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S3Q2 · Print Pattern - N Shape

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Type: Full I/O - pattern printing

Core idea: each row has | on both sides with \ moving one step right each row. Row i has i-1 spaces before \ and n-i spaces after.

n = int(input())
for i in range(1, n + 1):
    print("|" + " " * (i - 1) + "\\" + " " * (n - i) + "|")

Key rules: - | on both sides of every row - \ at position i within the inner area (1-indexed) - Leading spaces before \: i - 1 - Trailing spaces after \ (before closing |): n - i - No trailing spaces after the final |

Problem Statement

Problem (I/O type)

Given n, print an N-shaped pattern. | appears on both sides of every row, and \ moves diagonally from top-left to bottom-right.

Examples:

Input 
3
Output 
|\  |
| \ |
|  \|
Input 
2
Output 
|\ |
| \|

Deriving the pattern

For n=3, the inner area between the | bars has width n:

Row 1:  |  \  ·  ·  |     leading=0, trailing=2
Row 2:  |  ·  \  ·  |     leading=1, trailing=1
Row 3:  |  ·  ·  \  |     leading=2, trailing=0
Row i Leading spaces Trailing spaces
1 i-1 = 0 n-i = 2
2 i-1 = 1 n-i = 1
3 i-1 = 2 n-i = 0

Row structure: | + (i-1) spaces + \ + (n-i) spaces + |

The trailing spaces (between \ and |) are inside the pattern - they're needed to keep the right | properly aligned.

No trailing spaces after the closing |

The problem says no spaces to the right of the pattern. The spaces inside the bars are part of the pattern. The closing | is always the last character - nothing follows it.

Tracing n = 3

i i-1 spaces \ n-i spaces Row
1 `` \ ·· |\··|
2 · \ · |·\·|
3 ·· \ `` |··\|

(dots = spaces)

Solution approaches

n = int(input())
for i in range(1, n + 1):
    print(f"|{' '*(i-1)}\\{' '*(n-i)}|")

n = int(input())
for i in range(1, n + 1):
    leading  = " " * (i - 1)
    trailing = " " * (n - i)
    print("|" + leading + "\\" + trailing + "|")

n = int(input())
for i in range(1, n + 1):
    # Place \ at position i-1 (0-indexed) in a field of width n
    inner = " " * (i - 1) + "\\" + " " * (n - i)
    print("|" + inner + "|")

n = int(input())
row = lambda i: "|" + " " * (i-1) + "\\" + " " * (n-i) + "|"
for i in range(1, n+1):
    print(row(i))

Edge case: n = 1

i=1: leading=0, trailing=0 → | + + `\` + + | = |\|

Key takeaways

01

Diagonal \ at row i: leading = i-1

The backslash moves one step right per row. At row i (1-indexed), it has i-1 spaces before it and n-i spaces after - both decreasing and increasing symmetrically.

02

Spaces inside bars are part of the pattern

The trailing spaces between \ and the closing | keep the right bar aligned. They're inside the pattern, not trailing whitespace - so they must stay.

03

Always escape backslash: \\

In Python strings, \ is an escape character. Write "\\" to get a literal backslash in output. In f-strings, place "\\" outside the braces.