Working Examples by Hand and Finding Edge Cases

18 min · Intermediate

A solution that works on the sample can still fail on a hidden test. The fix is a habit: trace a small example by hand, then hunt the edge cases on purpose.

Trace one example before you code

Dry run is working through code or an idea by hand on a small input, writing down every variable's value after each step, to see what it really does.

Take the smallest example that still shows the rule. Follow it step by step on paper and write each variable's value. The trace becomes the loop you will write.

Python

1sales = [5, 3, 8, 8, 2, 9]2best = sales[0]3for day, amount in enumerate(sales):4    if amount > best:5        best = amount6    print(day, amount, best)

Example 1 · A hand trace turned into prints

Python

1marks = [40, 75, 75, 60]2passed = 03for mark in marks:4    if mark >= 50:5        passed += 16    print(mark, passed)

Output

What this code does

Each printed line matches one row you would write on paper.
The count moves only when the mark is 50 or more, so 40 leaves it at 0.

Each line is one row of the trace table. You can see the highest sale so far change only on days 2 and 5.

Try the small and strange inputs

Edge case is an input at the limit of what the problem allows, such as an empty list, one item or all equal values, where code most often breaks.

Bugs hide at the edges: an empty input, one item, all items equal, negative numbers, the largest size allowed. Write these cases down before the code and run each one.

Python

1def average(values):2    if not values:3        return 0.04    return sum(values) / len(values)5
6
7for case in ([], [5], [2, 2, 2], [-3, 3]):8    print(case, average(case))

Example 2 · max() of an empty list

Python

1orders = []2try:3    print(max(orders))4except ValueError as error:5    print("ValueError:", error)6print(max(orders, default=0))

Output

ValueError: max() iterable argument is empty
0

What this code does

An empty input is the classic edge case: max has nothing to compare and raises ValueError.
default=0 gives a safe answer when the list may be empty.

Without the first check, the empty list would divide by zero. The edge-case list found that before a judge did.

Write the checks first with assert

assert statement stops the program with AssertionError when its condition is false. It is a one-line test for a case whose answer you already know.

Once you know the answers for your cases, turn them into assert lines. If the code is wrong, the program stops at the first failing line.

Python

1def second_largest(values):2    distinct = sorted(set(values))3    return distinct[-2] if len(distinct) >= 2 else None4
5
6assert second_largest([4, 9, 7]) == 77assert second_largest([9, 9, 4]) == 48assert second_largest([5]) is None9assert second_largest([]) is None10print("all checks passed")

Example 3 · Second largest with repeats

Python

values = [9, 9, 4]print(sorted(values)[-2])print(sorted(set(values))[-2])

Output

9
4

What this code does

The first line returns 9 because the top value appears twice. It is the wrong answer for "second largest value".
A set first removes the repeats and gives 4.

Example 4 · Starting the best value at zero

Python

1changes = [-5, -2, -9]2wrong = 03right = changes[0]4for change in changes:5    wrong = max(wrong, change)6    right = max(right, change)7print(wrong, right)

Output

0 -2

What this code does

Starting at 0 hides the fact that every value is negative, so wrong stays 0 although no change is 0.
Starting from the first item always works when the list is not empty.

The set removes repeats, so [9, 9, 4] gives 4, not 9. The one-item and empty cases return None.

Watch the boundaries

Off-by-one mistakes come from ranges and slices. With 6 days and windows of 3 days there are 4 windows, not 3 and not 6. Count by hand, then compare with the code.

Python

days = [3, 1, 4, 1, 5, 9]k = 3for start in range(len(days) - k + 1):    window = days[start:start + k]    print(start, window, sum(window))

Example 5 · Windows longer than the list

Python

days = [3, 1, 4]for k in [2, 3, 5]:    print(k, max(0, len(days) - k + 1))

Output

2 2
3 1
5 0

What this code does

With k = 5 the formula gives -1, which is meaningless. max(0, ...) turns it into 0 windows.
Check the case where the window is bigger than the data.

The number of windows is len(days) - k + 1. The slice days[start:start + k] stops before start + k, so it holds exactly 3 items.

Next step

Check your understanding

10 quick questions

Trace one example before you code

Dry run is working through code or an idea by hand on a small input, writing down every variable's value after each step, to see what it really does.

Take the smallest example that still shows the rule. Follow it step by step on paper and write each variable's value. The trace becomes the loop you will write.

Try the small and strange inputs

Edge case is an input at the limit of what the problem allows, such as an empty list, one item or all equal values, where code most often breaks.

Bugs hide at the edges: an empty input, one item, all items equal, negative numbers, the largest size allowed. Write these cases down before the code and run each one.

1def second_largest(values):2 distinct = sorted(set(values))3 return distinct[-2] if len(distinct) >= 2 else None4 5 6assert second_largest([4, 9, 7]) == 77assert second_largest([9, 9, 4]) == 48assert second_largest([5]) is None9assert second_largest([]) is None10print("all checks passed")