Sync LeetCode submission Runtime - 661 ms (24.05%), Memory - 21.2 MB (91.83%)

Author: jimit105

1170-shortest-common-supersequence-/README.md

@@ -0,0 +1,30 @@
+<p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the shortest string that has both </em><code>str1</code><em> and </em><code>str2</code><em> as <strong>subsequences</strong></em>. If there are multiple valid strings, return <strong>any</strong> of them.</p>
+
+<p>A string <code>s</code> is a <strong>subsequence</strong> of string <code>t</code> if deleting some number of characters from <code>t</code> (possibly <code>0</code>) results in the string <code>s</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> str1 = &quot;abac&quot;, str2 = &quot;cab&quot;
+<strong>Output:</strong> &quot;cabac&quot;
+<strong>Explanation:</strong> 
+str1 = &quot;abac&quot; is a subsequence of &quot;cabac&quot; because we can delete the first &quot;c&quot;.
+str2 = &quot;cab&quot; is a subsequence of &quot;cabac&quot; because we can delete the last &quot;ac&quot;.
+The answer provided is the shortest such string that satisfies these properties.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> str1 = &quot;aaaaaaaa&quot;, str2 = &quot;aaaaaaaa&quot;
+<strong>Output:</strong> &quot;aaaaaaaa&quot;
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= str1.length, str2.length &lt;= 1000</code></li>
+	<li><code>str1</code> and <code>str2</code> consist of lowercase English letters.</li>
+</ul>

1170-shortest-common-supersequence-/solution.py

@@ -0,0 +1,36 @@
+# Approach 3: Bottom-Up Dynamic Programming
+
+# n = len(str1), m = len(str2)
+# Time: O(n * m * (n + m))
+# Space: O(m * (n + m))
+
+class Solution:
+    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
+        str1_len = len(str1)
+        str2_len = len(str2)
+
+        # Initialize the first row (when str1 is empty, the supersequence is str2's prefix)
+        prev_row = [str2[:col] for col in range(str2_len + 1)]
+
+        for row in range(1, str1_len + 1):
+            # Initialize the first column (when str2 is empty, the supersequence is str1's prefix)
+            curr_row = [str1[:row]] + [None for _ in range(str2_len)]
+
+            for col in range(1, str2_len + 1):
+                # If characters match, extend the supersequence from the diagonal value
+                if str1[row - 1] == str2[col - 1]:
+                    curr_row[col] = prev_row[col - 1] + str1[row - 1]
+                else:
+                    # If characters do not match, choose the shorter supersequence 
+                    # From previous row (exclude current str1 char)
+                    pick_s1 = prev_row[col]
+                    # From previous column (exclude current str2 char)
+                    pick_s2 = curr_row[col - 1]
+
+                    curr_row[col] = (pick_s1 + str1[row - 1] if len(pick_s1) < len(pick_s2) 
+                    else pick_s2 + str2[col - 1])
+
+            # Move to next row
+            prev_row = curr_row
+        
+        return prev_row[str2_len]