Next greater element ii

Given a circular integer array  (i.e., the next element of  is ), return the next greater number for every element in .

Here’s the [] to begin with.

The next greater number of a number  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  for this number.

Example 1:

Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number.
The second 1's next greater number needs to search circularly, which is also 2.

Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

Constraints:

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Optimal Approach: Monotonic Stack for Circular Arrays

Intuition

• In a non-circular Next Greater Element problem, you process elements from right to left using a stack and can stop once you reach the beginning.
• In the circular case, an element’s NGE might only appear after the end, so you must symbolically “loop” through the array twice.
• This is achieved by iterating from 2*n – 1 down t
  
  

  503. Next Greater Element II- Detailed Explanation

  Problem Statement

  Description:
  You are given a circular array of integers. For each element in the array, you need to find its next greater element. The next greater element of a number x is the first greater number to its right when traversing the array in a circular fashion (i.e. after the last element, the search continues from the first element). If no such greater number exists, return -1 for that element.

  Constraints:

  • The length of the array is at least 1.
  • The array is circular, meaning the search for the next greater element wraps around to the beginning.

  Example 1:

  • Input:
  • Output:
  • Explanation:
    • For index 0 (value 1), the next greater element is 2.
    • For index 1 (value 2), there is no greater element.
    • For index 2 (value 1), we wrap around to find that the next greater element is 2.

  Example 2:

  • Input:
  • Output:
  • Explanation:
    • For index 0 (value 3), the next greater element is 8.
    • For index 1 (value 8), there is no greater element.
    • For index 2 (value 4), there is no greater element to its right or when wrapping around.
    • For index 3 (value 1), the next greater element is 3 (after wrapping

      LeetCode 503: Next Greater Element II Solution in Python – A Step-by-Step Guide

      Imagine you’re spinning a wheel of numbers—like [1, 2, 1]—and for each number, you need to peek ahead, even wrapping around to the start, to find the next bigger value. That’s the clever twist of LeetCode 503: Next Greater Element II, a medium-level problem that’s a fantastic way to practice stacks in Python. We’ll explore two solutions: the Best Solution, a monotonic stack with circular handling that’s fast and elegant, and an Alternative Solution, a brute-force approach that’s simpler but slower. With detailed examples, clear code, and a friendly tone—especially for the stack magic—this guide will help you spot those greater elements, whether you’re new to coding or leveling up. Let’s spin that wheel and find some bigger numbers!

      In LeetCode 503: Next Greater Element II, you’re given a circular array of integers, meaning after the last element, you loop back to the first. For each number, you need to find the next greater element ahead of it—or return -1 if none exists. For example, in [1, 2, 1], the next greater element for 1 is 2, for 2 is -1 (nothing bigger), and for the last 1 is 2 (c

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      Code Implementation

      Problem Description

      You are given a circular array of integers called . For each element in , you depend on to find the next greater element, which is the first element to its right in the array (considering the array as circular) that is strictly greater than it. If there is no such element, restore for that position.

      The array is circular, meaning after the last element, you carry on from the beginning. Each element's "next" can wrap around. You must not reuse elements out of order, and there is only one correct next greater element (or ) for each position.

      Constraints:

      Thought Process

      At first glance, you might consider checking for each element in by looping through all subsequent elements (wrapping around if needed) to find the next greater element. However, this brute-force method would require examining up to elements for each position, leading to a time complexity of . For large arrays, this is inefficient.

      To improve, we recall the "Next Greater Element" problem for a linear array, which can be efficiently solved using a stack. The stack keeps track of indices whose next greater element hasn't been found yet.

      Problem Statement

      Given a circular integer array , return the array containing the for each element in nums.

      A of a number is the first greater number than the current number in its traversing-order in the array, which means you could search circularly to find its next greater number. If the next greater element doesn't exist, return -1 for the particular number number.

      Examples

      • Example 1:

        • Input: nums =
        • Expected Output:
        • Justification: For , the next greater number is . For , it's (the next number in the array). For the second , is again the next greater. For , there's no greater number, hence . For , looping over, is the next greater number.

      • Example 2:

        • Input: nums =
        • Expected Output:
        • Justification: The next greater for is , for is , for is again, and for there's no greater number (). For , considering the circular nature, is the next greater number.

      • Example 3:

        • Input: nums =
        • Expected Output:
        • Justification: For , as it's the greatest, no number is greater, so . For and , is the next greater number. For , , and , the next greater number is , considering the circular array. For , is the next greater number