Replace each element with its Next Larger Element on the right

Problem:
Given an Array, replace each element in the Array with its Next Element(To its RHS) which is Larger than it. If no such element exists, then no need to replace.
Ex:

i/p: {2,12,8,6,5,1,2,10,3,2}
o/p:{12,12,10,10,10,2,10,10,3,2}

Solution:

Naive solution is O(n^2) as it requires two nested loops.

A more optimal solution is to traverse the array from the right and put all elements in a BST, replacing each element in the array with its immediate in-order successor in the BST. This will take O(n log n) time to complete and require O(n) space.

The best solution also requires O(n) space but runs in O(n) time. We traverse the array from the left and keep pushing elements' indexes to a stack, ensuring that all the elements in the stack are always stored in decreasing order. Once the order is not decreasing, we pop the element from the stack (remember, it's an index) and replace the corresponding element in the array with an element currently being processed in the main loop:

void ReplaceNextLargest ( int * Arr, int N ) {
    stack<int> stk;
    stk.push(0);

    for( int idx = 1; idx < N; idx++ ) {
        while( !stk.empty() && (Arr[idx] > Arr[stk.top()]) ) {
            Arr[stk.top()] = Arr[idx];
            stk.pop();
        }
        stk.push(idx);
    }
}

Complexity:

time - O(n)
space - O(n)

Links and credits:
http://www.careercup.com/question?id=6311825561878528
http://www.geeksforgeeks.org/next-greater-element/