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Modernize Python 2 code to get ready for Python 3

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This commit is contained in:
cclauss
2017-11-25 10:23:50 +01:00
parent a03b2eafc0
commit 4e06949072
95 changed files with 580 additions and 521 deletions
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5
dynamic_programming/coin_change.py
View File

@ -5,6 +5,7 @@ Can you determine number of ways of making change for n units using
the given types of coins?
https://www.hackerrank.com/challenges/coin-change/problem
"""
from __future__ import print_function
def dp_count(S, m, n):
table = [0] * (n + 1)
@ -21,5 +22,5 @@ def dp_count(S, m, n):
return table[n]
if __name__ == '__main__':
print dp_count([1, 2, 3], 3, 4) # answer 4
print dp_count([2, 5, 3, 6], 4, 10) # answer 5
print(dp_count([1, 2, 3], 3, 4)) # answer 4
print(dp_count([2, 5, 3, 6], 4, 10)) # answer 5

13
dynamic_programming/edit_distance.py
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@ -51,11 +51,10 @@ class EditDistance:
return self.__solveDP(len(A)-1, len(B)-1)
if __name__ == '__main__':
import sys
if sys.version_info.major < 3:
input_function = raw_input
else:
input_function = input
try:
raw_input # Python 2
except NameError:
raw_input = input # Python 3
solver = EditDistance()
@ -63,10 +62,10 @@ if __name__ == '__main__':
print()
print("Enter the first string: ", end="")
S1 = input_function()
S1 = raw_input().strip()
print("Enter the second string: ", end="")
S2 = input_function()
S2 = raw_input().strip()
print()
print("The minimum Edit Distance is: %d" % (solver.solve(S1, S2)))

1
dynamic_programming/fastfibonacci.py
View File

@ -2,6 +2,7 @@
This program calculates the nth Fibonacci number in O(log(n)).
It's possible to calculate F(1000000) in less than a second.
"""
from __future__ import print_function
import sys

16
dynamic_programming/fibonacci.py
View File

@ -27,26 +27,22 @@ class Fibonacci:
if __name__ == '__main__':
import sys
print("\n********* Fibonacci Series Using Dynamic Programming ************\n")
# For python 2.x and 3.x compatibility: 3.x has no raw_input builtin
# otherwise 2.x's input builtin function is too "smart"
if sys.version_info.major < 3:
input_function = raw_input
else:
input_function = input
try:
raw_input # Python 2
except NameError:
raw_input = input # Python 3
print("\n Enter the upper limit for the fibonacci sequence: ", end="")
try:
N = eval(input())
N = eval(raw_input().strip())
fib = Fibonacci(N)
print(
"\n********* Enter different values to get the corresponding fibonacci sequence, enter any negative number to exit. ************\n")
while True:
print("Enter value: ", end=" ")
try:
i = eval(input())
i = eval(raw_input().strip())
if i < 0:
print("\n********* Good Bye!! ************\n")
break

9
dynamic_programming/longest_common_subsequence.py
View File

@ -3,6 +3,13 @@ LCS Problem Statement: Given two sequences, find the length of longest subsequen
A subsequence is a sequence that appears in the same relative order, but not necessarily continious.
Example:"abc", "abg" are subsequences of "abcdefgh".
"""
from __future__ import print_function
try:
xrange # Python 2
except NameError:
xrange = range # Python 3
def lcs_dp(x, y):
# find the length of strings
m = len(x)
@ -27,4 +34,4 @@ def lcs_dp(x, y):
if __name__=='__main__':
x = 'AGGTAB'
y = 'GXTXAYB'
print lcs_dp(x, y)
print(lcs_dp(x, y))

1
dynamic_programming/longest_increasing_subsequence.py
View File

@ -7,6 +7,7 @@ The problem is :
Given an ARRAY, to find the longest and increasing sub ARRAY in that given ARRAY and return it.
Example: [10, 22, 9, 33, 21, 50, 41, 60, 80] as input will return [10, 22, 33, 41, 60, 80] as output
'''
from __future__ import print_function
def longestSub(ARRAY): #This function is recursive

3
dynamic_programming/longest_increasing_subsequence_O(nlogn).py
View File

@ -1,3 +1,4 @@
from __future__ import print_function
#############################
# Author: Aravind Kashyap
# File: lis.py
@ -37,4 +38,4 @@ def LongestIncreasingSubsequenceLength(v):
v = [2, 5, 3, 7, 11, 8, 10, 13, 6]
print LongestIncreasingSubsequenceLength(v)
print(LongestIncreasingSubsequenceLength(v))

