Please correct my code sir

#1

I have scored 66 but to get reamining i need your help;

#2

@shudhanshu20092001_af6f20d24c617008 Your logic is quite complicated. So instead of using HashMap Do one thing :
You know that at any given instance your stack is storing Elements whose Next greater is yet to be found but you have traversed that element. So instead of storing Element in stack store the index of that element so that you will have index of the element and you can get the element using index(which makes it possible without using hashmap). Below is the exact approach of the solution I am talking about : (Note even after using this approach if you still have the problem then Let me know I will provide you my code for this problem.)
This approach makes use of a stack. This stack stores the indices of the appropriate elements from nums array. The top of the stack refers to the index of the Next Greater Element found so far. We store the indices instead of the elements since there could be duplicates in the nums array. The description of the method will make the above statement clearer.

We start traversing the numsnums array from right towards the left. For an element nums[i] encountered, we pop all the elements stack[top] from the stack such that nums[stack[top]] ≤ nums[i]. We continue the popping till we encounter a stack[top] satisfying nums[stack[top]]>nums[i]. Now, it is obvious that the current stack[top] only can act as the Next Greater Element for nums[i](right now, considering only the elements lying to the right of nums[i]).

If no element remains on the top of the stack, it means no larger element than nums[i] exists to its right. Along with this, we also push the index of the element just encountered(nums[i]), i.e. ii over the top of the stack, so thatnums[i](or stack[topstack[top) now acts as the Next Greater Element for the elements lying to its left.

We go through two such passes over the complete nums array. This is done so as to complete a circular traversal over the nums array. The first pass could make some wrong entries in the res array since it considers only the elements lying to the right of nums[i], without a circular traversal. But, these entries are corrected in the second pass.

#3

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