https://www.hackerrank.com/challenges/designer-door-mat/problem
Below is my submission:
n = input().split(' ')
rows, columns = int(n[0]), int(n[1])
if(rows % 2 == 1 and columns == 3*rows):
printale = ''
j = 1
k = rows - 7
for i in range(int(rows/2), -1, -1):
printale = '---'*i + '.|.'*j + '---'*i
if(i == 0):
print('-'*int(j+k/2) + 'Welcome' + '-'*int(j+k/2))
j -= 2
else:
print(printale)
j += 2
for l in range(1, int(rows/2)+1):
printale = '---'*l + '.|.'*j + '---'*l
print(printale)
j -= 2
Is there anything wrong with the code?
Yes, there is. The "WELCOME" in the problem statement is all-caps.
Here's my code:
def ispalindrome(p):
temp = p
rev = 0
while temp != 0:
rev = (rev * 10) + (temp % 10)
temp = temp // 10
if num == rev:
return True
else:
return False
num = int(input("Enter a number: "))
i = 1
count = 0
sum = 0
while (count <= num - 1):
if (palindrome(i) == True):
sum = sum + i
count = count + 1
i = i + 1
print("Sum of first", num, "palindromes is", sum)
I believe my ispalindrome() function works. I'm trying to figure out what's wrong inside my while loop.
here's my output so far:
n = 1 answer = 1,
n = 2 answer = 22,
n = 3 answer = 333 ...
I also think the runtime on this really sucks
Please help
i belive the problem is with your ispalindrom functon it returns 200 as palindrome number
def ispalindrome(p):
rev = int(str(p)[::-1])
if p == rev:
return True
else:
return False
num = int(input("Enter a number: "))
i = 1
count = 0
sum = 0
while (count <= num - 1):
if (ispalindrome(i) == True):
print(i)
sum = sum + i
count = count + 1
i = i + 1
print("Sum of first", num, "palindromes is", sum)
def is_palindrome(number):
return str(number) == str(number)[::-1]
num = int(input("Enter a number: "))
palindromes = [i for i in range(1, num) if is_palindrome(i)]
print(f"Sum of the {len(palindromes)} palindromes in range {num} is {sum(palindromes)}")
I'm having a lot of trouble converting infix notation to postfix.
For instance, I want to convert this
test(a(b+c), d()) - 3
into this
b c + a , d test 3 -
I tried this solution,
def composition(s):
i = 0
rpnstack = []
stack = []
ret = []
count = 0
while i < len(s) :
if i + 1 < len(s) and s[i + 1] == "(":
stack.append([count, rpnstack, s[i]])
i += 2
count = 1
rpnstack = []
elif s[i] == "(":
count += 1
rpnstack.append(s[i])
i += 1
elif s[i] == ")":
count -= 1
if count == 0:
for a in rpn(rpnstack):
ret.append(a)
a = stack.pop()
count = a[0]
rpnstack = a[1]
ret.append(a[2])
else:
rpnstack.append(s[i])
i += 1
else:
rpnstack.append(s[i])
i += 1
for a in rpn(rpnstack):
ret.append(a)
return ret
where RPN is the standard algorithm for the reverse polish notation and is is the infix string splitted with this regex
(\+|\-|\*|\/|\>|\<|\(|\)|\,)
But it only works sometimes.
This is the full implementation of the rpn function
operator = -10
operand = -20
leftparentheses = -30
rightparentheses = -40
empty = -50
operands = ["+", "-", "*", "/", ">", "<", "=", ","]
def precedence(s):
if s is '(':
return 0
elif s is '+' or '-':
return 1
elif s is '*' or '/' or '%':
return 2
else:
return 99
def typeof(s):
if s is '(':
return leftparentheses
elif s is ')':
return rightparentheses
elif s in operands:
return operator
elif s is ' ':
return empty
else :
return operand
def rpn(infix):
postfix = []
temp = []
for i in infix :
type = typeof(i)
if type is leftparentheses :
temp.append(i)
elif type is rightparentheses :
next = temp.pop()
while next is not '(' or skip > 0:
postfix.append(next)
next = temp.pop()
elif type is operand:
postfix.append(i)
elif type is operator:
p = precedence(i)
while len(temp) is not 0 and p <= precedence(temp[-1]) :
postfix.append(temp.pop())
temp.append(i)
elif type is empty:
continue
while len(temp) > 0 :
postfix.append(temp.pop())
return postfix
if i try to use the code against this infix expression:
i < test.func()
i get:
[' test.func', 'i ', '<']
and against this
i < 10
i get:
['i ', ' 10', '<']
How can I fix this?
