From 265aa30b156c5ca7b117b87208a9db592750c28d Mon Sep 17 00:00:00 2001
From: Jimit Dholakia <25223469+jimit105@users.noreply.github.com>
Date: Sun, 29 Dec 2024 07:27:00 +0000
Subject: [PATCH] Sync LeetCode submission Runtime - 14 ms (44.74%), Memory -
 19.7 MB (47.58%)

---
 0498-diagonal-traverse/README.md   | 27 ++++++++++++++++++++
 0498-diagonal-traverse/solution.py | 40 ++++++++++++++++++++++++++++++
 2 files changed, 67 insertions(+)
 create mode 100644 0498-diagonal-traverse/README.md
 create mode 100644 0498-diagonal-traverse/solution.py

diff --git a/0498-diagonal-traverse/README.md b/0498-diagonal-traverse/README.md
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+<p>Given an <code>m x n</code> matrix <code>mat</code>, return <em>an array of all the elements of the array in a diagonal order</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/04/10/diag1-grid.jpg" style="width: 334px; height: 334px;" />
+<pre>
+<strong>Input:</strong> mat = [[1,2,3],[4,5,6],[7,8,9]]
+<strong>Output:</strong> [1,2,4,7,5,3,6,8,9]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> mat = [[1,2],[3,4]]
+<strong>Output:</strong> [1,2,3,4]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == mat.length</code></li>
+	<li><code>n == mat[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>5</sup> &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
+</ul>
diff --git a/0498-diagonal-traverse/solution.py b/0498-diagonal-traverse/solution.py
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+# Approach 1: Diagonal Iteration and Reversal
+
+# Time: O(n * m)
+# Space: O(min(n, m))
+
+class Solution:
+    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
+        if not mat or not mat[0]:
+            return []
+
+        n, m = len(mat), len(mat[0])
+
+        result, intermediate = [], []
+
+        for d in range(n + m - 1):
+            intermediate.clear()
+
+            # We need to figure out the "head" of this diagonal
+            # The elements in the first row and the last column
+            # are the respective heads.
+            r, c = 0 if d < m else d - m + 1, d if d < m else m - 1
+
+            # Iterate until one of the indices goes out of scope
+            # Take note of the index math to go down the diagonal
+            while r < n and c > -1:
+                intermediate.append(mat[r][c])
+                r += 1
+                c -= 1
+
+            # Reverse even numbered diagonals. The
+            # article says we have to reverse odd 
+            # numbered articles but here, the numbering
+            # is starting from 0 :P
+            if d % 2 == 0:
+                result.extend(intermediate[::-1])
+            else:
+                result.extend(intermediate)
+
+        return result
+