Heap / Priority Queue — Reasoning Check

1. What is the third-largest element of [9,4,9,2,7], counting duplicate occurrences as separate ranks? Enter one integer.

2. A frequency table is {10:4, 20:2, 30:2, 40:1}, and k=2. Which acceptance rule correctly handles the cutoff tie?

Select one answer.

Require all of 10,20,30, even though k is 2. Require 10 and exactly one of 20 or 30, in either output order. Accept any two distinct values, because ties make all selections valid. Require [10,20] exactly, because the numerically smaller tied value must win.

3. KSmallestPairs receives A=[1,1], B=[0,0,2], and k=6. A solution deduplicates output pairs by their displayed values and returns only [1,0] and [1,2]. What is wrong?

Select one answer.

Only two input values are distinct, so the two-row result is required. The contract counts index pairs: four [1,0] rows and two [1,2] rows must be retained. The result must contain six different displayed value pairs. Identical pair values should be combined by adding their sums together.

4. For sorted A=[-5,-1,4,8], sorted B=[-2,3], and k=2, the pair-sum heap seeds only rows i=0 and i=1. Why may rows i>=2 be omitted?

Select one answer.

Every later pair in a seeded row is smaller than every pair in an omitted row. Two seeded index pairs already have sums no larger than any pair in an omitted row. Omitted rows must contain duplicate values that can be discarded. Each array contributes only its first two elements to every possible answer.

5. A MedianFinder receives 8, then 2, then 5, then 11. What median should it return after the fourth insertion? Enter the numeric value.

6. IPO starts with capital w=1 and permits at most three projects. The (required capital, profit) pairs are (1,2), (3,-5), and (4,4). What should the optimal process do?

Select one answer.

Take (1,2), reach capital 3, then stop because the best remaining affordable profit is negative. Take (1,2) and then (3,-5), because the algorithm must use every allowed project slot. Take (1,2) and then (4,4), reaching 7 even though the second project is not affordable. Take no projects because it is impossible to complete all three.