Shortening Solutions for USACO

Problems and Contests
#1

Hi, I was working on some USACO Bronze Problems in Python and realized that my solution was way too slow. How could I shorten my solution and still solve the problem at the same time?
Problem - http://www.usaco.org/index.php?page=viewproblem2&cpid=1107
My Solution -
https://drive.google.com/file/d/1LDu3qTI5iu7gG-zfgEE0fuzMXs65vpzm/view?usp=sharing
N = input()
N = int(N)
nums = []
Person1 = []
Person2 = []
Person3 = []
for i in range(0,N,1):
nums.append(input())

def Determine_Names(a):
Names = []
Name = ‘’
for i in range(0,len(a), 1):
if (a[i] == ’ '):
return Name.join(Names)
else:
Names.append(a[i])
def Determine_Year(a):
Years = []
Year = ‘’
x = 0
for i in range(0, len(a), 1):
if (a[i] == ’ '):
x = x+1
if (x == 4):
i = i+1
for i in range(i, len(a), 1):
if (a[i] == ’ '):
return Year.join(Years)
else:
Years.append(a[i])
def Determine_Other(a):
Names = []
Name = ‘’
for i in range(-1,-(len(a)),-1):
if (a[i] == ’ '):
Names.reverse()
return Name.join(Names)
else:
Names.append(a[i])

def Determine_Prefix(a):
Prefixes = []
Prefix = ‘’
x = 0
for i in range(0, len(a), 1):
if (a[i] == ’ '):
x = x+1
if (x == 3):
i = i+1
for i in range(i, len(a), 1):
if (a[i] == ’ '):
return Prefix.join(Prefixes)
else:
Prefixes.append(a[i])
def Years_Apart(a, b, c):
if (c == ‘previous’):
Years = a - b + 12
if (c == ‘next’):
Years = (b - a + 12)*-1

return Years

Person1.append(Determine_Names(nums[0]))
Person1.append(Determine_Prefix(nums[0]))
Person1.append(Determine_Other(nums[0]))
Person1.append(Determine_Year(nums[0]))

Person2.append(Determine_Names(nums[1]))
Person2.append(Determine_Prefix(nums[1]))
Person2.append(Determine_Other(nums[1]))
Person2.append(Determine_Year(nums[1]))

Person3.append(Determine_Names(nums[2]))
Person3.append(Determine_Prefix(nums[2]))
Person3.append(Determine_Other(nums[2]))
Person3.append(Determine_Year(nums[2]))
def Year_Number(a):
if (a == ‘Ox’):
return 1
if (a == ‘Dragon’):
return 4
if (a == ‘Tiger’):
return 2
if (a == ‘Rabbit’):
return 3
if (a == ‘Snake’):
return 5
if (a == ‘Horse’):
return 6
if (a == ‘Goat’):
return 7
if (a == ‘Monkey’):
return 8
if (a == ‘Rooster’):
return 9
if (a == ‘Dog’):
return 10
if (a == ‘Pig’):
return 11
if (a == ‘Rat’):
return 12
def Find_Year(a):
if (a == 1):
return ‘Ox’
if (a == 2):
return ‘Tiger’
if (a == 3):
return ‘Rabbit’
if (a == 4):
return ‘Dragon’
if (a == 5):
return ‘Snake’
if (a == 6):
return ‘Horse’
if (a == 7):
return ‘Goat’
if (a == 8):
return ‘Monkey’
if (a == 9):
return ‘Rooster’
if (a == 10):
return ‘Dog’
if (a == 11):
return ‘Pig’
if (a == 12):
return ‘Rat’
def Find_Birth(a,b,c,d):
if (b[0] == d):
return b[3]
if (c[0] == d):
return c[3]
if (a[0] == d):
return a[3]

Person1.append(Years_Apart(1, Year_Number(Person1[3]), Person1[1]))
Person2.append(Years_Apart(Year_Number(Find_Birth(Person1, Person2, Person3, Person2[2])), Year_Number(Person2[3]), Person2[1]))
Person3.append(Years_Apart(Year_Number(Find_Birth(Person1, Person2, Person3, Person3[2])), Year_Number(Person3[3]), Person3[1]))

def Find_People(a,b,c,d):
if (b[0] == d):
return [b[2], b[4]]
if (c[0] == d):
return [c[2], c[4]]
if (a[0] == d):
return [a[2], a[4]]
Sum = 0
for i in range(0, N, 1):
if (i == 0):
Place_Hldr = Find_People(Person1, Person2, Person3, ‘Elsie’)
Sum = Sum + int(Place_Hldr[1])
if (Place_Hldr[0] == ‘Bessie’):
print(Sum)
break
continue
Place_Hldr = Find_People(Person1, Person2, Person3, Place_Hldr[0])
Sum = Sum + int(Place_Hldr[1])
if (Place_Hldr[0] == ‘Bessie’):
print(Sum)
break

#2

Well, instead of this, you can use a dictionary and that will shorten the code by a lot. This function basically does the same thing as a dictionary.

#3

Please format your code. It’s extremely unreadable like this.

#4

Sorry, I copy and pasted from IDLE. There’s a separate file on the google drive link which is formatted and on python 3.6.9

#5

@77Panda When possible, use lists and dictionaries instead of repeating yourself. There are also not necessarily only 3 different people. You can try to emulate the analysis at http://www.usaco.org/current/data/sol_prob1_bronze_feb21.html It’s in C++, but try to convert it into python.

Your code is also not particularly pythonic, string.split() and list comprehensions can help you reduce some of your for loops.

Code in python (try doing the above before looking): https://ideone.com/WKVp9p

#6

Please read the pinned post about formatting code.