diff --git a/Array1.java b/Array1.java
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+// Approach:
+// 1. Create an answer array to store the prefix products.
+// 2. Traverse from left to right and store the product of all elements before the current index.
+// 3. Traverse from right to left while maintaining a suffix product.
+// 4. Multiply the prefix product with the suffix product to get the final answer.
+// 5. This avoids using division and computes the result in one extra pass.
+
+// Time Complexity: O(n)
+// Space Complexity: O(1) (excluding the output array)
+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        int n=nums.length;
+        int[] ans=new int[n];
+        ans[0]=1;
+        for(int i=1;i<n;i++){
+            ans[i]=nums[i-1]*ans[i-1];
+        }
+        int res=1;
+        for(int i=n-1;i>=0;i--){
+            ans[i]=res*ans[i];
+            res*=nums[i];
+        }
+        return ans;
+    }
+}
+//problem 2
+// Approach:
+// 1. Start from the top-left cell of the matrix.
+// 2. Traverse diagonally upward when (row + col) is even.
+// 3. Traverse diagonally downward when (row + col) is odd.
+// 4. Whenever a boundary is reached, change the direction accordingly.
+// 5. Continue until all matrix elements are added to the result array.
+
+// Time Complexity: O(m * n)
+// Space Complexity: O(1)
+class Solution {
+    public int[] findDiagonalOrder(int[][] mat) {
+        int m=mat.length;
+        int n=mat[0].length;
+        int [] result=new int[m*n];
+        int row=0,col=0;
+        int idx=0;
+        while(idx<m*n){
+            result[idx++]=mat[row][col];
+            if((row+col)%2==0){
+                if(col==n-1){
+                    row++;
+                }else if(row==0){
+                    col++;
+                }else{
+                    row--;
+                    col++;
+                }
+            } else{
+                if(row==m-1){
+                    col++;
+                }else if(col==0){
+                    row++;
+                }else{
+                    row++;
+                    col--;
+                }
+            }
+        }
+        return result;
+    }
+}
+//problem 3
+// Approach:
+// 1. Maintain four boundaries: top, bottom, left, and right.
+// 2. Traverse from left to right, then move the top boundary down.
+// 3. Traverse from top to bottom, then move the right boundary left.
+// 4. If rows remain, traverse from right to left and move the bottom boundary up.
+// 5. If columns remain, traverse from bottom to top and move the left boundary right.
+// 6. Repeat until all elements are visited.
+
+// Time Complexity: O(m * n)
+// Space Complexity: O(1) (excluding the output list)
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> res=new ArrayList<>();
+        int m=matrix.length;
+        int n=matrix[0].length;
+        int left=0;
+        int right=n-1;
+        int top=0;
+        int bottom=m-1;
+        while(left<=right && top<=bottom){
+        for(int i=left;i<=right;i++){
+            res.add(matrix[top][i]);
+        }top++;
+        for(int i=top;i<=bottom;i++){
+            res.add(matrix[i][right]);
+        }right--;
+        if(top<=bottom){
+        for(int i=right;i>=left;i--){
+            res.add(matrix[bottom][i]);
+        }
+        bottom--;
+        }
+        if(left<=right){
+        for(int i=bottom;i>=top;i--){
+            res.add(matrix[i][left]);
+        }left++;
+        }
+        }
+        return res;
+    }
+}