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codeparrot
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xlcost-text-to-code

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Possible value of Q
val_Q . append ( X + i ) NEW_LINE print ( len ( val_Q ) ) NEW_LINE
Print all the numbers
for i in range ( len ( val_Q ) ) : NEW_LINE INDENT print ( val_Q [ i ] , end = " ▁ " ) NEW_LINE DEDENT
Driver Code
X = 3 NEW_LINE values_of_Q ( X ) NEW_LINE
Function that finds gcd of 2 strings
def gcd ( str1 , str2 ) : NEW_LINE
If str1 length is less than that of str2 then recur with gcd ( str2 , str1 )
if ( len ( str1 ) < len ( str2 ) ) : NEW_LINE INDENT return gcd ( str2 , str1 ) NEW_LINE DEDENT
If str1 is not the concatenation of str2
elif ( not str1 . startswith ( str2 ) ) : NEW_LINE INDENT return " " NEW_LINE DEDENT elif ( len ( str2 ) == 0 ) : NEW_LINE
GCD string is found
return str1 NEW_LINE else : NEW_LINE
Cut off the common prefix part of str1 & then recur
return gcd ( str1 [ len ( str2 ) : ] , str2 ) NEW_LINE
Function to find GCD of array of strings
def findGCD ( arr , n ) : NEW_LINE INDENT result = arr [ 0 ] NEW_LINE for i in range ( 1 , n ) : NEW_LINE INDENT result = gcd ( result , arr [ i ] ) NEW_LINE DEDENT DEDENT
Return the GCD of strings
return result NEW_LINE
Given array of strings
arr = [ " GFGGFG " , " GFGGFG " , " GFGGFGGFGGFG " ] NEW_LINE n = len ( arr ) NEW_LINE
Function Call
print ( findGCD ( arr , n ) ) NEW_LINE
Define the size of the matrix
N = 4 NEW_LINE M = 4 NEW_LINE
Function to check given matrix balanced or unbalanced
def balancedMatrix ( mat ) : NEW_LINE
Flag for check matrix is balanced or unbalanced
is_balanced = True NEW_LINE
Iterate row until condition is true
i = 0 NEW_LINE while i < N and is_balanced : NEW_LINE
Iterate cols until condition is true
j = 0 NEW_LINE while j < N and is_balanced : NEW_LINE
Check for corner edge elements
if ( ( i == 0 or i == N - 1 ) and ( j == 0 or j == M - 1 ) ) : NEW_LINE INDENT if mat [ i ] [ j ] >= 2 : NEW_LINE INDENT isbalanced = False NEW_LINE DEDENT DEDENT
Check for border elements
elif ( i == 0 or i == N - 1 or j == 0 or j == M - 1 ) : NEW_LINE INDENT if mat [ i ] [ j ] >= 3 : NEW_LINE INDENT is_balanced = False NEW_LINE DEDENT DEDENT
Check for the middle ones
else : NEW_LINE INDENT if mat [ i ] [ j ] >= 4 : NEW_LINE INDENT is_balanced = False NEW_LINE DEDENT DEDENT j += 1 NEW_LINE i += 1 NEW_LINE
Return balanced or not
if is_balanced : NEW_LINE INDENT return " Balanced " NEW_LINE DEDENT else : NEW_LINE INDENT return " Unbalanced " NEW_LINE DEDENT
Given matrix mat [ ] [ ]
mat = [ [ 1 , 2 , 3 , 4 ] , [ 3 , 5 , 2 , 6 ] , [ 5 , 3 , 6 , 1 ] , [ 9 , 5 , 6 , 0 ] ] NEW_LINE
Function call
print ( balancedMatrix ( mat ) ) NEW_LINE
Function to convert unix time to Human readable format
def unixTimeToHumanReadable ( seconds ) : NEW_LINE
Save the time in Human readable format
ans = " " NEW_LINE
Number of days in month in normal year
daysOfMonth = [ 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 ] NEW_LINE ( currYear , daysTillNow , extraTime , extraDays , index , date , month , hours , minutes , secondss , flag ) = ( 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 ) NEW_LINE
Calculate total days unix time T
daysTillNow = seconds // ( 24 * 60 * 60 ) NEW_LINE extraTime = seconds % ( 24 * 60 * 60 ) NEW_LINE currYear = 1970 NEW_LINE
Calculating current year
while ( daysTillNow >= 365 ) : NEW_LINE INDENT if ( currYear % 400 == 0 or ( currYear % 4 == 0 and currYear % 100 != 0 ) ) : NEW_LINE INDENT daysTillNow -= 366 NEW_LINE DEDENT else : NEW_LINE INDENT daysTillNow -= 365 NEW_LINE DEDENT currYear += 1 NEW_LINE DEDENT
