Python program to right rotate n-numbers by 1
Last Updated :
07 Feb, 2023
Given a number n. The task is to print n-integers n-times (starting from 1) and right rotate the integers by after each iteration.
Examples:
Input: 6
Output :
1 2 3 4 5 6
2 3 4 5 6 1
3 4 5 6 1 2
4 5 6 1 2 3
5 6 1 2 3 4
6 1 2 3 4 5
Input : 3
Output :
1 2 3
2 3 1
3 1 2
Method 1:
Below is the implementation.
Python3
def print_pattern(n):
for i in range(1, n+1, 1):
for j in range(1, n+1, 1):
if i == j:
print(j, end=" ")
if i <= j:
for k in range(j+1, n+1, 1):
print(k, end=" ")
for p in range(1, j, 1):
print(p, end=" ")
print()
print_pattern(3)
|
Method 2: Using pop(),append() and loops
Python3
def print_pattern(n):
x = []
for i in range(1, n+1):
x.append(i)
print(i, end=" ")
print()
j = 0
while(j < n-1):
a = x[0]
x.pop(0)
x.append(a)
for k in x:
print(k, end=" ")
print()
j += 1
print_pattern(6)
|
Output
1 2 3 4 5 6
2 3 4 5 6 1
3 4 5 6 1 2
4 5 6 1 2 3
5 6 1 2 3 4
6 1 2 3 4 5
Using modulus operator: In this approach, we are using the modulus operator and adding i to the loop variable j to get the current number in each iteration. The time complexity of this approach is O(n^2) as it requires two nested loops. The space complexity is O(1) as we are only using a few variables and not using any extra data structures.
Python3
def print_pattern(n):
for i in range(n):
for j in range(n):
print((j+i)%n+1, end=" ")
print()
print_pattern(6)
|
Output
1 2 3 4 5 6
2 3 4 5 6 1
3 4 5 6 1 2
4 5 6 1 2 3
5 6 1 2 3 4
6 1 2 3 4 5
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