3sum problem created by vetional - https://repl.it/MS4f/0

main.py

# Determine if any 3 integers in an array sum to 0.

# For example, for array [4, 3, -1, 2, -2, 10], the function should return true since 3 + (-1) + (-2) = 0. To make things simple, each number can be used at most once.

def solution(A):
  A.sort() # O(nlogn)
  print A
  for i, x in enumerate(A): # O(n**2)
    
    start = 0
    end = len(A) - 1
    
    while start < end:
      # make sure the start or end 
      # is not the currrent number
      if start == i:
          start += 1
      if end == i:
          end -= 1
          
      s = A[start] + A[end]
      print start, end, s, -1 * x
      
      if s == -1 * x:
        print "found: ", A[start] , A[end], x
        return True
      
      if s >  -1 * x:
        end -=  1
        
      elif s <  -1 * x:
        start += 1
  
  return False
  
A = [4, 3, -1, 2, -2, 10]
print solution(A)