{"id":864,"date":"2025-08-11T11:52:43","date_gmt":"2025-08-11T06:22:43","guid":{"rendered":"https:\/\/codeanddebug.in\/blog\/?p=864"},"modified":"2025-08-11T11:52:44","modified_gmt":"2025-08-11T06:22:44","slug":"next-greater-element-ii","status":"publish","type":"post","link":"https:\/\/codeanddebug.in\/blog\/next-greater-element-ii\/","title":{"rendered":"Next Greater Element II | Leetcode 503 | Imaginary List Optimal Solution"},"content":{"rendered":"\n

Given a circular integer array nums<\/code> (i.e., the next element of nums[nums.length - 1]<\/code> is nums[0]<\/code>), return the next greater number<\/strong> for every element in<\/em> nums<\/code>.<\/p>\n\n\n\n

Here’s the [Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n\n

The next greater number<\/strong> of a number x<\/code> 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<\/code> for this number.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> nums = [1,2,1]
Output:<\/strong> [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.<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [1,2,3,4,3]
Output:<\/strong> [2,3,4,-1,4]<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n