Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 18.3 MB (7.20%)

Author: jimit105

0033-search-in-rotated-sorted-array/README.md

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+<p>There is an integer array <code>nums</code> sorted in ascending order (with <strong>distinct</strong> values).</p>
+
+<p>Prior to being passed to your function, <code>nums</code> is <strong>possibly rotated</strong> at an unknown pivot index <code>k</code> (<code>1 &lt;= k &lt; nums.length</code>) such that the resulting array is <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code> (<strong>0-indexed</strong>). For example, <code>[0,1,2,4,5,6,7]</code> might be rotated at pivot index <code>3</code> and become <code>[4,5,6,7,0,1,2]</code>.</p>
+
+<p>Given the array <code>nums</code> <strong>after</strong> the possible rotation and an integer <code>target</code>, return <em>the index of </em><code>target</code><em> if it is in </em><code>nums</code><em>, or </em><code>-1</code><em> if it is not in </em><code>nums</code>.</p>
+
+<p>You must write an algorithm with <code>O(log n)</code> runtime complexity.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<pre><strong>Input:</strong> nums = [4,5,6,7,0,1,2], target = 0
+<strong>Output:</strong> 4
+</pre><p><strong class="example">Example 2:</strong></p>
+<pre><strong>Input:</strong> nums = [4,5,6,7,0,1,2], target = 3
+<strong>Output:</strong> -1
+</pre><p><strong class="example">Example 3:</strong></p>
+<pre><strong>Input:</strong> nums = [1], target = 0
+<strong>Output:</strong> -1
+</pre>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
+	<li>All values of <code>nums</code> are <strong>unique</strong>.</li>
+	<li><code>nums</code> is an ascending array that is possibly rotated.</li>
+	<li><code>-10<sup>4</sup> &lt;= target &lt;= 10<sup>4</sup></code></li>
+</ul>

0033-search-in-rotated-sorted-array/solution.py

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+# Approach 1: Find Pivot Index + Binary Search
+
+# Time: O(log n)
+# Space: O(1)
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        n = len(nums)
+        left, right = 0, n - 1
+
+        # Find the index of the pivot element (the smallest element)
+        while left <= right:
+            mid = (left + right) // 2
+            if nums[mid] > nums[-1]:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        # Binary search over an inclusive range [left_boundary ~ right_boundary]
+        def binary_search(left_boundary, right_boundary, target):
+            left, right = left_boundary, right_boundary
+            while left <= right:
+                mid = (left + right) // 2
+                if nums[mid] == target:
+                    return mid
+                elif nums[mid] > target:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+            return -1
+
+        # Binary search over elements on the pivot element's left
+        if (answer := binary_search(0, left - 1, target)) != -1:
+            return answer
+
+        # Binary search over elements on the pivot element's right
+        return binary_search(left, n - 1, target)
+