Conveyor Batch Sequence
In Conveyor Batch Sequence, you are given a list of integers and need to return those same values in ascending order.
This is a straightforward sorting task. The important thing is to keep all original values. If a number appears more than once, both copies should still be there after sorting. Negative values also count as normal numbers, so they should appear before zero and positive values when they are smaller.
For example, nums = [14,5,12,5] should return [5,5,12,14]. Both copies of 5 stay in the list, just moved into sorted positions. Another example is nums = [10,-5,10], which should become [-5,10,10]. That shows both how negatives are handled and how duplicates are preserved.
If the array is empty, return an empty array. If it has one value, return that same one value. So the task is simply to sort the numbers numerically in ascending order and return the result.
Example Input & Output
The negative value moves to the front because it is the smallest number.
Both copies of 5 remain while the list is reordered from low to high.
An empty input stays empty because there is nothing to reorder.
Algorithm Flow
Solution Approach
The cleanest solution is to apply an ascending numeric sort and return the sorted array. Since the problem only asks for the values in low-to-high order, a normal sort already captures the full requirement.
The main implementation detail is making sure the comparison is numeric rather than string-based. In JavaScript, for example, the default sort() can compare values as text. That would produce incorrect results such as putting 12 before 5. To avoid that, you should pass a comparator that compares numbers by their actual value.
A simple version looks like this:
The comparator a - b is enough. If a is smaller, the result is negative, so a is placed before b. If the values are equal, both remain in the array, which is exactly what we want for repeated weights like [5, 5].
This method handles every common case in one consistent rule. Empty arrays stay empty. A single-element array remains unchanged. Negative values move to the front when they are smaller than the rest. Repeated numbers stay repeated, because sorting only reorders elements and does not drop any of them.
Using the built-in sort also keeps the code easy to read. You do not need nested loops, counters, or special-case logic. For an easy sorting problem like this, the best answer is usually the one that states the sorting rule as directly as possible.
The time complexity is O(n log n), which is the standard cost of comparison sorting. The space complexity depends on the language's implementation. If you want to avoid mutating the original array, you can make a copy before sorting, but the algorithmic idea stays the same.
So the whole strategy is: take the batch weights, sort them numerically in ascending order with an explicit comparator, and return the sorted list with every original value still included.
Best Answers
import java.util.*;
class Solution {
public int[] conveyor_batch_sequence(int[] nums) {
int[] result = nums.clone();
Arrays.sort(result);
return result;
}
}Comments (0)
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