Sync LeetCode submission Runtime - 3 ms (44.72%), Memory - 18.1 MB (13.03%)

Author: jimit105

2048-build-array-from-permutation/README.md

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+<p>Given a <strong>zero-based permutation</strong> <code>nums</code> (<strong>0-indexed</strong>), build an array <code>ans</code> of the <strong>same length</strong> where <code>ans[i] = nums[nums[i]]</code> for each <code>0 &lt;= i &lt; nums.length</code> and return it.</p>
+
+<p>A <strong>zero-based permutation</strong> <code>nums</code> is an array of <strong>distinct</strong> integers from <code>0</code> to <code>nums.length - 1</code> (<strong>inclusive</strong>).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,2,1,5,3,4]
+<strong>Output:</strong> [0,1,2,4,5,3]<strong>
+Explanation:</strong> The array ans is built as follows: 
+ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
+    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
+    = [0,1,2,4,5,3]</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5,0,1,2,3,4]
+<strong>Output:</strong> [4,5,0,1,2,3]
+<strong>Explanation:</strong> The array ans is built as follows:
+ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
+    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
+    = [4,5,0,1,2,3]</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>0 &lt;= nums[i] &lt; nums.length</code></li>
+	<li>The elements in <code>nums</code> are <strong>distinct</strong>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow-up:</strong> Can you solve it without using an extra space (i.e., <code>O(1)</code> memory)?</p>

2048-build-array-from-permutation/solution.py

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+# Approach 1: Build As Required
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def buildArray(self, nums: List[int]) -> List[int]:
+        return [nums[i] for i in nums]
+