lB]);
+                    counter++;
+                }
+                lB++;
+            }
+        }
+        return list;
+    }
+}
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diff --git a/ProductExceptSelf.java b/ProductExceptSelf.java
new file mode 100644
index 00000000..47ad0bd2
--- /dev/null
+++ b/ProductExceptSelf.java
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+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
+// Three line explanation of solution in plain english
+//Here we move left to right keeping the track of product till the previous element
+//same we can do from right to left and keep track of element till the next element
+//we will multiply those to get the product
+//Instead fo keeping track of right to left, we just had the rlProduct which has the value of products till next element
+// we mutiply that to existing lrProduct to get answer
+class ProductExceptSelf {
+    public int[] productExceptSelf(int[] nums) {
+        int[] ans = new int[nums.length];
+        //always the product next to 0th element 1
+        int lrProduct = 1;
+        ans[0] = lrProduct;
+        //left to right product except self
+        for(int i = 1;i<nums.length;i++)
+        {
+            lrProduct = nums[i-1] * lrProduct;
+            ans[i] = lrProduct;
+        }
+        //same as before product next to right of len is 1
+        int rlProduct = 1;
+        //insated of having separate array, update the ans array to multply right to left product
+        for(int j = nums.length-2;j>=0;j--)
+        {
+            rlProduct = nums[j+1]*rlProduct;
+            ans[j] = ans[j] * rlProduct;
+        }
+        return ans;
+    }
+}
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