Largest of Three Numbers in Python

Problem statement

Given three numbers, print the largest. Equal values may appear — any of the tied largest values is a correct answer.

Approach 1: simple loop

Compare with an if/elif ladder. Each condition asks "is this number at least as large as both others?" and the first true branch wins.

a, b, c = 12, 45, 23

if a >= b and a >= c:
    largest = a
elif b >= a and b >= c:
    largest = b
else:
    largest = c

print(largest)
# Output: 45

Note the >=: with ties like 7, 7, 3, the first branch still fires and the answer is correct.

Approach 2: reusable function

Keep a running champion: start with the first number, replace it whenever a later number beats it. This shape scales to ten numbers without changing its logic.

def largest_of(numbers):
    best = numbers[0]
    for n in numbers[1:]:
        if n > best:
            best = n
    return best

print(largest_of([12, 45, 23]))       # 45
print(largest_of([5, 5, 5]))          # 5
print(largest_of([-2, -9, -4]))       # -2

Only two ideas: remember the best so far, and update on a strict win. The negative-numbers case works because every comparison is relative.

Approach 3: Pythonic alternative

Python has a built-in for "largest": max(). It accepts any number of arguments, or a single list, and even a key function for custom ordering.

a, b, c = 12, 45, 23

print(max(a, b, c))          # 45
print(max([12, 45, 23]))     # 45 — same thing via a list

For exams, write the ladder; for code you will keep, max() says exactly what it means.

Dry run

Trace the champion version with [12, 45, 23]. Start: best = 12. Compare 45 > 12 → yes, best = 45. Compare 23 > 45 → no, best stays 45. Answer: 45 — two comparisons, exactly what is needed for three numbers. For the ladder version, the first test 12 >= 45 fails immediately, the second 45 >= 12 and 45 >= 23 passes, and b is printed.

Common errors

Writing a > b or a > c — "or" makes the condition far too easy to satisfy; a number larger than only one of the others slips through. Use and.
Using = in the comparison instead of ==/>=.
Assuming the numbers are positive — starting "largest" at 0 breaks on all-negative input; start with the first number instead.
Forgetting ties: strict > everywhere can leave the answer unset in the ladder version.
Printing all three branches' results — only one comparison chain should win.

Practice on PyDebug

Turn the idea into a debugging habit with these free browser exercises: