next-greater-element-i.py

# 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.
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# You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.
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# 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.
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# Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.
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#  
# Example 1:
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# 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.
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#
# Example 2:
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# 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.
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#
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# Constraints:
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# 	1 <= nums1.length <= nums2.length <= 1000
# 	0 <= nums1[i], nums2[i] <= 104
# 	All integers in nums1 and nums2 are unique.
# 	All the integers of nums1 also appear in nums2.
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#
#  
# 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]:
        d = {}
        ans = []
        # Store index of numbers in nums2
        for i in range(0, len(nums2)):
            d[nums2[i]] = i
        for i in range(0, len(nums1)):
            index = d[nums1[i]] # 4
            found = False
            for j in range (index, len(nums2)):
                if nums2[j] > nums1[i]:
                    ans.append(nums2[j])
                    found = True
                    break
            if found == False:
                ans.append(-1)
        return ans