Python Program to Print Odd Numbers from 1 to n

Use a for loop with a step of 2 starting from 1 up to n like for i in range(1, n+1, 2): print(i) to print odd numbers from 1 to n.

Examples

Input5

Output1 3 5

Input10

Output1 3 5 7 9

Input1

Output1

How to Think About It

To print odd numbers from 1 to n, start counting from 1 and move forward by adding 2 each time. This way, you only get odd numbers. Stop when you reach or pass n.

Algorithm

1

Get the input number n.

2

Start a loop from 1 to n, increasing by 2 each time.

3

Print the current number in the loop.

4

Stop when the current number is greater than n.

Code

python

n = int(input("Enter the value of n: "))
for i in range(1, n + 1, 2):
    print(i)

Output

Enter the value of n: 10
1
3
5
7
9

Dry Run

Let's trace the program with input n = 5 to see how it prints odd numbers.

1

Input

User enters n = 5

2

Loop start

Start loop with i = 1

3

Print i

Print 1

4

Next i

Increase i by 2, i = 3

5

Print i

Print 3

6

Next i

Increase i by 2, i = 5

7

Print i

Print 5

8

Next i

Increase i by 2, i = 7 (greater than n, stop loop)

i
1
3
5

Why This Works

Step 1: Start from 1

We start counting from 1 because it is the first odd number.

Step 2: Increment by 2

Adding 2 each time ensures we only get odd numbers (1, 3, 5, ...).

Step 3: Stop at n

We stop the loop when the number exceeds n to avoid printing numbers beyond the limit.

Alternative Approaches

Using if condition inside a loop

python

n = int(input("Enter the value of n: "))
for i in range(1, n + 1):
    if i % 2 != 0:
        print(i)

This checks every number from 1 to n and prints it only if it is odd; less efficient but easy to understand.

Using list comprehension and join

python

n = int(input("Enter the value of n: "))
print('\n'.join(str(i) for i in range(1, n + 1) if i % 2 != 0))

Creates a list of odd numbers and prints them all at once; concise but uses more memory.

Complexity: O(n) time, O(1) space

Time Complexity

The loop runs roughly n/2 times because it skips even numbers, so time complexity is O(n).

Space Complexity

The program uses constant extra space, O(1), as it prints numbers directly without storing them.

Which Approach is Fastest?

Using range(1, n+1, 2) is faster than checking each number with if because it skips even numbers entirely.

Approach	Time	Space	Best For
Range with step 2	O(n)	O(1)	Efficient and simple for printing odd numbers
If condition inside loop	O(n)	O(1)	Easy to understand but less efficient
List comprehension	O(n)	O(n)	Concise but uses extra memory for list