Problem Description
Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums. The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number.
Key Insights
- The problem involves a circular array, which means the array wraps around.
- A brute-force approach would involve checking each element against every other element, leading to O(n^2) complexity.
- A more efficient approach uses a monotonic stack to keep track of indices of elements in a decreasing order, allowing us to find the next greater element in O(n) time.
Space and Time Complexity
Time Complexity: O(n)
Space Complexity: O(n)
Solution
To solve the problem, we can use a monotonic stack. We will iterate through the circular array twice (effectively treating it as a linear array for the purposes of finding the next greater elements). For each element, we will check if it is greater than the element represented by the index stored at the top of the stack. If it is, we pop from the stack and assign the current element as the next greater element for the index popped. If we go through all elements and the stack is not empty, we assign -1 for the remaining indices. This ensures each index is processed efficiently.