Python Program for Print Number series without using any loop
Problem – Givens Two number N and K, our task is to subtract a number K from N until number(N) is greater than zero, once the N becomes negative or zero then we start adding K until that number become the original number(N). Note : Not allow to use any loop. Examples :
Input : N = 15 K = 5
Output : 15 10 5 0 1 5 10 15
Input : N = 20 K = 6
Output : 20 14 8 2 -4 2 8 14 20
Explanation – We can do it using recursion idea is that we call the function again and again until N is greater than zero (in every function call we subtract N by K). Once the number becomes negative or zero we start adding K in every function call until the number becomes the original number. Here we use a single function for both addition and subtraction but to switch between addition or subtraction function we used a Boolean flag.
Python3
def PrintNumber(N, Original, K, flag):
print (N, end = " ")
if (N < = 0 ):
if (flag = = 0 ):
flag = 1
else :
flag = 0
if (N = = Original and ( not (flag))):
return
if (flag = = True ):
PrintNumber(N - K, Original, K, flag)
return
if ( not (flag)):
PrintNumber(N + K, Original, K, flag);
return
N = 20
K = 6
PrintNumber(N, N, K, True )
|
Output :
20 14 8 2 -4 2 8 14 20
The time complexity is O(N/K), where N is the input number and K is the step value.
The auxiliary space is O(N/K)
Approach Name: Recursive Mathematical Approach using Function Calls.
Steps:
- Define a function print_series(N, K) that takes two parameters, N and K.
- Check if N is less than or equal to zero, if yes, print N and return.
- Print N and call the print_series(N-K, K) function recursively.
- Print N again.
Python3
def print_series(N, K):
if N < = 0 :
print (N, end = " " )
return
print (N, end = " " )
print_series(N - K, K)
print (N, end = " " )
N = 20
K = 6
print_series(N, K)
|
Output
20 14 8 2 -4 2 8 14 20
It has a time complexity of O(log(N/K)) and an auxiliary space of O(log(N/K)).
Please refer complete article on Print Number series without using any loop for more details!
Last Updated :
23 Apr, 2023
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