Updating extradays because it will give days till previous day and we have include current day
extraDays = daysTillNow + 1 NEW_LINE if ( currYear % 400 == 0 or ( currYear % 4 == 0 and currYear % 100 != 0 ) ) : NEW_LINE INDENT flag = 1 NEW_LINE DEDENT
Calculating MONTH and DATE
month = 0 NEW_LINE index = 0 NEW_LINE if ( flag == 1 ) : NEW_LINE INDENT while ( True ) : NEW_LINE INDENT if ( index == 1 ) : NEW_LINE INDENT if ( extraDays - 29 < 0 ) : NEW_LINE INDENT break NEW_LINE DEDENT month += 1 NEW_LINE extraDays -= 29 NEW_LINE DEDENT else : NEW_LINE INDENT if ( extraDays - daysOfMonth [ index ] < 0 ) : NEW_LINE INDENT break NEW_LINE DEDENT month += 1 NEW_LINE extraDays -= daysOfMonth [ index ] NEW_LINE DEDENT index += 1 NEW_LINE DEDENT DEDENT else : NEW_LINE INDENT while ( True ) : NEW_LINE INDENT if ( extraDays - daysOfMonth [ index ] < 0 ) : NEW_LINE INDENT break NEW_LINE DEDENT month += 1 NEW_LINE extraDays -= daysOfMonth [ index ] NEW_LINE index += 1 NEW_LINE DEDENT DEDENT
Current Month
if ( extraDays > 0 ) : NEW_LINE INDENT month += 1 NEW_LINE date = extraDays NEW_LINE DEDENT else : NEW_LINE INDENT if ( month == 2 and flag == 1 ) : NEW_LINE INDENT date = 29 NEW_LINE DEDENT else : NEW_LINE INDENT date = daysOfMonth [ month - 1 ] NEW_LINE DEDENT DEDENT
Calculating HH : MM : YYYY
hours = extraTime // 3600 NEW_LINE minutes = ( extraTime % 3600 ) // 60 NEW_LINE secondss = ( extraTime % 3600 ) % 60 NEW_LINE ans += str ( date ) NEW_LINE ans += " / " NEW_LINE ans += str ( month ) NEW_LINE ans += " / " NEW_LINE ans += str ( currYear ) NEW_LINE ans += " ▁ " NEW_LINE ans += str ( hours ) NEW_LINE ans += " : " NEW_LINE ans += str ( minutes ) NEW_LINE ans += " : " NEW_LINE ans += str ( secondss ) NEW_LINE
Return the time
return ans NEW_LINE
Driver code
if __name__ == " _ _ main _ _ " : NEW_LINE
Given unix time
T = 1595497956 NEW_LINE
Function call to convert unix time to human read able
ans = unixTimeToHumanReadable ( T ) NEW_LINE
Print time in format DD : MM : YYYY : HH : MM : SS
print ( ans ) NEW_LINE
Function to find area of rectangle inscribed another rectangle of length L and width W
def AreaofRectangle ( L , W ) : NEW_LINE
Area of rectangle
INDENT area = ( W + L ) * ( W + L ) / 2 NEW_LINE DEDENT
Return the area
INDENT return area NEW_LINE DEDENT
Driver Code
if __name__ == " _ _ main _ _ " : NEW_LINE
Given Dimensions
INDENT L = 18 NEW_LINE W = 12 NEW_LINE DEDENT
Function Call
INDENT print ( AreaofRectangle ( L , W ) ) NEW_LINE DEDENT
Python3 program to implement the above approach
import math NEW_LINE import sys NEW_LINE
Function to count the minimum steps required to reduce n
def downToZero ( n ) : NEW_LINE
Base case
if ( n <= 3 ) : NEW_LINE INDENT return n NEW_LINE DEDENT
Allocate memory for storing intermediate results
dp = [ - 1 ] * ( n + 1 ) NEW_LINE
Store base values
dp [ 0 ] = 0 NEW_LINE dp [ 1 ] = 1 NEW_LINE dp [ 2 ] = 2 NEW_LINE dp [ 3 ] = 3 NEW_LINE
Stores square root of each number
for i in range ( 4 , n + 1 ) : NEW_LINE
Compute square root
sqr = ( int ) ( math . sqrt ( i ) ) NEW_LINE best = sys . maxsize NEW_LINE
Use rule 1 to find optimized answer
while ( sqr > 1 ) : NEW_LINE
Check if it perfectly divides n
if ( i % sqr == 0 ) : NEW_LINE INDENT best = min ( best , 1 + dp [ sqr ] ) NEW_LINE DEDENT sqr -= 1 NEW_LINE
Use of rule 2 to find the optimized answer
best = min ( best , 1 + dp [ i - 1 ] ) NEW_LINE
Store computed value
dp [ i ] = best NEW_LINE
Return answer
return dp [ n ] NEW_LINE
Driver Code
if __name__ == " _ _ main _ _ " : NEW_LINE INDENT n = 4 NEW_LINE print ( downToZero ( n ) ) NEW_LINE DEDENT
Function to find the minimum steps required to reduce n
