In a heap with n elements with the smallest element at the root, the 7th…

Key idea: explore only the nodes near the root using a small auxiliary min-heap of candidate nodes.

• Initialize an auxiliary min-heap and insert the root of the given min-heap (the smallest element).

• Repeat the following k−1 times (here k = 7, so repeat 6 times): extract the minimum element from the auxiliary heap and insert its children from the original heap (if they exist) into the auxiliary heap.

• After these extractions, the minimum element remaining at the root of the auxiliary heap is the k-th (7th) smallest element.

Time complexity: each heap operation on the auxiliary heap costs O(log k), and we do O(k) operations overall, so the total time is O(k log k).

Conclusion: for k = 7 this is O(7 log 7), which is Θ(1) with respect to n, so the 7th smallest element can be found in Θ(1) time (treating k as a constant).