The 420 Exam Marks Before the Deadline

The suggestion contains a conceptual error. Once a list is sorted, the median is found by position, not by searching for a value. For an odd number of valid marks, the median is the middle element; for an even number, it is the average of the two middle elements. Binary search is for locating a known target in sorted data, so using it to 'find the median' is unnecessary. The teacher should simply index the sorted list.

Sorting 420 items is inexpensive. A comparison sort such as merge sort does on the order of n log n comparisons, roughly 420 × 9 ≈ 3,800, which is trivial. The real issue is handling 'ABS'. If 'ABS' is left in the list as text, sorting may place it inconsistently or fail in a numeric sort, and including it as zero would distort the median. The teacher must filter absent students out of the median calculation while keeping them in the rank order with a dash. A separate list of valid numeric marks is the simplest approach.

The colleague's method is therefore not harmful at this size, but it is logically wrong and adds a step. The correct reasoning is: filter, sort, index the middle. At 420 records, clarity and correctness matter more than algorithmic sophistication.

One suitable algorithm is:

valid = []
counts = empty dictionary
duplicates = empty set
for each record in the 420 records:
    if record.mark == 'ABS':
        continue
    valid.append(record.mark)
    if record.mark in counts:
        duplicates.add(record.mark)
    counts[record.mark] = counts.get(record.mark, 0) + 1
sort valid
n = length(valid)
if n is odd:
    median = valid[n // 2]
else:
    median = (valid[n // 2 - 1] + valid[n // 2]) / 2

'ABS' is skipped rather than converted to 0, because converting it would pull the median downwards and misrepresent the cohort. The absent students remain in the separate rank-order output with a dash. Duplicate detection uses a dictionary, so each mark is checked in average constant time and the whole pass is O(n). At 420 records this is trivial and avoids a second sort. An alternative is to sort the valid marks and compare adjacent values, which is also acceptable at this size. The dictionary approach is chosen because it detects duplicates during the same pass that builds the median list, reducing the chance of inconsistency before the deadline.

Linear search checks names one by one. With 420 records, the average successful lookup examines about 210 names and the worst case 420. For a handful of lookups this is negligible. It also works on the data exactly as entered and carries no risk of separating names from marks.

Sorting by name and using binary search costs O(n log n) once, roughly 3,800 comparisons, and then about 9 comparisons per lookup. The break-even point is several hundred lookups; below that, the sort dominates. Sorting also introduces a practical risk in a spreadsheet: if only the name column is sorted, marks become attached to the wrong students. Any sort must move whole records together. 'ABS' entries also need to remain aligned.

For this teacher, with 15 minutes and a one-off results task, linear search or a spreadsheet filter/lookup is the better recommendation. It is adequate at n = 420, easier to verify, and less error-prone under time pressure. Sorting by name is justified only if the teacher expects to perform many repeated lookups, for example when entering late results over several days. The decision should be based on the number of expected searches, not on the theoretical superiority of binary search.