Day25_Next Greater Element I.py

# Next Greater Element I                                 Difficulty = Easy



# 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.



class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        stack = []
        nextGreater = {}
        
        i = 0
        while i < len(nums2):
            while len(stack) and stack[-1] < nums2[i]:
                k = stack.pop(-1)
                nextGreater[k] = nums2[i]
            stack.append(nums2[i])
            i+=1
        
        while len(stack):
            k = stack.pop(-1)
            nextGreater[k] = -1
        
        result = []
        for i in nums1:
            result.append(nextGreater[i])
        
        return result