Sync LeetCode submission Runtime - 44 ms (54.52%), Memory - 28.7 MB (45.55%)

Author: jimit105

0041-first-missing-positive/README.md

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+<p>Given an unsorted integer array <code>nums</code>. Return the <em>smallest positive integer</em> that is <em>not present</em> in <code>nums</code>.</p>
+
+<p>You must implement an algorithm that runs in <code>O(n)</code> time and uses <code>O(1)</code> auxiliary space.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,0]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The numbers in the range [1,2] are all in the array.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,4,-1,1]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 1 is in the array but 2 is missing.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [7,8,9,11,12]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The smallest positive integer 1 is missing.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>

0041-first-missing-positive/solution.py

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+# Approach 3: Cycle Sort
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def firstMissingPositive(self, nums: List[int]) -> int:
+        n = len(nums)
+
+        i = 0
+        while i < n:
+            correct_idx = nums[i] - 1
+            if 0 < nums[i] <= n and nums[i] != nums[correct_idx]:
+                nums[i], nums[correct_idx] = nums[correct_idx], nums[i]
+            else:
+                i += 1
+
+        # Iterate through nums
+        # return smallest missing positive integer
+        for i in range(n):
+            if nums[i] != i + 1:
+                return i + 1
+
+        # If all elements are at the correct index
+        # the smallest missing positive number is n + 1
+        return n + 1
+