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import pandas as pd
import pymysql
from sqlalchemy import create_engine

engine = create_engine('mysql+pymysql://<user>:<password>@<host>[:<port>]/<dbname>')

df = pd.read_sql_query('SELECT * FROM table', engine)
df.head()

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# Print all possible n pairs of balanced parentheses.
# For example, when n is 2, the function should print “(())” and “()()”. 
# When n is 3, we should get “((()))”, “(()())”, “(())()”, “()(())”, “()()()”.

def fn(l,r):
  #print ('call : fn(' +  str(l) + ',' + str(r) + ')')
  if l < 0 or r < 0 or l > r:
    return 0
  if l == 0 and r == 0:

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class node:
  def __init__(self, data):
    print "creating node: ",  data
    self.data = data
    self.left = None
    self.right = None
  
  def insert(self, data):
    if self.data > data:

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# 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

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# Given a string, find the longest substring without repeating characters.
from collections import defaultdict

def solution(s):
  if not s:
    return None
  
  tmp = ""
  longest = ""
  dict = defaultdict(int)

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# Problem Statement : Given a cost matrix Cost[][] where Cost[i][j] denotes the Cost 
# of visiting cell with coordinates (i,j), find a min-cost path to reach a cell (x,y)
# from cell (0,0) under the condition that you can only travel one step right or one
# step down. (We assume that all costs are positive integers)


def paths(Cost, row, col, dr, dc, memo):

  if row == dr and col == dc:
    return 1

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def maxprofit(price, year, memo):
  print price, year
  profit = 0
  if len(price) == 1:
    return price[0] * year
    
  key = '-'.join([str(p) for p in price]) + '--' +str(year)
  if key in memo:

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def caterpillarMethod(A, s):
  n = len(A)
  front, total = 0, 0
  
  for back in xrange(n):
    print "back: ",back
    while (front < n and total + A[front] <= s):
      total += A[front]
      front += 1
      print "total:", total, "back:", back, "front:",front

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'''
Problem: 
You are given a non-empty, zero-indexed array A of n (1 ¬ n ¬ 100 000) integers a0, a1, . . . , an−1 (0 ¬ ai ¬ 1 000). This array represents number of mushrooms growing on the consecutive spots along a road. You are also given integers k and m (0 ¬ k, m < n).

A mushroom picker is at spot number k on the road and should perform m moves. In
one move she moves to an adjacent spot. She collects all the mushrooms growing on  spots she visits. The goal is to calculate the maximum number of mushrooms that the mushroom picker can collect in m moves.
'''

import math

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from functools import reduce

def golden_max_slice(A):
  max_ending = max_slice = 0
  
  if all(a > 0 for a in A):
    return reduce((lambda x, y: x + y), A)
  if all(a < 0 for a in A):
    return max(A)