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

Given an array of unsorted integers, return the largest 3 integers in the array. Can you solve this in linear runtime?

Comments:

• mark - 10 years, 3 months ago

  var unsortedIntArray = new[] { 2, 3, 77, 33, 12, 75, 2, 9, 34, 59, 21, 17 };

  var top3 = unsortedIntArray.OrderByDescending(x => x).Take(3);

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• wobblebucket - 10 years, 3 months ago

  def top(unsorted, num):
      r = []
      for x in range(num):
          top_idx = 0
          for idx, i in enumerate(unsorted):    
              if (i > unsorted[top_idx]):
                  top_idx = idx
  
          r.append(unsorted[top_idx])
          unsorted.pop(top_idx)
      return r
  
  print top([5,9,2,7,6,4,1,3,8], 3)
  

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• asheehan - 10 years, 3 months ago

  in Ruby

  unsorted_array = 100.times.map{ Random.rand(100) }
  top_array = [0, 0, 0]
  unsorted_array.each do |x|
    for y in 0..2 do
      if top_array[y] < x
        top_array[y] = x
        top_array.sort!
        break
      end
    end
  end
  puts top_array
  

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• person - 10 years, 3 months ago

  Erlang:

  -module(top3).
  -export([find_top3/1]).
  
  find_top3(L) ->
      find_top3(L, {0,0,0}).
  find_top3([], {Max, Med, Min})->
      {Max, Med, Min};
  find_top3([H|T], {Max, Med, Min}) when H >= Max ->
      find_top3(T, {H, Max, Med});
  find_top3([H|T], {Max, Med, Min}) when H =< Min ->
      find_top3(T, {Max, Med, Min});
  find_top3([H|T], {Max, Med, Min}) when H >= Med ->
      find_top3(T, {Max, H, Med});
  find_top3([H|T], {Max, Med, Min}) when H < Med ->
      find_top3(T, {Max, Med, H}).
  

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• Foppe - 10 years, 3 months ago

  #! /usr/bin/python
  
  ints = [8, 445, 74, 85, 65, 90, 5, 102, 105, 558, 54, 2, 98, 26, 75]
  
  no1 = 0
  no2 = 0
  no3 = 0
  for i in ints:
      if i < no3:
          continue
      elif i > no1:
          no3 = no2
          no2 = no1
          no1 = i
          continue
      elif i > no2:
          no3 = no2
          no2 = i
          continue
      else:
          no3 = i
  
  print("%d %d %d" % (no1, no2, no3))
  

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