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Increasing itertools.permutations performance

I am completing a problem where I have create a function that takes a positive integer and returns the next bigger number that can be formed by rearranging its digits. For example: 12 --> 21, 513 --> 531, 12435 --> 12453, 9817121211 --> 9817122111.
I've recompiled my code over and over increasing performance but have eventually come unto a stop where I can't get it any faster. Does anyone have any advice? Its the itertools.permutations line which is taking the vast majority of the time.
def next_bigger(n):
num = str(n)
num1 = set(int(x) for x in str(num))
if num == num[0] *len(num):
return -1
#full_set = set(num)
lis = set(int(''.join(nums)) for nums in itertools.permutations(num, len(num)))
lis = sorted(lis)
try:
return int(lis[lis.index(n)+1])
except Exception:
return -1
Link to problem: https://www.codewars.com/kata/55983863da40caa2c900004e/train/python

If you are looking for better performance "time complexity wise", The approach would be to find the "key" of the algorithm. In this case you should ask yourself, what does it means to create the next bigger nummber? The answer is just as simple as a swap between two adjacent numbers. The code would be like this.
def next_bigger(n):
num_string = list(str(n))
for i in range(1, len(num_string)):
if i == len(num_string):
return -1
#find two the two numbers one bigger than the other with the minimun order
if num_string[-i] > num_string[-i-1]:
compare_reference = num_string[-i]
index_reference = -i
#check if the current number is smaller than any of the tail
for k, current in enumerate(num_string[-i:]):
if num_string[-i-1] < current and current < compare_reference:
compare_reference = current
index_reference = -i+k
#interchange the locations:
num_string[index_reference] = num_string[-i-1]
num_string[-i-1] = compare_reference
#check if the tail is larger than one digit
if i > 1:
#order the rest of the vector to create the smaller number (ordering it).
lower_part_ordered = sort_ascendant(num_string[-i:])
else:
lower_part_ordered = [num_string[-i]]
# create a string from the list
return int("".join(num_string[:-i] + lower_part_ordered))
# no match found means a number like 65311
return -1

While not a way to increase the permutations function performance per se, this was the method I found to increase performance of the code. many thanks to all that offered help!
def next_bigger(n):
num_string = list(str(n))
a = []
for i in range(1, len(num_string)):
if i == len(num_string):
return -1
p = int(num_string[-i])
q = int (num_string[-(i+1)])
if p > q:
a.append(num_string[:-(i+1)])
lis = list(num_string[-(i+1):])
if len(lis) > 1:
lis2 = list(set(lis))
lis2.sort()
qindex = lis2.index(str(q))
first = lis2[qindex+1]
a[0].append(first)
lis.remove(first)
lis.sort()
for j in range (len(lis)):
a[0].append(lis[j])
return int("".join(a[0]))
return -1

python (passing parameters to functions)

I'm not really new to python but I came across this problem that has just puzzled me.
So I was solving the maze runner problem, using A* and then was finding the hardest possible maze for a given dimension. For this purpose, I created a function called generateHardMaze() that is called from the main function and takes an attribute newMaze.
Now here is where things get weird, when I change the value of newMaze in the if condition within the while loop the hardMaze value changes without the code entering the second if condition. I'm not really sure why this happening was hoping someone could help me.
I'm using pycharm as my IDE and python3.6.* if that makes any difference.
I'm sure this isn't how oops works but I'm thinking this is a python thing. Has anyone ever come across anything like this? If yes please sympathize.
Thanks in advance.
def solveMazeAManH(newMaze,rows,cols):
startTime = time.time()
backTrackPriority = []
setup_cells(rows, cols)
# start and end points of the maze
start = (0, 0)
end = (rows - 1, cols - 1)
current = start
print("The path to be take is: ")
print(current)
frinLength = 0
# traversing the neighbours
while current != end:
unvisited.remove(current)
neighboursDFSandA(newMaze, current, rows, cols)
heuristic = calManhattanDis(current, end) # finding the heuristic for every traversal
try:
if not currentNeighbours:
if not backTrackPriority:
print("No path available!")
return 0
else:
while not currentNeighbours:
current = nextPopMan(backTrackPriority, end)
backTrackPriority.remove(current)
neighboursDFSandA(newMaze, current, rows, cols)
neighbor = leastPathChildMan(heuristic, current, end)
backTrackPriority.append(current)
current = neighbor
print(current)
frinLength += 1
except:
print("No path Found!")
return 0
return frinLength
endTime = time.time()
print("The time taken to solve the maze using A* with manhattan distance: ")
print(startTime - endTime)
def generateHardMaze(newMazes):
rows = len(newMazes)
cols = len(newMazes[0])
hardMaze = newMaze
print("Solving the original maze!")
fringLength = solveMazeAManH(newMazes, rows, cols)
print("Creating new harder Maze:")
pFlag = True
pCout = 0
while pFlag:
count = 0
flag = True
while flag:
point = choice(setup_cells(rows, cols))
if (newMazes[point[0]][point[1]] == 1):
newMazes[point[0]][point[1]] = 0
else:
newMazes[point[0]][point[1]] = 1
if (fringLength < solveMazeAManH(newMazes, rows, cols)):
print("Harder Maze--------------------")
hardMaze = newMaze
fringLength = solveMazeAManH(newMazes, rows, cols)
count = 0
else:
count += 1
if count >= 10:
flag = False
print("one")
newMazes = creatMaze(rows)
pCout += 1
if pCout >= 100:
pFlag = False
print(hardMaze)

