Sync LeetCode submission Runtime - 858 ms (52.27%), Memory - 18 MB (93.18%)

Author: jimit105

3142-longest-unequal-adjacent-groups-subsequence-ii/README.md

@@ -0,0 +1,72 @@
+<p>You are given a string array <code>words</code>, and an array <code>groups</code>, both arrays having length <code>n</code>.</p>
+
+<p>The <strong>hamming distance</strong> between two strings of equal length is the number of positions at which the corresponding characters are <strong>different</strong>.</p>
+
+<p>You need to select the <strong>longest</strong> <span data-keyword="subsequence-array">subsequence</span> from an array of indices <code>[0, 1, ..., n - 1]</code>, such that for the subsequence denoted as <code>[i<sub>0</sub>, i<sub>1</sub>, ..., i<sub>k-1</sub>]</code> having length <code>k</code>, the following holds:</p>
+
+<ul>
+	<li>For <strong>adjacent</strong> indices in the subsequence, their corresponding groups are <strong>unequal</strong>, i.e., <code>groups[i<sub>j</sub>] != groups[i<sub>j+1</sub>]</code>, for each <code>j</code> where <code>0 &lt; j + 1 &lt; k</code>.</li>
+	<li><code>words[i<sub>j</sub>]</code> and <code>words[i<sub>j+1</sub>]</code> are <strong>equal</strong> in length, and the <strong>hamming distance</strong> between them is <code>1</code>, where <code>0 &lt; j + 1 &lt; k</code>, for all indices in the subsequence.</li>
+</ul>
+
+<p>Return <em>a string array containing the words corresponding to the indices <strong>(in order)</strong> in the selected subsequence</em>. If there are multiple answers, return <em>any of them</em>.</p>
+
+<p><strong>Note:</strong> strings in <code>words</code> may be <strong>unequal</strong> in length.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">words = [&quot;bab&quot;,&quot;dab&quot;,&quot;cab&quot;], groups = [1,2,2]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[&quot;bab&quot;,&quot;cab&quot;]</span></p>
+
+<p><strong>Explanation: </strong>A subsequence that can be selected is <code>[0,2]</code>.</p>
+
+<ul>
+	<li><code>groups[0] != groups[2]</code></li>
+	<li><code>words[0].length == words[2].length</code>, and the hamming distance between them is 1.</li>
+</ul>
+
+<p>So, a valid answer is <code>[words[0],words[2]] = [&quot;bab&quot;,&quot;cab&quot;]</code>.</p>
+
+<p>Another subsequence that can be selected is <code>[0,1]</code>.</p>
+
+<ul>
+	<li><code>groups[0] != groups[1]</code></li>
+	<li><code>words[0].length == words[1].length</code>, and the hamming distance between them is <code>1</code>.</li>
+</ul>
+
+<p>So, another valid answer is <code>[words[0],words[1]] = [&quot;bab&quot;,&quot;dab&quot;]</code>.</p>
+
+<p>It can be shown that the length of the longest subsequence of indices that satisfies the conditions is <code>2</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">words = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;], groups = [1,2,3,4]</span></p>
+
+<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;]</span></p>
+
+<p><strong>Explanation: </strong>We can select the subsequence <code>[0,1,2,3]</code>.</p>
+
+<p>It satisfies both conditions.</p>
+
+<p>Hence, the answer is <code>[words[0],words[1],words[2],words[3]] = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;]</code>.</p>
+
+<p>It has the longest length among all subsequences of indices that satisfy the conditions.</p>
+
+<p>Hence, it is the only answer.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == words.length == groups.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>1 &lt;= groups[i] &lt;= n</code></li>
+	<li><code>words</code> consists of <strong>distinct</strong> strings.</li>
+	<li><code>words[i]</code> consists of lowercase English letters.</li>
+</ul>

3142-longest-unequal-adjacent-groups-subsequence-ii/solution.py

@@ -0,0 +1,43 @@
+# Approach: Dynamic Programming
+
+class Solution:
+    def getWordsInLongestSubsequence(self, words: List[str], groups: List[int]) -> List[str]:
+        n = len(groups)
+        dp = [1] * n
+        prev = [-1] * n
+        max_idx = 0
+
+        for i in range(1, n):
+            for j in range(i):
+                if(
+                    self.check(words[i], words[j])
+                    and dp[j] + 1 > dp[i]
+                    and groups[i] != groups[j]
+                ):
+                    dp[i] = dp[j] + 1
+                    prev[i] = j
+
+            if dp[i] > dp[max_idx]:
+                max_idx = i
+
+        ans = []
+        i = max_idx
+
+        while i >= 0:
+            ans.append(words[i])
+            i = prev[i]
+        ans.reverse()
+        return ans
+
+    def check(self, s1, s2):
+        if len(s1) != len(s2):
+            return False
+        diff = 0
+        for c1, c2 in zip(s1, s2):
+            if c1 != c2:
+                diff += 1
+                if diff > 1:
+                    return False
+
+        return diff == 1
+