Please review my code for the problem given in the title, it stops after showing correct output for the first test case and shows Runtime error if I put t > 1 in Hackerearth IDE.
t = int(input())
while t > 0:
t = t - 1
n, k = map(int, input().split())
A = [int(i) for i in input().split()]
p = []
if k > n:
k = k - n
elif k == n:
k = 0
else:
k = k
for j in range(k):
x = A.pop()
p.append(x)
p = p[::-1]
for j in range(len(A)):
p.append(A[j])
print(*p)
Related
if __name__ == '__main__':
x = int(input())
y = int(input())
z = int(input())
n = int(input())
print[[i, j, k] for i in range( x + 1) for j in range( y + 1) for k in range( z + 1) if( i + j + k ) != n ]
The above code shows Syntax Error: invalid syntax
please help me understanding the error
You are missing parenthesis for print. Call it like so: print(...).
I come up a solution for leetcode "5. Longest Palindromic Substring" with parts of duplicate codes. One of good ways to solve duplicate code is to make a function. How do I write my check here to a function? I am confused what I should return to make both variables - longest and ans - being updated. Thanks!
The part of duplicate code:
if len(s[l:r+1]) > longest:
longest = len(s[l:r+1])
ans = s[l:r+1]
Full code:
class Solution:
def longestPalindrome(self, s: str) -> str:
if len(s) == 0:
return ''
if len(s) == 1:
return s
longest = 0
ans = ''
for pos in range(len(s)-1):
l, r = pos, pos
if pos > 0 and pos < len(s) - 1 and s[pos-1] == s[pos+1]:
l, r = pos-1, pos+1
while l > 0 and r < len(s) - 1 and s[l-1] == s[r+1]:
l -= 1
r += 1
# duplicate code 1
if len(s[l:r+1]) > longest:
longest = len(s[l:r+1])
ans = s[l:r+1]
if s[pos] == s[pos+1]:
l, r = pos, pos+1
while l > 0 and r < len(s) - 1 and s[l-1] == s[r+1]:
l -= 1
r += 1
# duplicate code 2
if len(s[l:r+1]) > longest:
longest = len(s[l:r+1])
ans = s[l:r+1]
if ans == '' and len(s) > 0:
return s[0]
return ans
The if statements and while loops before the duplicate code blocks are mostly duplicated as well, as is using the longest variable to keep track of the length of ans when you already have ans -- here's one way you could simplify things via another function:
class Solution:
def find_longest(self, s, left, right):
if s[left] == s[right]:
if right - left + 1 > len(self.ans):
self.ans = s[left:right + 1]
if left > 0 and right < len(s) - 1:
self.find_longest(s, left - 1, right + 1)
def longestPalindrome(self, s: str) -> str:
if len(s) == 1:
return s
self.ans = ''
for pos in range(len(s) - 1):
self.find_longest(s, pos, pos)
self.find_longest(s, pos, pos + 1)
return self.ans
I'm trying to write a code for calculating the number of factors for an arbitrary integer number but unfortunately when I run that I receive a false answer
I have tried for loop without defining function in this case and I got the result.Contrary to that, when I define a function I can't see the proper result
r = 0
def factor(a):
global r
for i in range(1, a + 1):
if a % i == 0:
r += 1
return r
a = int(input())
factor(a)
for example 18 has 6 factors but I receive just 1.
Use print to check your code. Indentation in Python matters. Also, global is not needed.
def factor(a):
r = 0
for i in range(1, a + 1):
if a % i == 0:
print('i', i)
r += 1
return r
a = int(input())
print(factor(a))
It was an indentation problem: the function should only return after the loop has finished iterating.
r = 0
def factor(a):
global r
for i in range(1, a + 1):
if a % i == 0:
r += 1
return r
a = 18 # int(input())
factor(a)
output:
6
This code works but it is not very efficient is there any help on a faster code in python to find n knowing that n is an integer above 0 and that n has no upper bound, how(x) will return you 1 if x>n, 0 if x = n, and -1 if x
def how(x):
if x > n:
return 1
elif x < n:
return -1
else:
return 0
def find(how):
if how(1) == 1:
return 1
x = 2
while how(x) != 1:
x = x**x
v = x
while how(x) != 0:
if how(x) == 1:
v = x
x = (x+1)//2
else:
x += (v-x+1)//2
return x
Rebecca, I've added some print statements so you can see where goes what wrong. As Patrick Artner said... its a bit confusing which way to go so I've tried to clean-up some things that enable you to continue exploring comparison of two variables against each other (and fake error catching (0).
Lets start and remove the confusing lingo and produce something workable code. With current below script it runs and with value = 1, reference = 1 you get the below print result in a continues loop until YOU stop the script manually:
v1 = n: error
loop1 1
def selector(v1, n):
if v1 > n:
print 'v1 > n', v1, n
return 1
elif v1 < n:
print 'v1 < n', v1, n
return -1
else:
print 'v1 == n: error'
return 0
def find(value, reference):
if selector(value, reference) == 1:
return 1
while selector(value, reference) != 1:
x = value**value
print 'loop1', x
v = x
while selector(value, reference) != 0:
print 'loop2'
if selector(value, reference) == 1:
v = value
x = (value+1)/2
print 'loop2-if', v, x
else:
x += (v-(value+1))/2
print 'loop2-else', x
print ' Almost done...'
return x
if __name__ == '__main__':
n = 1
print find(1, 1)
Happy exploring,....
I've tested to see if the list is empty and I'm sure it's not empty. So why this error continues?
The code is to find the smallest amount of factorial to a sum is equal to N.
def fat(n):
if n == 0 or n == 1:
return 1
else:
return n * fat(n - 1)
n = int(input())
ns = [x+1 for x in range(n)]
listaFat = []
l = []
for x in ns:
if fat(x) < n:
listaFat.append(x)
maior = max(listaFat)
listaFat.remove(maior)
print(listaFat)
print(len(listaFat))
while (fat(listaFat[-1]) + fat(maior)) < n:
l.append(listaFat[-1])
if len(listaFat) > 0:
listaFat.remove(listaFat[-1])
continue
else: break
print(l)
print(listaFat)