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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)
1var nextGreaterElements = function(nums) {
2 let n = nums.length;
3 let res = new Array(n).fill(-1);
4 let stack = [];
5 for (let i = 0; i < 2 * n; i++) {
6 while (stack.length && nums[stack[stack.length - 1]] < nums[i % n]) {
7 res[stack.pop()] = nums[i % n];
8 }
9 if (i < n) stack.push(i);
10 }
11 return res;
12};This JavaScript solution implements a similar algorithm using a stack to track indices, iterating over the array twice to capture cyclic nature. The use of a modulo operator is key in managing indices.
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 Python solution performs direct iterative comparisons on pairs of elements. For each index, it tests succeeding elements using a nested loop construct, leveraging modulo to achieve wrap-around.