Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Adam - 10 years, 3 months ago

  expected_sum = (1 + 1000000) * (500000) array_sum = sum(xs) missing = expected_sum - array_sum

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• Max Burstein - 10 years, 3 months ago

  I like this approach. Good stuff

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• Karl - 10 years, 3 months ago

  Very nice for this problem yes (elegant, even), but what about two missing numbers?

  The codified solutions could be easily modified to find both.

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• Foppe - 10 years, 3 months ago

  #! /usr/bin/pyton
  
  import random
  
  # Sum of integers 1 ... 1,000,000
  sum_of_all_integers = (1000000 + 1) * (1000000 / 2)
  
  # Create array with one item missing
  array_of_integers = []
  for i in range(1, 1000001):
      array_of_integers.append(i)
  random_number_missing = random.randint(1,1000000)
  array_of_integers.remove(random_number_missing)
  random.shuffle(array_of_integers)
  
  # Sum elements in array
  sum_of_integers = 0
  for i in array_of_integers:
      sum_of_integers += i
  
  # Subtract sum of array with missing number from
  # sum of array wth all numbers
  missing_number = sum_of_all_integers - sum_of_integers
  
  # Test
  print("Expected: %d" % random_number_missing)
  print("Calculated: %d" % missing_number)
  

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• Beginner - 10 years, 3 months ago

  Would anyone care to explain what linear run time is? All the websites I've looked at assume I know too much.

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• Foppe - 10 years, 3 months ago

  Big Oh notation states "how well an algorithm scales as data(ponts) increases". Linear being almost optimal as run-time means you run only one time over the data as shown in my solution above. The array is traversed only once for adding all integers. A less optimal solution (n log n) would start with sorting the array then picking the missing number. You'd need to traverse the array more than once to sort it en then traverse it again to find the missing integer. Optimisation of algorithms is important when dealing with lhuge data-sets. One million integers is relatively not large.

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• Anonymous - 10 years, 3 months ago

  Basically only looking at each number in the array once

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• wobblebucket - 10 years, 3 months ago

  import random
  total = 499999500000L
  mystery = range(1, 1000000)
  mystery.remove(random.randint(1,1000000))
  print total - sum(mystery)
  

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