From 733c71d36464b2d6b00b7298b2ef9aebe5fd8ed9 Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Tue, 6 Oct 2026 12:53:56 -0400
Subject: [PATCH] feat(work): add solution for Next Greater Element I

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 .../Easy/Array/496.next-greater-element-i.py  | 75 +++++++++++++++++++
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 create mode 100644 work/1/Easy/Array/496.next-greater-element-i.py

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+"""
+496. Next Greater Element I
+Difficulty: Easy
+https://leetcode.com/problems/next-greater-element-i/
+
+──────────────────────────────────────────────────
+
+The next greater element of some element x in an array is the first
+greater element that is to the right of x in the same array.
+
+You are given two distinct 0-indexed integer arrays nums1 and nums2,
+where nums1 is a subset of nums2.
+
+For each 0 <= i < nums1.length, find the index j such that nums1[i]
+== nums2[j] and determine the next greater element of nums2[j] in
+nums2. If there is no next greater element, then the answer for this
+query is -1.
+
+Return an array ans of length nums1.length such that ans[i] is the
+next greater element as described above.
+
+
+
+Example 1:
+
+Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
+Output: [-1,3,-1]
+Explanation: The next greater element for each value of nums1 is as
+follows:
+- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater
+element, so the answer is -1.
+- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
+- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater
+element, so the answer is -1.
+
+Example 2:
+
+Input: nums1 = [2,4], nums2 = [1,2,3,4]
+Output: [3,-1]
+Explanation: The next greater element for each value of nums1 is as
+follows:
+- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
+- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater
+element, so the answer is -1.
+
+
+
+Constraints:
+
+        • 1 <= nums1.length <= nums2.length <= 1000
+
+        • 0 <= nums1[i], nums2[i] <= 10^4
+
+        • All integers in nums1 and nums2 are unique.
+
+        • All the integers of nums1 also appear in nums2.
+
+
+
+Follow up: Could you find an O(nums1.length + nums2.length) solution?
+"""
+
+
+class Solution:
+    def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
+        mono_stack = []
+        next_greater = {}
+
+        for cur in nums2:
+            while mono_stack and cur > mono_stack[-1]:
+                val = mono_stack.pop()
+                next_greater[val] = cur
+            mono_stack.append(cur)
+
+        return [next_greater.get(x, -1) for x in nums1]