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This approach utilizes a stack to keep track of the indices of the elements for which we are finding the next greater element. We traverse the array twice (length times two) to account for the circular nature of the array. For each element, we repeatedly pop elements from the stack while the current element is greater than the element represented by the index at the top of the stack. As we pop elements, we update their next greater element in the resulting array. The stack helps ensure that each element is processed efficiently, leading to an O(n) complexity.
Time Complexity: O(n)
Space Complexity: O(n) (due to the stack and result array)
1#include <vector>
2#include <stack>
3using namespace std;
4
5vector<int> nextGreaterElements(vector<int>& nums) {
6 int n = nums.size();
7 vector<int> res(n, -1);
8 stack<int> s;
9 for (int i = 0; i < 2 * n; ++i) {
10 while (!s.empty() && nums[s.top()] < nums[i % n]) {
11 res[s.top()] = nums[i % n];
12 s.pop();
13 }
14 if (i < n) {
15 s.push(i);
16 }
17 }
18 return res;
19}In this C++ solution, a stack is used to efficiently find the next greater elements by storing indices. The array is traversed twice using a modulo operation to accommodate the circular property. Elements whose indices are stored in the stack are continuously checked and resolved based on the current element.
This approach uses two separate traversals to find the next greater element by checking for each element's greater counterpart. We double-iterate through the array copying elements to support cyclic checking. Though simpler conceptually, its efficiency isn't as optimal due to direct comparisons, and it could degrade to an O(n^2) complexity in worst cases. It's educational for understanding direct exhaustive search in circular constructs.
Time Complexity: O(n^2) in the worst case
This Java solution uses a doubly nested loop to determine each element's next greater element in the array. This approach clearly elucidates a direct solution mechanism for circular arrays.