the story of a tree hackerrank solution error

Found this problem in hackerrank and have failed to pass some testcases.
One day Bob drew a tree, with n nodes and n-1 edges on a piece of paper. He soon discovered that parent of a node depends on the root of the tree. The following images shows an example of that:
Learning the fact, Bob invented an exciting new game and decided to play it with Alice. The rules of the game is described below:
Bob picks a random node to be the tree's root and keeps the identity of the chosen node a secret from Alice. Each node has an equal probability of being picked as the root.
Alice then makes a list of g guesses, where each guess is in the form u v and means Alice guesses that parent(v) = u is true. It's guaranteed that an undirected edge connecting u and v exists in the tree.
For each correct guess, Alice earns one point. Alice wins the game if she earns at least k points (i.e., at least k of her guesses were true).
Alice and Bob play q games. Given the tree, Alice's guesses, and the value of k for each game, find the probability that Alice will win the game and print it on a new line as a reduced fraction in the format p/q.
Solution:
There is a tree with some edges marked with arrows. For every vertex in a tree you have to count how many arrows point towards it.For one fixed vertex this may be done via one DFS. Every arrow that was traversed during DFS in direction opposite to its own adds 1.If you know the answer for vertex v, you can compute the answer for vertex u adjacent to v in O(1).
It's almost the same as for v, but if there are arrows u->v or v->u, their contributions are reversed.Now you can make the vertex u crawl over the whole graph by moving to adjacent vertices in the second DFS.
Problem: It is not able to pass all the test cases. I did sanity testing of the code and found no problem but I am not getting any clue why this is not working in hackerrank platform.
import sys
def gcd(a, b):
if not b:
return a
return gcd(b, a%b)
def dfs1(m, guess, root, seen):
'''keep 1 node as root and calculate how many arrows are pointing towards it'''
count = 0
for i in m[root]:
if seen[i][root] != 1 and seen[root][i] != 1:
seen[i][root] = 1
seen[root][i] = 1
count += (1 if guess[root][i] == 1 else 0) + dfs1(m, guess, i, seen)
return count
def dfs2(m, guess, root, seen, cost, k):
'''now make every node as root and calculate how many nodes
are pointed towards it; If u is the root node for which
dfs1 calculated n (number of arrows pointed towards the root)
then for v (adjacent node of u), it would be n-1 as v is the
made the parent now in this step (only if there is a guess, if
there is no guess then it would be not changed)'''
win = cost >= k
for i in m[root]:
if seen[i][root] != 1 and seen[root][i] != 1:
seen[i][root] = 1
seen[root][i] = 1
win += dfs2(m, guess, i, seen, cost - (1 if guess[root][i] == 1 else -guess[i][root]), k)
return win
q = int(raw_input().strip())
for a0 in xrange(q):
n = int(raw_input().strip())
m = {}
guess = [[0 for i in range(n+1)] for i in range(n+1)]
seen = [[0 for i in range(n+1)] for i in range(n+1)]
for a1 in xrange(n-1):
u,v = raw_input().strip().split(' ')
u,v = [int(u),int(v)]
if u not in m:
m[u] = []
m[u].append(v)
if v not in m:
m[v] = []
m[v].append(u)
g,k = raw_input().strip().split(' ')
g,k = [int(g),int(k)]
for a1 in xrange(g):
u,v = raw_input().strip().split(' ')
u,v = [int(u),int(v)]
guess[u][v] = 1
cost = dfs1(m, guess, 1, seen)
seen = [[0 for i in range(n+1)] for i in range(n+1)]
win = dfs2(m, guess, 1, seen, cost, k)
g = gcd(win, n)
print("{0}/{1}".format(win/g, n/g))

