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SQP7 Next Greater Element II

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.

Example 1:
Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation:
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:
- 1 <= nums.length <= 10^4
- -109 <= nums[i] <= 10^9

This Problem is the Extended Version of the Next Gratest Element I this is the Circular Array List

We used a Prevesoly Concept but Different Manner

Basically in this problem add one more Same array at the end of the amin array for Double the Size of the Array

                    Example: arr = [2,10,12,1,11]
                    add Same array at the End of the array
                    updateArray = [2,10,12,1,11,2,10,12,1,11] 
                    {Same array Copy at the End of the Array}

After adding the end of the Same array, we used the Preveselly Same Concept in Different Manner

In this approch we Iterate at the Forword -> direction of the array and we prevessly we Iterate in the <- Backword Direction in the Array

Every ith Iteration wh take the i % n for the Accurate and Right Index of the Array becouse our array Sized is Doubled and Circular Array that why we take the Mod Every Time.

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