Second Largest Number in a List in Python

Problem statement

Approach 1: simple loop

Remove the maximum, then take the maximum of what remains — the two-step version that reads exactly like the problem statement.

nums = [12, 45, 2, 41]

copy = nums[:]          # keep the original untouched
copy.remove(max(copy))  # drop ONE occurrence of the largest

print(max(copy))
# Output: 41

remove() deletes only the first matching occurrence, so duplicates of the maximum survive — which is what the "duplicates count" reading wants. Work on a copy so the original list is intact for later use.

Approach 2: reusable function

For the distinct reading, collapse duplicates with a set first, sort, and pick the second from the end — the version most exams and interviews mean.

nums = [12, 45, 45, 2]

unique = sorted(set(nums))
print(unique[-2])
# Output: 12

Build it defensively and the edge cases say their names: a list with fewer than two distinct values has no second largest at all.

def second_largest(nums):
    unique = sorted(set(nums))
    if len(unique) < 2:
        return None          # explicitly: there is no answer
    return unique[-2]

Approach 3: Pythonic alternative

One pass, two variables: walk the list once, tracking the largest and second largest seen so far. This is the version that scales — it also works streaming over data you cannot sort.

nums = [12, 45, 2, 41]

first = second = float("-inf")
for n in nums:
    if n > first:
        second = first     # old champion demoted
        first = n
    elif first > n > second:
        second = n

print(second)
# Output: 41

Starting both at negative infinity means any real number beats them — even a list of all-negative numbers works. Note the chained comparison first > n > second, which skips duplicates of the champion.

Dry run

Trace the one-pass version over [12, 45, 2, 41]. Start: first = -inf, second = -inf. n=12: 12 > -inf → second stays -inf, first = 12. n=45: 45 > 12 → second = 12, first = 45. n=2: not > 45; is 45 > 2 > 12? No — 2 does not beat 12, skip. n=41: not > 45; is 45 > 41 > 12? Yes → second = 41. Answer: 41. Every number had exactly one comparison path — that is what "one pass" buys.

Common errors

• Sorting the original list in place (nums.sort()) — the surrounding program may still need the original order; sort a copy or use sorted().
• Picking sorted(nums)[-2] when duplicates of the maximum exist — with [12, 45, 45, 2] that returns 45, not the second distinct value. Decide which definition you need.
• Initialising second largest to 0 — breaks on all-negative lists; use float("-inf") or the first two elements.
• Forgetting the too-short list: [7] or [7, 7] (distinct reading) has no second largest — return None instead of crashing.
• Demoting without saving — second = first must happen before first = n, or the old champion is lost.

Practice on PyDebug

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

• Range boundary practice
• Operator precedence practice
• Variables and assignment practice