Unlock the Circular Puzzle: Find Next Greater Elements!
Intuition:\nThe approach to solving this problem involves using a stack to find the next greater element for each value in the circular array nums. We traverse 
# Intuition:\nThe approach to solving this problem involves using a stack to find the next greater element for each value in the circular array nums. We traverse the array in reverse order (from right to left) to account for the circular nature. For each element, we pop elements from the stack until we find an element that is greater than the current element. If there is no such element, we assign -1 as the next greater element for the current element. Otherwise, we assign the greater element as the next greater element for the current element. Finally, we reverse the result to maintain the original order of elements in the array.\n\n# Algorithm:\n\nInitialize an empty stack st to store elements while traversing the circular array in reverse order and an empty vector nqe to store the next greater elements for each value in nums.\nTraverse the circular array nums in reverse order (from right to left) using a loop starting from 2*nums.size()-1 and decrementing by 1 in each iteration. This is done to handle the circular nature of the array, as the last element\'s next greater element can be at the beginning of the array.\nWhile traversing, for each element nums[i], do the following:\na. While the stack is not empty and the current element nums[i%n] is greater than or equal to the top of the stack, pop elements from the stack. This step ensures that the stack only contains elements that are greater than nums[i%n], as we want to find the next greater element.\nb. If the index i is less than n, it means we are in the original array, and we can directly access the top of the stack to find the next greater element. If the stack becomes empty after the while loop, it means there is no next greater element for the current element nums[i%n], so we assign -1 as the next greater element for this element in the vector nqe.\nc. Otherwise, we are in the circular part of the array, and we need to search circularly for the next greater element. In this case, we push the next greater element\'s value (found at the top of the stack) into the vector nqe.\nd. Push the current element nums[i%n] onto the stack.\nAfter completing the loop, the vector nqe will contain the next greater elements for each value in the circular array, taking the circular nature into account.\nReverse the vector nqe to maintain the original order of elements in the array.\nReturn the modified vector nqe as the final result.\nComplexity Analysis:\n\n# Time Complexity:\n The algorithm traverses the circular array once, which takes O(n) time, where n is the length of nums.\n# Space Complexity:\n The algorithm uses a stack to store at most n elements and a vector nqe to store the next greater elements, each of which takes O(n) extra space.\n\nOverall, the algorithm has a linear time complexity of O(n) and a linear space complexity of O(n). It efficiently finds the next greater element for each element in the circular array nums using a stack.\n\n\n\n\n\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> nextGreaterElements(vector<int>& nums) {\n        vector<int> nqe;\n        stack<int> st;\n        \n          int n=nums.size();\n        for(int i=2*nums.size()-1; i>=0; i--){\n            while(!st.empty() && nums[i%n]>=st.top()){\n                st.pop();\n            }\n            if(i<n){\n                if(st.empty()==false){   \n                nqe.push_back(st.top());\n               }     else nqe.push_back(-1);\n            }\n            st.push(nums[i%n]);  \n        }\n        reverse(nqe.begin(),nqe.end());\n            return nqe;\n    }\n};\n```\nIf you found my solution helpful, please consider upvoting it to show your appreciation! Your support motivates me to provide more helpful answers. Thank you!\n
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