Pls check its giving error in first test case

#1
#2

hello @sneha23
u missed the fact that given array is circular.

consider this case->
3
4 2 1
answer should be -1 4 4 (element greater than 2 is 4 in its right ) but urs will give -1 -1 -1.

follow 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

hy i tried but i m not able to build the code for circular portion can help?

#4

pls refer this->

vector<int> nextGreaterElements(vector<int> &nums)
{
    int n = nums.size();
    vector<int> res(n);

    stack<int> s;

    for (int i = 2 * n - 1; i >= 0; i--)
    {

        while (!s.empty() && nums[s.top()] <= nums[i % n])
        {
            s.pop();
        }

        res[i % n] = (s.empty() ? -1 : nums[s.top()]);
        s.push(i % n);
    }

    return res;
}

basically we are repeating the procedure that we use to solve
for linear array again to account circular array.
thats the reason why the loop is running 2*n times.
inside logic is iterate from last element of array and maintain an inscresing order in stack. this will help us to answer next greater element in array efficiently.

#5

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