From bc7b80a0a5e11c35f33be1c1e435bdf4d9f00cb6 Mon Sep 17 00:00:00 2001
From: subbu4061 <vuppalabunny1@gmail.com>
Date: Mon, 12 Jan 2026 09:09:25 -0800
Subject: [PATCH] Adding the Array-1 Solutions

---
 diagonal-traverse.java            | 44 ++++++++++++++++++++++++
 product-of-array-except-self.java | 28 +++++++++++++++
 spiral-matrix.java                | 57 +++++++++++++++++++++++++++++++
 3 files changed, 129 insertions(+)
 create mode 100644 diagonal-traverse.java
 create mode 100644 product-of-array-except-self.java
 create mode 100644 spiral-matrix.java

diff --git a/diagonal-traverse.java b/diagonal-traverse.java
new file mode 100644
index 00000000..77aff4e6
--- /dev/null
+++ b/diagonal-traverse.java
@@ -0,0 +1,44 @@
+// 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 use a boolean variable dir to decide whether to move up-right or down-left along the diagonals. When the traversal hits a boundary (top row, bottom row, left column, or right column), it changes direction and adjusts the row or column accordingly until all elements are visited.
+
+class Solution {
+    public int[] findDiagonalOrder(int[][] mat) {
+        int m = mat.length;
+        int n = mat[0].length;
+        int[] output = new int[m*n];
+        int row =0;
+        int col =0;
+        // If this is true we go up and if this is false we go down
+        boolean dir = true;
+        for(int i=0; i<output.length; i++) {
+            output[i] = mat[row][col];
+            if(dir) {
+                if(col == n-1) {
+                    row++;
+                    dir= false;
+                } else if (row == 0) {
+                    col ++;
+                    dir = false;
+                } else {
+                    row --;
+                    col ++;
+                }
+            } else {
+                if (row == m-1) {
+                    col++;
+                    dir = true;
+                } else if (col ==0) {
+                    row++;
+                    dir = true;
+                } else {
+                    row++;
+                    col --;
+                }
+            }
+        }
+        return output;
+        
+    }
+}
diff --git a/product-of-array-except-self.java b/product-of-array-except-self.java
new file mode 100644
index 00000000..cac2c417
--- /dev/null
+++ b/product-of-array-except-self.java
@@ -0,0 +1,28 @@
+// Time Complexity :O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Three line explanation of solution in plain english : First, the code initializes an output array of the same size as the input array nums to store the result, and a variable runMul to keep track of running products. 
+// The first loop goes left to right through the array, setting each output[i] to the product of all elements before index i.
+// runMul is updated at each step by multiplying it with the current number.The second loop goes right to left, multiplying each output[i] by the product of all elements after index i. 
+// Again, runMul is updated at each step to accumulate the running product from the right.Finally, the output array contains the product of all elements of nums except the element itself at each index, and it is returned.
+
+
+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        int n = nums.length;
+        int[] output = new int[n];
+        int runMul =1;
+        for(int i=0; i<n; i++) {
+            output[i] = runMul;
+            runMul *= nums[i];
+            
+        }
+        
+        runMul = 1;
+        for(int i=n-1; i>=0; i--) {
+            output[i] *= runMul;
+            runMul *= nums[i]; 
+        }
+        return output;
+    }
+}
\ No newline at end of file
diff --git a/spiral-matrix.java b/spiral-matrix.java
new file mode 100644
index 00000000..6efa0921
--- /dev/null
+++ b/spiral-matrix.java
@@ -0,0 +1,57 @@
+// 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 keep track of the current direction (RIGHT, DOWN, LEFT, UP) using an enum. We use bounds (rightBound, downBound, leftBound, upperBound) to know when to turn and shrink the spiral. In each iteration, we add the current element to the output list and moves the row/column according to the direction. When a bound is reached, it changes direction and updates the corresponding bound until all elements are added.
+
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> output = new ArrayList<>();
+        int m = matrix.length;
+        int n = matrix[0].length;
+        enum Direction {UP, DOWN, LEFT, RIGHT}
+        int rightBound = n-1;
+        int downBound = m-1;
+        int leftBound = 0;
+        int upperBound = 1;
+        int row = 0;
+        int col = 0;
+        Direction dir = Direction.RIGHT;
+        for(int i=0; i<m*n; i++) {
+            output.add(matrix[row][col]);
+            if (dir == Direction.RIGHT) {
+                if(col == rightBound) {
+                    row++;
+                    rightBound--;
+                    dir = Direction.DOWN;
+                } else {
+                    col++;
+                }
+            } else if (dir == Direction.DOWN) {
+                if (row == downBound) {
+                    col--;
+                    downBound--;
+                    dir = Direction.LEFT;
+                } else {
+                    row++;
+                }
+            } else if (dir == Direction.LEFT) {
+                if (col == leftBound) {
+                    row--;
+                    leftBound++;
+                    dir = Direction.UP;
+                } else {
+                    col--;
+                }
+            } else {
+                if (row == upperBound) {
+                    col++;
+                    upperBound++;
+                    dir = Direction.RIGHT;
+                } else {
+                    row--;
+                }
+            }
+        }
+        return output;
+    }
+}