5
dynamic_programming/longest_sub_array.py
View File

@ -6,6 +6,7 @@ This is a pure Python implementation of Dynamic Programming solution to the long
The problem is :
Given an array, to find the longest and continuous sub array and get the max sum of the sub array in the given array.
'''
from __future__ import print_function
class SubArray:
@ -13,7 +14,7 @@ class SubArray:
def __init__(self, arr):
# we need a list not a string, so do something to change the type
self.array = arr.split(',')
print("the input array is:", self.array)
print(("the input array is:", self.array))
def solve_sub_array(self):
rear = [int(self.array[0])]*len(self.array)
@ -28,5 +29,5 @@ if __name__ == '__main__':
whole_array = input("please input some numbers:")
array = SubArray(whole_array)
re = array.solve_sub_array()
print("the results is:", re)
print(("the results is:", re))

119
dynamic_programming/max_sub_array.py
View File

@ -1,59 +1,60 @@
"""
author : Mayank Kumar Jha (mk9440)
"""
import time
import matplotlib.pyplot as plt
from random import randint
def find_max_sub_array(A,low,high):
if low==high:
return low,high,A[low]
else :
mid=(low+high)//2
left_low,left_high,left_sum=find_max_sub_array(A,low,mid)
right_low,right_high,right_sum=find_max_sub_array(A,mid+1,high)
cross_left,cross_right,cross_sum=find_max_cross_sum(A,low,mid,high)
if left_sum>=right_sum and left_sum>=cross_sum:
return left_low,left_high,left_sum
elif right_sum>=left_sum and right_sum>=cross_sum :
return right_low,right_high,right_sum
else:
return cross_left,cross_right,cross_sum
def find_max_cross_sum(A,low,mid,high):
left_sum,max_left=-999999999,-1
right_sum,max_right=-999999999,-1
summ=0
for i in range(mid,low-1,-1):
summ+=A[i]
if summ > left_sum:
left_sum=summ
max_left=i
summ=0
for i in range(mid+1,high+1):
summ+=A[i]
if summ > right_sum:
right_sum=summ
max_right=i
return max_left,max_right,(left_sum+right_sum)
if __name__=='__main__':
inputs=[10,100,1000,10000,50000,100000,200000,300000,400000,500000]
tim=[]
for i in inputs:
li=[randint(1,i) for j in range(i)]
strt=time.time()
(find_max_sub_array(li,0,len(li)-1))
end=time.time()
tim.append(end-strt)
print("No of Inputs Time Taken")
for i in range(len(inputs)):
print(inputs[i],'\t\t',tim[i])
plt.plot(inputs,tim)
plt.xlabel("Number of Inputs");plt.ylabel("Time taken in seconds ")
plt.show()
"""
author : Mayank Kumar Jha (mk9440)
"""
from __future__ import print_function
import time
import matplotlib.pyplot as plt
from random import randint
def find_max_sub_array(A,low,high):
if low==high:
return low,high,A[low]
else :
mid=(low+high)//2
left_low,left_high,left_sum=find_max_sub_array(A,low,mid)
right_low,right_high,right_sum=find_max_sub_array(A,mid+1,high)
cross_left,cross_right,cross_sum=find_max_cross_sum(A,low,mid,high)
if left_sum>=right_sum and left_sum>=cross_sum:
return left_low,left_high,left_sum
elif right_sum>=left_sum and right_sum>=cross_sum :
return right_low,right_high,right_sum
else:
return cross_left,cross_right,cross_sum
def find_max_cross_sum(A,low,mid,high):
left_sum,max_left=-999999999,-1
right_sum,max_right=-999999999,-1
summ=0
for i in range(mid,low-1,-1):
summ+=A[i]
if summ > left_sum:
left_sum=summ
max_left=i
summ=0
for i in range(mid+1,high+1):
summ+=A[i]
if summ > right_sum:
right_sum=summ
max_right=i
return max_left,max_right,(left_sum+right_sum)
if __name__=='__main__':
inputs=[10,100,1000,10000,50000,100000,200000,300000,400000,500000]
tim=[]
for i in inputs:
li=[randint(1,i) for j in range(i)]
strt=time.time()
(find_max_sub_array(li,0,len(li)-1))
end=time.time()
tim.append(end-strt)
print("No of Inputs Time Taken")
for i in range(len(inputs)):
print((inputs[i],'\t\t',tim[i]))
plt.plot(inputs,tim)
plt.xlabel("Number of Inputs");plt.ylabel("Time taken in seconds ")
plt.show()