Sync LeetCode submission Runtime - 3 ms (93.57%), Memory - 20.8 MB (27.05%)

Author: jimit105

0056-merge-intervals/README.md

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+<p>Given an array&nbsp;of <code>intervals</code>&nbsp;where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, merge all overlapping intervals, and return <em>an array of the non-overlapping intervals that cover all the intervals in the input</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> intervals = [[1,3],[2,6],[8,10],[15,18]]
+<strong>Output:</strong> [[1,6],[8,10],[15,18]]
+<strong>Explanation:</strong> Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> intervals = [[1,4],[4,5]]
+<strong>Output:</strong> [[1,5]]
+<strong>Explanation:</strong> Intervals [1,4] and [4,5] are considered overlapping.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>intervals[i].length == 2</code></li>
+	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
+</ul>

0056-merge-intervals/solution.py

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+# Approach 2: Sorting
+
+# Time: O(n log n)
+# Space: O(n)
+
+class Solution:
+    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
+        intervals.sort(key=lambda x: x[0])
+
+        merged = []
+        for interval in intervals:
+            # if the list of merged intervals is empty or if the current
+            # interval does not overlap with the previous, simply append it.
+            if not merged or merged[-1][1] < interval[0]:
+                merged.append(interval)
+
+            else:
+                merged[-1][1] = max(merged[-1][1], interval[1])
+
+        return merged
+