Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.6 MB (12.04%)

Author: jimit105

0496-next-greater-element-i/README.md

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+<p>The <strong>next greater element</strong> of some element <code>x</code> in an array is the <strong>first greater</strong> element that is <strong>to the right</strong> of <code>x</code> in the same array.</p>
+
+<p>You are given two <strong>distinct 0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, where <code>nums1</code> is a subset of <code>nums2</code>.</p>
+
+<p>For each <code>0 &lt;= i &lt; nums1.length</code>, find the index <code>j</code> such that <code>nums1[i] == nums2[j]</code> and determine the <strong>next greater element</strong> of <code>nums2[j]</code> in <code>nums2</code>. If there is no next greater element, then the answer for this query is <code>-1</code>.</p>
+
+<p>Return <em>an array </em><code>ans</code><em> of length </em><code>nums1.length</code><em> such that </em><code>ans[i]</code><em> is the <strong>next greater element</strong> as described above.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [4,1,2], nums2 = [1,3,4,2]
+<strong>Output:</strong> [-1,3,-1]
+<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:
+- 4 is underlined in nums2 = [1,3,<u>4</u>,2]. There is no next greater element, so the answer is -1.
+- 1 is underlined in nums2 = [<u>1</u>,3,4,2]. The next greater element is 3.
+- 2 is underlined in nums2 = [1,3,4,<u>2</u>]. There is no next greater element, so the answer is -1.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [2,4], nums2 = [1,2,3,4]
+<strong>Output:</strong> [3,-1]
+<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:
+- 2 is underlined in nums2 = [1,<u>2</u>,3,4]. The next greater element is 3.
+- 4 is underlined in nums2 = [1,2,3,<u>4</u>]. There is no next greater element, so the answer is -1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length &lt;= nums2.length &lt;= 1000</code></li>
+	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>
+	<li>All integers in <code>nums1</code> and <code>nums2</code> are <strong>unique</strong>.</li>
+	<li>All the integers of <code>nums1</code> also appear in <code>nums2</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Could you find an <code>O(nums1.length + nums2.length)</code> solution?

0496-next-greater-element-i/solution.py

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+# Approach 2: Better Brute Force
+
+# Time: O(m * n)
+# Space: O(n)
+
+class Solution:
+    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        hash_table = {num: i for i, num in enumerate(nums2)}
+
+        res = [0] * len(nums1)
+
+        for i, num in enumerate(nums1):
+            j = hash_table[num] + 1
+            while j < len(nums2):
+                if nums2[j] > num:
+                    res[i] = nums2[j]
+                    break
+                j += 1
+            else:
+                res[i] = -1
+
+        return res
+