def downToZero ( n ) : NEW_LINE
Base case
if ( n <= 3 ) : NEW_LINE INDENT return n ; NEW_LINE DEDENT
Return answer based on parity of n
if ( n % 2 == 0 ) : NEW_LINE INDENT return 3 ; NEW_LINE DEDENT else : NEW_LINE INDENT return 4 ; NEW_LINE DEDENT
Driver Code
if __name__ == ' _ _ main _ _ ' : NEW_LINE INDENT n = 4 ; NEW_LINE print ( downToZero ( n ) ) ; NEW_LINE DEDENT
Function to find minimum number of elements required to form A [ ] by performing mirroring operation
def minimumrequired ( A , N ) : NEW_LINE
Initialize K
K = N NEW_LINE while ( K > 0 ) : NEW_LINE
Odd length array cannot be formed by mirror operation
if ( K % 2 ) == 1 : NEW_LINE INDENT ans = K NEW_LINE break NEW_LINE DEDENT ispalindrome = 1 NEW_LINE
Check if prefix of length K is palindrome
for i in range ( 0 , K // 2 ) : NEW_LINE
Check if not a palindrome
if ( A [ i ] != A [ K - 1 - i ] ) : NEW_LINE INDENT ispalindrome = 0 NEW_LINE DEDENT
If found to be palindrome
if ( ispalindrome == 1 ) : NEW_LINE INDENT ans = K // 2 NEW_LINE K = K // 2 NEW_LINE DEDENT
Otherwise
else : NEW_LINE INDENT ans = K NEW_LINE break NEW_LINE DEDENT
Return the final answer
return ans NEW_LINE
Driver code
A = [ 1 , 2 , 2 , 1 , 1 , 2 , 2 , 1 ] NEW_LINE N = len ( A ) NEW_LINE print ( minimumrequired ( A , N ) ) NEW_LINE
Function to find the sum of the product of all the integers and their positive divisors up to N
def sumOfFactors ( N ) : NEW_LINE INDENT ans = 0 NEW_LINE DEDENT
Iterate for every number between 1 and N
for i in range ( 1 , N + 1 ) : NEW_LINE
Find the first multiple of i between 1 and N
first = i NEW_LINE
Find the last multiple of i between 1 and N
last = ( N // i ) * i NEW_LINE
Find the total count of multiple of in [ 1 , N ]
factors = ( last - first ) // i + 1 NEW_LINE
Compute the contribution of i using the formula
totalContribution = ( ( ( factors * ( factors + 1 ) ) // 2 ) * i ) NEW_LINE
Add the contribution of i to the answer
ans += totalContribution NEW_LINE
Return the result
return ans NEW_LINE
Given N
N = 3 NEW_LINE
Function call
print ( sumOfFactors ( N ) ) NEW_LINE
Function to return M < N such that N ^ M - N & M is maximum
def getMaxDifference ( N ) : NEW_LINE
Initialize variables
M = - 1 ; NEW_LINE maxDiff = 0 ; NEW_LINE
Iterate for all values < N
for i in range ( N ) : NEW_LINE
Find the difference between Bitwise XOR and AND
diff = ( N ^ i ) - ( N & i ) ; NEW_LINE
Check if new difference is greater than previous maximum
if ( diff >= maxDiff ) : NEW_LINE
Update variables
maxDiff = diff ; NEW_LINE M = i ; NEW_LINE
Return the answer
return M ; NEW_LINE
Driver Code
if __name__ == ' _ _ main _ _ ' : NEW_LINE
Given number N
N = 6 ; NEW_LINE
Function call
print ( getMaxDifference ( N ) ) ; NEW_LINE
Python3 program for the above approach
import math NEW_LINE
Function to flip all bits of N
def findM ( N ) : NEW_LINE INDENT M = 0 ; NEW_LINE DEDENT
Finding most significant bit of N
MSB = int ( math . log ( N ) ) ; NEW_LINE
Calculating required number
for i in range ( MSB ) : NEW_LINE INDENT if ( ( N & ( 1 << i ) ) == 0 ) : NEW_LINE INDENT M += ( 1 << i ) ; NEW_LINE DEDENT DEDENT
Return the answer
return M ; NEW_LINE
Driver Code
if __name__ == ' _ _ main _ _ ' : NEW_LINE
Given number
N = 6 ; NEW_LINE
Function call
print ( findM ( N ) ) ; NEW_LINE
Matrix of containers
cont = [ [ 0 for i in range ( 1000 ) ] for j in range ( 1000 ) ] NEW_LINE
Function to find the number of containers that will be filled in X seconds
def num_of_containers ( n , x ) : NEW_LINE INDENT count = 0 NEW_LINE DEDENT
Container on top level
cont [ 1 ] [ 1 ] = x NEW_LINE for i in range ( 1 , n + 1 ) : NEW_LINE INDENT for j in range ( 1 , i + 1 ) : NEW_LINE DEDENT