I'm trying to make a factoring program, but it doesn't seem to work with negative number a-, b- and c-inputs.
from fractions import gcd
factor = -1
opp = 0
number = 1
success = 0
a = int(input("a-value: "))
b = int(input("b-value: "))
c = int(input("c-value: "))
factors = []
d = 0
e = 0
while number <= abs(a*c):
#Checking for multiples
if abs(a*c) % number == 0:
factor += 1
factors.append(number)
number += 1
while (factor-opp) >= 0:
#Checking for actual factors
d = int(factors[factor])
e = int(factors[opp])
if (abs(d+e) or abs(d-e)) == abs(b):
success += 1
break
else:
factor -= 1
opp += 1
if success > 0:
if (d+e) == b:
e = e
elif (d-e) == b:
e -= 2*e
elif (e-d) == b:
d -= 2*d
elif (-d-e) == b:
d -= 2*d
e -= 2*e
#Figuring out the equation
if d % a == 0:
d /= a
f = 1
else:
f = a/gcd(d,a)
d /= gcd(d,a)
if e % a == 0:
e /= a
g = 1
else:
g = a/gcd(e,a)
e /= gcd(e,a)
#Displaying the answer
if d >= 0:
d = str("+" + str(int(d)))
if e >= 0:
e = str("+" + str(int(e)))
elif e < 0:
e = str(int(e))
else:
d = str(int(d))
if e >= 0:
e = str("+" + str(int(e)))
elif e < 0:
e = str(int(e))
if f == 1:
if g == 1:
print ("(x" + d + ")(x" + e + ")")
else:
g = str(int(g))
print ("(x" + d + ")(" + g + "x" + e + ")")
elif g == 1:
f = str(int(f))
print ("(" + f + "x" + d + ")(x" + e + ")")
else:
f = str(int(f))
g = str(int(g))
print ("(" + f + "x" + d + ")(" + g + "x" + e + ")")
else:
print("This equation cannot be factored into integers.")
More specifically, the problem is somewhere within this block, I think. I've tested it out with print statements:
while (factor-opp) >= 0:
#Checking for actual factors
d = int(factors[factor])
e = int(factors[opp])
if (abs(d+e) or abs(d-e)) == abs(b):
success += 1
break
else:
factor -= 1
opp += 1
I've searched everywhere: my programming textbook, online searches about inputting negatives, everything. What am I doing wrong here?
Ok I am able to reproduce your issue for a simple testcase like - a=1 , b=0, c=-4 .
The issue is in the line -
if (abs(d+e) or abs(d-e)) == abs(b):
This does not check whether abs(b) is equal to abs(d+e) or abs(d-e) , instead it first evaluates the result of (abs(d+e) or abs(d-e)) , which would return the first non-zero result , and then compare that against abs(b) , so for negative numbers this does not evaluate the result correctly. Change that condition to -
if abs(d+e) == abs(b) or abs(d-e) == abs(b):
or you can also use a set -
if abs(b) in {abs(d+e), abs(d-e)}: #Though I doubt if using set would give any performance improvement because of the overhead of creating a set.
Demo after changes -
a-value: 1
b-value: 0
c-value: -4
(x+2)(x-2)
a-value: 1
b-value: -1
c-value: -6
(x-3)(x+2)
One more thing, there is something you have not considered , when a=-1 , b=-4 , c=-4 , the result should come to -(x+2)(x+2) , but the current program results in (x+2)(x+2) .
If I have 2 numbers in binary form as a string, and I want to add them I will do it digit by digit, from the right most end. So 001 + 010 = 011
But suppose I have to do 001+001, how should I create a code to figure out how to take carry over responses?
bin and int are very useful here:
a = '001'
b = '011'
c = bin(int(a,2) + int(b,2))
# 0b100
int allows you to specify what base the first argument is in when converting from a string (in this case two), and bin converts a number back to a binary string.
This accepts an arbitrary number or arguments:
>>> def bin_add(*bin_nums: str) -> str:
... return bin(sum(int(x, 2) for x in bin_nums))[2:]
...
