Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.9 MB (5.19%)

Author: jimit105

0162-find-peak-element/README.md

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+<p>A peak element is an element that is strictly greater than its neighbors.</p>
+
+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, find a peak element, and return its index. If the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>
+
+<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.</p>
+
+<p>You must write an algorithm that runs in <code>O(log n)</code> time.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,1]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 3 is a peak element and your function should return the index number 2.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,1,3,5,6,4]
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>nums[i] != nums[i + 1]</code> for all valid <code>i</code>.</li>
+</ul>

0162-find-peak-element/solution.py

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+# Approach 3: Iterative Binary Search
+
+# Time: O(log n)
+# Space: O(1)
+
+class Solution:
+    def findPeakElement(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+
+        while l < r:
+            mid = (l + r) // 2
+            if nums[mid] > nums[mid + 1]:
+                r = mid
+            else:
+                l = mid + 1
+
+        return l