One possibility is that the code is correct but you are getting a stack overflow.
There can be 100,000 nodes and if these are all connected in a line your depth first search recursion will fail.
If this is true, then converting the DFS code from a recursive to an iterative formulation (by keeping a stack of things to try in an array) should help.
Another possibility is that there could be a guess such as 1,2 and a guess such as 2,1. In this case I am not sure that the score updating code would work:
win += dfs2(m, guess, i, seen, cost - (1 if guess[root][i] == 1 else -guess[i][root]), k)
perhaps this would work better:
win += dfs2(m, guess, i, seen, cost - guess[root][i] + guess[i][root], k)

Project Euler Project 67 - Python

I am doing the Project Euler #67 in Python. My program, which worked for Project 18, does not work for Project 67.
Code (excludes the opening of the file and the processing of information):
for i in range(len(temp)):
list1 = temp[i]
try:
list2 = temp[i+1]
trynum1 = list1[lastinput] + max(list2[lastinput],list2[lastinput+1])
try:
trynum2 = list1[lastinput+1] + max(list2[lastinput+1],list2[lastinput+2])
if trynum1 > trynum2:
outputlist.append(list1[lastinput])
else:
outputlist.append(list1[lastinput+1])
lastinput += 1
except IndexError:
outputlist.append(list1[0])
except IndexError:
if list1[lastinput] > list1[lastinput+1]:
outputlist.append(list1[lastinput])
else:
outputlist.append(list1[lastinput+1])
Variables:
temp is the triangle of integers
outputlist is a list which stores the numbers chosen by the program
I know the answer is 7273, but my program finds 6542. I cannot find an error which causes the situation. Please may you help me on it.
Logic
My approach to this program is to find one number (list1[lastinput]) and add it up with the larger number of the two below it (trynum1), compare with the number to the right of the first number (list1[lastinput+1]), adding the larger number of two below it (trynum2). I append the larger one to the output list.

This approach is logically flawed. When you're in row 1, you don't have enough information to know whether moving right or left will lead you to the largest sum, not with only a 2-row lookahead. You would need to look all the way to the bottom to ensure getting the best path.
As others have suggested, start at the bottom and work up. Remember, you don't need the entire path, just the sum. At each node, add the amount of the better of the two available paths (that's the score you get in taking that node to the bottom). When you get back to the top, temp[0][0], that number should be your final answer.

I thought day and night about problem 18 and I solved it, the same way I solved this one.
P.S. 100_triangle.txt is without 1st string '59'.
# Maximum path sum II
import time
def e67():
start = time.time()
f=open("100_triangle.txt")
summ=[59]
for s in f:
slst=s.split()
lst=[int(item) for item in slst]
for i in range(len(lst)):
if i==0:
lst[i]+=summ[i]
elif i==len(lst)-1:
lst[i]+=summ[i-1]
elif (lst[i]+summ[i-1])>(lst[i]+summ[i]):
lst[i]+=summ[i-1]
else:
lst[i]+=summ[i]
summ=lst
end = time.time() - start
print("Runtime =", end)
f.close()
return max(summ)
print(e67()) #7273

