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  1. Practice
  2. Python
  3. Basic Data Types
  4. Find the Runner-Up Score!
  5. Discussions

Find the Runner-Up Score!

  • bicole + 2 comments

    gtorrecilla. What do you think of this? Its not the best by line count, but I believe it finds the 2nd largest within O(n).

    max = max2 = -100 - int(raw_input())
nums = map(int,raw_input().split(' '))
for i in range(len(nums)):
    if nums[i] > max2:
        if nums[i] > max:
            max,max2 = nums[i],max
        elif nums[i] < max:
            max2 = nums[i]
print(max2)
    • omnius42 + 0 comments

      You're doing a bit more work than you need to. Instead of iterating over a range, just iterate over the number list.

      raw_input()
max = max2 = -101
for n in map(int,raw_input().split()):
    if n > max2:
        if n > max:
            max,max2 = n,max
        elif n < max:
            max2 = n
print(max2)
    • evandam92 + 1 comment

      I think this is the best approach too to stay within O(n). My solution was something similar:

      n = int(input())
a = [int(x) for x in input().split()]
largest = secondlargest = -100
for x in a:
    if x > largest:
        tmp = largest
        largest = x
        secondlargest = tmp
    elif x > secondlargest and x != largest:
        secondlargest = x
print(secondlargest)
      • somewhatabstract + 1 comment

        You don't need that tmp var, just reorder your first condition:

        n = int(input())
a = [int(x) for x in input().split()]
largest = secondlargest = -100
for x in a:
    if x > largest:
        secondlargest = largest
        largest = x
    elif x > secondlargest and x != largest:
        secondlargest = x
print(secondlargest)
        • youngshawzju + 0 comments

          works!but a little improvement is needed