Find the greater element question

#1

I know how to solve this question if array is not circular but its being circular i am facing problem.
Please explain or give the code.

Find the greater element
#2

Hi @aayuushh, let me share my approach with you, i am using stack to solve this question you will know why,
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 nums 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.

dry run to understand this approach !!
try to implement it yourself, if you get into any trouble refer the pseudo-code below

pseudo-code:-

void nextGreaterElements(int nums[], int n)
{
    stack<int> s;
    int res[n];
    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);
    }
    for (int i = 0; i < n; i++) {
        cout << res[i] << " ";
    }
    cout << endl;
    return;
}

Sorry for long post , In case of any doubt feel free to ask :slight_smile:
Mark your doubt as RESOLVED if you got the answer

#3

I hope I’ve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.

On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.