BinaryHeap is a max-heap, and Reverse is the fix

Implement two priority-queue operations:

pub fn k_smallest(nums: Vec<i64>, k: usize) -> Vec<i64>
pub fn merge_sorted(lists: Vec<Vec<i64>>) -> Vec<i64>

k_smallest returns the smallest min(k, nums.len()) values in ascending order. Count duplicates separately. Return an empty vector for zero k or empty input. A large k does not require storage for nonexistent elements.

merge_sorted combines lists that are individually in nondecreasing order. Return every value, including duplicates, in nondecreasing order. Empty inner lists and an empty outer list are valid. Inputs and the resulting output fit in memory.

The starter compiles and passes Clippy, but chooses the wrong heap ordering. Fix both functions; the grader calls both through its supplied harness.

Choose which value should come out first

BinaryHeap<T> returns the greatest value according to T's ordering. Wrapping values in Reverse reverses that ordering, making the smallest original value come out first:

use std::cmp::Reverse;
use std::collections::BinaryHeap;

fn main() {
    let mut largest_first = BinaryHeap::from([3, 9, 1]);
    assert_eq!(largest_first.pop(), Some(9));

    let mut smallest_first = BinaryHeap::from([Reverse(3), Reverse(9), Reverse(1)]);
    assert_eq!(smallest_first.pop(), Some(Reverse(1)));
}

Reverse<T> is a transparent wrapper around T. It changes comparison behavior without adding a heap allocation of its own. This does not make every heap operation free: comparisons, moving elements and growing storage still have costs.

Unwrap a popped value with a pattern such as Some(Reverse(value)). Negating integers is not an equivalent general solution: negating i64::MIN cannot produce a representable positive i64.

Keep the smallest values in a bounded max-heap

For k_smallest, a max-heap is useful because its root is the largest candidate you currently hold. That is the candidate to discard when a smaller value arrives.

Set the effective limit to min(k, nums.len()), and return early if it is zero. Push each input value. Whenever the heap exceeds that limit, pop its largest value. After each iteration, it contains the smallest values seen so far, up to the limit. Finish by returning those values in sorted order.

For [5, 1, 4, 2, 3] and k = 2, the final candidates are 1 and 2. Their order inside the heap is not the required output order.

Do not reserve k + 1 elements directly. k can be much larger than the input length, and adding one can overflow. The reference starts with an empty heap and lets it grow with the candidates actually encountered.

Let n be the input length and m = min(k, n). For positive m, the push-then-evict method takes O(n log(m + 1)) time in the worst case; sorting the retained values adds O(m log(m + 1)). Its heap storage is O(m), with room for one temporary extra candidate. This is additional storage: the function already receives an allocated input vector.

Sorting the owned input is also a valid way to meet the output contract. A bounded heap is especially useful when retaining only a small fraction of the input or adapting the algorithm to a stream. Which approach is faster depends on the workload; k == n does not imply a universal timing result.

Heap iteration is not sorted output

into_sorted_vec() returns ascending order according to the heap's element type. into_vec() returns its storage without sorting it. Iteration also does not promise sorted order. It is fine to iterate for an order-independent calculation or diagnostic display; do not treat that sequence as sorted.

use std::cmp::Reverse;
use std::collections::BinaryHeap;

fn main() {
    let heap = BinaryHeap::from([5, 1, 8, 3]);
    assert_eq!(heap.peek(), Some(&8));
    assert_eq!(heap.into_sorted_vec(), vec![1, 3, 5, 8]);

    let reversed = BinaryHeap::from([Reverse(1), Reverse(3), Reverse(2)]);
    let values: Vec<_> = reversed.into_sorted_vec().into_iter()
        .map(|Reverse(value)| value).collect();
    assert_eq!(values, vec![3, 2, 1]);
}

Notice the final result: sorting Reverse values in their ascending order gives descending order for the original numbers. Repeatedly popping that same min-heap would instead yield the original numbers in ascending order.

Merge using one candidate per list

For merge_sorted, put the first value of each nonempty list into a min-heap. Store Reverse((value, list_index, position)) so each entry also identifies where to find its successor.

Repeatedly pop the smallest candidate, append its value to the output, and insert the next value from the same list if one exists. Because each input list is sorted, no unseen value in that list belongs before its current candidate.

Tuples compare lexicographically. The value is the primary key; list index and position break ties. This keeps duplicate values while making candidate selection well-defined. The output contains values only, so their source identity is not part of the result.

With r input lists, q nonempty lists and N total values, this approach takes O(r + N log(q + 1)) time and O(q) candidate storage. The output needs O(N) space, and this function also retains the provided input lists during the merge.

Keep custom ordering consistent

If you implement ordering for your own type, Ord, PartialOrd and equality must agree. In particular, partial_cmp(a, b) must return Some(a.cmp(b)). Deriving one comparison trait while manually implementing the other can make them inconsistent; Clippy catches that pattern, but cannot prove every custom ordering correct.

Violating ordering requirements is a logic error with unspecified behavior, not a promise of one particular wrong result. Do not change an element's relative ordering through interior mutability while it is in a heap. See the BinaryHeap contract and Ord requirements.