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codehouseindia
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Merge pull request codehouseindia#350 from sanskarsri/master
pattern codes
2 parents 47dc1ee + 7a6231d commit bf25001

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count-no-of-square-in-range.py

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#CODE 3
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# Python program to count squares between a and b
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def CountSquares(n,a, b):
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cnt = 0 # initialize result
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# Traverse through all numbers
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for i in range (a, b + 1):
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j = 1;
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while j ** n <= i:
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if j ** n == i:
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cnt = cnt + 1
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j = j + 1
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i = i + 1
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return cnt
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# Driver Code
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n=2
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a = 49
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b = 65
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print ("Count of squares is:", CountSquares(n,a, b) )

hollow-triangle-pattern.py

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# CODE 1
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# pyramid pattern
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def printPattern(n) :
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k = 0
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for i in range(1,n) : #row 6
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# Print spaces
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for j in range(i,n-1) :
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print(' ', end='')
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# Print #
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while (k != (2 * i - 1)) :
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if (k == 0 or k == 2 * i - 2) :
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print('*', end='')
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else :
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print(' ', end ='')
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k = k + 1
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k = 0;
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print ("") # print next row
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# print last row
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for i in range(0, 2 * n -3) :
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print ('*', end = '')
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# Driver code
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n =input()
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n=int(n)
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printPattern(n)

new.py

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if 5>3:
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print("correct")
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x = "awesome"
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def myfunc():
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print("Python is " + x)
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myfunc()
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x=1
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y = 2.8 # float
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z = 1j # complex
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#convert from int to float:
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a = float(x)
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#convert from float to int:
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b = int(y)
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#convert from int to complex:
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c = complex(x)
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print(a)
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print(b)
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print(c)
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print(type(a))
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print(type(b))
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print(type(c))
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import random
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print(random.randrange(1, 10))
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b = "Hello, World!"
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print(b[2:5])
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b = "Hello, World!"
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print(b[-5:-2])
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# Check if the phrase "ain" is present in the following text:
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txt = "The rain in Spain stays mainly in the plain"
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x = "ain" in txt
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print(x)
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# Check if the phrase "ain" is NOT present in the following text:
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txt = "The rain in Spain stays mainly in the plain"
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x = "ain" not in txt
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print(x)
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# The format() method takes unlimited number of arguments, and are placed into the respective placeholders:
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# Example
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quantity = 3
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itemno = 567
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price = 49.95
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myorder = "I want {} pieces of item {} for {} dollars."
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print(myorder.format(quantity, itemno, price))
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txt = "We are the so-called \"Vikings\" from the north."
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print(txt)
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# # CODE 1
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# # pyramid pattern
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# def printPattern(n) :
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# k = 0
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# for i in range(1,n) : #row 6
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# # Print spaces
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# for j in range(i,n-1) :
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# print(' ', end='')
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# # Print #
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# while (k != (2 * i - 1)) :
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# if (k == 0 or k == 2 * i - 2) :
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# print('*', end='')
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# else :
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# print(' ', end ='')
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# k = k + 1
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# k = 0;
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# print ("") # print next row
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# # print last row
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# for i in range(0, 2 * n -3) :
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# print ('*', end = '')
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# # Driver code
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# n = 10
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# printPattern(n)
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# print("\n")
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#CODE 2
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# Returns sum of n + nn + nnn + .... (m times)
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# def Series(n):
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# # Converting the number to string
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# str_n = str(n)
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# # Initializing result as number and string
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# sums = n
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# sum_str = str(n)
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# # Adding remaining terms
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# for i in range(1, n):
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# # Concatenating the string making n, nn, nnn...
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# sum_str = sum_str + str_n
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# # Before adding converting back to integer
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# sums = sums + int(sum_str)
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# return sums
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# # Driver Code
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# n = 2
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# total = Series(n)
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# print(total)
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# #CODE 3
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# # Python program to count squares between a and b
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# def CountSquares(n,a, b):
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# cnt = 0 # initialize result
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# # Traverse through all numbers
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# for i in range (a, b + 1):
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# j = 1;
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# while j ** n <= i:
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# if j ** n == i:
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# cnt = cnt + 1
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# j = j + 1
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# i = i + 1
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# return cnt
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# # Driver Code
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# n=2
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# a = 49
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# b = 65
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# print ("Count of squares is:", CountSquares(n,a, b) )
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#CODE 4
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def checkPattern(string, pattern):
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# len stores length of the given pattern
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l = len(pattern)
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# if length of pattern is more than length
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# of input string, return false;
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if len(string) < l:
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return False
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for i in range(l - 1):
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# x, y are two adjacent characters in pattern
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x = pattern[i]
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y = pattern[i + 1]
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# find index of last occurrence of
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# character x in the input string
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last = string.rindex(x)
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# find index of first occurrence of
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# character y in the input string
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first = string.index(y)
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# return false if x or y are not present
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# in the input string OR last occurrence of
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# x is after the first occurrence of y
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# in the input string
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if last == -1 or first == -1 or last > first:
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return False
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# return true if
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# string matches the pattern
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return True
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# Driver Code
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string = "a rabbit jumps joyfully"
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first="a"
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second="j"
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pattern = first + second
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print(checkPattern(string, pattern))
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#CODE 5
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# import datetime
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# def is_date_the_nth_friday_of_month(nth, date=None):
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# #nth is an integer representing the nth weekday of the month
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# #date is a datetime.datetime object, which you can create by doing datetime.datetime(2016,1,11) for January 11th, 2016
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# if not date:
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# #if date is None, then use today as the date
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# date = datetime.datetime.today()
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# if date.weekday() == 4:
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# #if the weekday of date is Friday, then see if it is the nth Friday
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# if (date.day - 1) // 7 == (nth - 1):
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# #We use integer division to determine the nth Friday
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# #if this integer division == 0, then date is the first Friday,
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# # if the integer division is...
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# # 1 == 2nd Friday
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# # 2 == 3rd Friday
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# # 3 == 4th Friday
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# # 4 == 5th Friday
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# return True
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# return False
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# date=2,2020
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# is_date_the_nth_friday_of_month(12, date)

return-series-sum.py

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#CODE 2
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# Returns sum of n + nn + nnn + .... (m times)
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def Series(n):
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# Converting the number to string
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str_n = str(n)
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# Initializing result as number and string
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sums = n
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sum_str = str(n)
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# Adding remaining terms
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for i in range(1, n):
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# Concatenating the string making n, nn, nnn...
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sum_str = sum_str + str_n
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# Before adding converting back to integer
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sums = sums + int(sum_str)
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return sums
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# Driver Code
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n = 2
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total = Series(n)
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print(total)

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