Sync LeetCode submission Runtime - 9 ms (12.52%), Memory - 17.9 MB (36.45%)

Author: jimit105

0768-partition-labels/README.md

@@ -0,0 +1,32 @@
+<p>You are given a string <code>s</code>. We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string <code>&quot;ababcc&quot;</code> can be partitioned into <code>[&quot;abab&quot;, &quot;cc&quot;]</code>, but partitions such as <code>[&quot;aba&quot;, &quot;bcc&quot;]</code> or <code>[&quot;ab&quot;, &quot;ab&quot;, &quot;cc&quot;]</code> are invalid.</p>
+
+<p>Note that the partition is done so that after concatenating all the parts in order, the resultant string should be <code>s</code>.</p>
+
+<p>Return <em>a list of integers representing the size of these parts</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;ababcbacadefegdehijhklij&quot;
+<strong>Output:</strong> [9,7,8]
+<strong>Explanation:</strong>
+The partition is &quot;ababcbaca&quot;, &quot;defegde&quot;, &quot;hijhklij&quot;.
+This is a partition so that each letter appears in at most one part.
+A partition like &quot;ababcbacadefegde&quot;, &quot;hijhklij&quot; is incorrect, because it splits s into less parts.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;eccbbbbdec&quot;
+<strong>Output:</strong> [10]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 500</code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>

0768-partition-labels/solution.py

@@ -0,0 +1,25 @@
+# Approach 1: Two Pointers
+
+# k = no. of unique chars in s
+# Time: O(n)
+# Space: O(k)
+
+class Solution:
+    def partitionLabels(self, s: str) -> List[int]:
+        last_occurrence = [0] * 26
+        for i, char in enumerate(s):
+            last_occurrence[ord(char) - ord('a')] = i
+
+        partition_end = 0
+        partition_start = 0
+        partition_sizes = []
+
+        for i, char in enumerate(s):
+            partition_end = max(partition_end, last_occurrence[ord(char) - ord('a')])
+
+            if i == partition_end:
+                partition_sizes.append(i - partition_start + 1)
+                partition_start = i + 1
+
+        return partition_sizes
+