Though starting from the bottom is more efficient, I wanted to see if I could implement Dijkstra's algorithm on this one; it works well and only takes a few seconds (didn't time it precisely):
from math import inf
f = open("p067_triangle.txt", "r")
tpyramid = f.read().splitlines()
f.close()
n = len(tpyramid)
pyramid = [[100 - int(tpyramid[i].split()[j]) for j in range(i+1)] for i in range(n)]
paths = [[inf for j in range(i+1)] for i in range(n)]
paths[0][0] = pyramid[0][0]
def mini_index(pyr):
m = inf
for i in range(n):
mr = min([i for i in pyr[i] if i >= 0]+[inf])
if mr < m:
m, a, b = mr, i, pyr[i].index(mr)
return m, a, b
counter = 0
omega = inf
while counter < n*(n+1)/2:
min_weight, i, j = mini_index(paths)
if i != n-1:
paths[i+1][j] = min( paths[i+1][j], min_weight + pyramid[i+1][j])
paths[i+1][j+1] = min( paths[i+1][j+1], min_weight + pyramid[i+1][j+1])
else:
omega = min(omega, min_weight)
paths[i][j] = -1
counter += 1
print(100*n - omega)

Here is my solution. Indeed you have to take the bottom - up approach.
Result confirmed with PE. Thanks!
def get_triangle(listLink):
triangle = [[int(number) for number in row.split()] for row in open(listLink)]
return triangle
listOfLists = get_triangle('D:\\Development\\triangle.txt')
for i in range(len(listOfLists) - 2, -1, -1):
for j in range(len(listOfLists[i])):
listOfLists[i][j] += max(listOfLists[i+1][j], listOfLists[i+1][j+1])
print(listOfLists[0][0])

What's wrong with my recursion implementation?

I have recently started learning programming, just completed a course on edX. I was trying to solve this problem on HackerRank and it is running out of time in each case. What am I doing wrong?
n,k = input().strip().split(' ')
n,k = [int(n),int(k)]
x = [int(x_temp) for x_temp in input().strip().split(' ')]
x.sort()
def transmitter(aList=[], target=0):
'''
accepts a list of house location, and a target location for the transmitter
returns the optimal number of transmitters required to cover all the houses
'''
List = aList[:]
start = target - k
end = target + k + 1
for i in range(start, end):
if i in List:
List.remove(i)
if not List:
return 1
m = max(List)
for e in List:
if transmitter(List, e) < m:
m = transmitter(List, e)
return 1 + m
m = max(x)
for e in x:
if transmitter(x, e) < m:
m = transmitter(x, e)
print(m)
I am pretty new to this. Sorry for making any obvious mistakes, or for posting this here in case this is not the suitable site. In that case, it will be really helpful if you can recommend a site where I can ask such question.
the screenshot of the question

I'm pretty sure a greedy algorithm solves this problem optimally in just O(N) time. There's not need for any recursion. Just place each transmitter in turn as far to the right as you can without leaving any houses to its left uncovered. Stop when the last house is covered.
Here's how I'd code that:
def hackerland(houses, k): # houses should be sorted list of locations
first = None # location of first uncovered house
last = 0 # last location covered by a previous transmitter
prev = None
count = 0 # transmitters = []
for x in houses:
if first is not None and x > first + k:
first = None
count += 1 # transmitters.append(prev)
last = prev + k
if last is not None and x > last:
last = None
first = x
prev = x
if first is not None:
count += 1 # transmitters.append(prev)
return count # return transmitters
I've included comments that show how this code could be easily modified to return a list of the transmitter locations, rather than just a count of how many are needed.

It is not necessary to take a recursive approach. In fact, you can just work forward, iterate over the houses, placing transmitters when the previously placed one does not reach far enough to cover the current house, etc.
It is a bit more complicated than that, but not much. See this code:
# input
n,k = input().strip().split(' ')
n,k = [int(n),int(k)]
x = [int(x_temp) for x_temp in input().strip().split(' ')]
# eliminate duplicate house x-xoordinates, they don't influence the result
houses = list(set(x))
houses.sort()
# add extreme far dummy house (will make the loop easier)
houses.append(100000)
reachedX = 0 # coordinate until where the previously placed transmitter reaches
unreachedX = -1 # coordinate that the next one needs to cover (to the left)
lastHouseId = -1 # index where previous transmitter was placed
transmitters = [] # coordinates of the placed transmitters
for houseId, houseX in enumerate(houses):
if reachedX > unreachedX: # we might still be in range of last transmitter
if houseX > reachedX: # we just went out of reach
unreachedX = houseX # this house must be covered by next one
elif houseX - k > unreachedX: # transmitter here wouldn't reach far enough back
lastHouseId = houseId - 1 # place it on previous house
reachedX = houses[lastHouseId] + k
transmitters.append(houses[lastHouseId])
print(transmitters)
print(len(transmitters))