>>> x = bin_add('1', '10', '100')
>>> x
'111'
>>> int(x, base = 2)
7
Here's an easy to understand version
def binAdd(s1, s2):
if not s1 or not s2:
return ''
maxlen = max(len(s1), len(s2))
s1 = s1.zfill(maxlen)
s2 = s2.zfill(maxlen)
result = ''
carry = 0
i = maxlen - 1
while(i >= 0):
s = int(s1[i]) + int(s2[i])
if s == 2: #1+1
if carry == 0:
carry = 1
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
elif s == 1: # 1+0
if carry == 1:
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
else: # 0+0
if carry == 1:
result = "%s%s" % (result, '1')
carry = 0
else:
result = "%s%s" % (result, '0')
i = i - 1;
if carry>0:
result = "%s%s" % (result, '1')
return result[::-1]
Can be simple if you parse the strings by int (shown in the other answer). Here is a kindergarten-school-math way:
>>> def add(x,y):
maxlen = max(len(x), len(y))
#Normalize lengths
x = x.zfill(maxlen)
y = y.zfill(maxlen)
result = ''
carry = 0
for i in range(maxlen-1, -1, -1):
r = carry
r += 1 if x[i] == '1' else 0
r += 1 if y[i] == '1' else 0
# r can be 0,1,2,3 (carry + x[i] + y[i])
# and among these, for r==1 and r==3 you will have result bit = 1
# for r==2 and r==3 you will have carry = 1
result = ('1' if r % 2 == 1 else '0') + result
carry = 0 if r < 2 else 1
if carry !=0 : result = '1' + result
return result.zfill(maxlen)
>>> add('1','111')
'1000'
>>> add('111','111')
'1110'
>>> add('111','1000')
'1111'
It works both ways
# as strings
a = "0b001"
b = "0b010"
c = bin(int(a, 2) + int(b, 2))
# as binary numbers
a = 0b001
b = 0b010
c = bin(a + b)
you can use this function I did:
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
#a = int('10110', 2) #(0*2** 0)+(1*2**1)+(1*2**2)+(0*2**3)+(1*2**4) = 22
#b = int('1011', 2) #(1*2** 0)+(1*2**1)+(0*2**2)+(1*2**3) = 11
sum = int(a, 2) + int(b, 2)
if sum == 0: return "0"
out = []
while sum > 0:
res = int(sum) % 2
out.insert(0, str(res))
sum = sum/2
return ''.join(out)
def addBinary(self, A, B):
min_len, res, carry, i, j = min(len(A), len(B)), '', 0, len(A) - 1, len(B) - 1
while i>=0 and j>=0:
r = carry
r += 1 if A[i] == '1' else 0
r += 1 if B[j] == '1' else 0
res = ('1' if r % 2 == 1 else '0') + res
carry = 0 if r < 2 else 1
i -= 1
j -= 1
while i>=0:
r = carry
r += 1 if A[i] == '1' else 0
res = ('1' if r % 2 == 1 else '0') + res
carry = 0 if r < 2 else 1
i -= 1
while j>=0:
r = carry
r += 1 if B[j] == '1' else 0
res = ('1' if r % 2 == 1 else '0') + res
carry = 0 if r < 2 else 1
j -= 1
if carry == 1:
return '1' + res
return res
#addition of two binary string without using 'bin' inbuilt function
numb1 = input('enter the 1st binary number')
numb2 = input("enter the 2nd binary number")
list1 = []
carry = '0'
maxlen = max(len(numb1), len(numb2))
x = numb1.zfill(maxlen)
y = numb2.zfill(maxlen)
for j in range(maxlen-1,-1,-1):
d1 = x[j]
d2 = y[j]
if d1 == '0' and d2 =='0' and carry =='0':
list1.append('0')
carry = '0'
elif d1 == '1' and d2 =='1' and carry =='1':
list1.append('1')
carry = '1'
elif (d1 == '1' and d2 =='0' and carry =='0') or (d1 == '0' and d2 =='1' and
carry =='0') or (d1 == '0' and d2 =='0' and carry =='1'):
list1.append('1')
carry = '0'
elif d1 == '1' and d2 =='1' and carry =='0':
list1.append('0')
carry = '1'
else:
list1.append('0')
if carry == '1':
list1.append('1')
addition = ''.join(list1[::-1])
print(addition)
Not an optimal solution but a working one without use of any inbuilt functions.
# two approaches
# first - binary to decimal conversion, add and then decimal to binary conversion
# second - binary addition normally
# binary addition - optimal approach
# rules
# 1 + 0 = 1
# 1 + 1 = 0 (carry - 1)
# 1 + 1 + 1(carry) = 1 (carry -1)
aa = a
bb = b
len_a = len(aa)
len_b = len(bb)
min_len = min(len_a, len_b)
carry = 0
arr = []
while min_len > 0:
last_digit_aa = int(aa[len(aa)-1])
last_digit_bb = int(bb[len(bb)-1])
add_digits = last_digit_aa + last_digit_bb + carry
carry = 0
if add_digits == 2:
add_digits = 0
carry = 1
if add_digits == 3:
add_digits = 1
carry = 1
arr.append(add_digits) # will rev this at the very end for output
aa = aa[:-1]
bb = bb[:-1]
min_len -= 1
a_len_after = len(aa)
b_len_after = len(bb)
if a_len_after > 0:
while a_len_after > 0:
while carry == 1:
if len(aa) > 0:
sum_digit = int(aa[len(aa) - 1]) + carry
if sum_digit == 2:
sum_digit = 0
carry = 1
arr.append(sum_digit)
aa = aa[:-1]
else:
carry = 0
arr.append(sum_digit)
aa = aa[:-1]
else:
arr.append(carry)
carry = 0
if carry == 0 and len(aa) > 0:
arr.append(aa[len(aa) - 1])
aa = aa[:-1]
a_len_after -= 1
if b_len_after > 0:
while b_len_after > 0:
while carry == 1:
if len(bb) > 0:
sum_digit = int(bb[len(bb) - 1]) + carry
if sum_digit == 2:
sum_digit = 0
carry = 1
arr.append(sum_digit)
bb = bb[:-1]
else:
carry = 0
arr.append(sum_digit)
bb = bb[:-1]
else:
arr.append(carry)
carry = 0
if carry == 0 and len(bb) > 0:
arr.append(bb[len(bb) - 1])
bb = bb[:-1]
b_len_after -= 1
if carry == 1:
arr.append(carry)
out_arr = reversed(arr)
out_str = "".join(str(x) for x in out_arr)
return out_str