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

Comments:

• yanik - 8 years, 10 months ago

  Remove from array and pop in at the end...

  $arr = [1,3,9,0,5,4,0,0,6,6,0,2,4,1,0,4,9,7,0,0,1,2,0]
  $zero_bucket = []
  
  $arr.delete_if do |x|
    if x == 0
      $zero_bucket << x
      true
    end
  end
  
  $zero_bucket.each {|x| $arr << x}
  
  p $arr
  

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• Anonymous - 8 years, 10 months ago

  #python
  def movezeros(za):
      for item in za:
          if item == 0:
              za.remove(item)
              za.append(item)
      return za
  

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• Daniel - 8 years, 9 months ago

  You can do it in-place in O(n+k) where k is the length of the largest consecutive run of zeros.

  int[] arr = {1,3,9,0,5,4,0,0,6,6,0,2,4,1,0,4,9,7,0,0,1,2,0};
  
  int c = 0;
  for(int i = 0; i < arr.length; i++) {
      while(i+c < arr.length && arr[i+c] == 0)
          c++;
      if(c == 0) continue;
  
      while(i+c < arr.length && arr[i+c] == 0)
          c++;
  
      if(i+c >= arr.length)
          arr[i] = 0;
      else
          arr[i] = arr[i + c];
  }
  

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• Daniel - 8 years, 9 months ago

  Sorry I meant to say O(n*k) not O(n+k)

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