S3Q2 · Pattern Printing - Z Pattern¶

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Type: Full I/O - pattern printing

Core idea: top and bottom rows are full stars; diagonal rows have a star at position n - i (0-indexed) with leading spaces only - no trailing spaces.

n = int(input())
for i in range(1, n + 1):
    if i == 1 or i == n:
        print("*" * n)
    else:
        print(" " * (n - i) + "*")

Key rules: - Row 1 and row n: full row of n stars - Rows 2 to n-1: n - i leading spaces then * (no trailing spaces) - i is 1-indexed

Problem Statement¶

Problem (I/O type)

Given n, print a Z-shaped pattern: - Top row: n stars - Diagonal: star moves from right to left as rows increase - Bottom row: n stars - No trailing spaces after the last * on any line

Examples:

Input

Output

*****
   *
  *
 *
*****

Input

Output

****
  *
 *
****

Deriving the pattern¶

For n = 5, mark where the * appears on each row (1-indexed):

Row `i`	Type	Leading spaces	Star position
1	Top border	-	All `n` stars
2	Diagonal	3	`n - i = 3`
3	Diagonal	2	`n - i = 2`
4	Diagonal	1	`n - i = 1`
5	Bottom border	-	All `n` stars

The diagonal star starts at the second-to-last column (n-2 in 0-indexed, which is n-i spaces for i=2) and moves one step left each row, reaching column 0 at row n-1.

Formula for diagonal rows: " " * (n - i) + "*"

No trailing spaces

The problem explicitly says no space after the last *. Since the diagonal rows have their star at a position left of the rightmost column, there would be trailing spaces if you tried to pad to width n. The formula " " * (n - i) + "*" naturally produces no trailing spaces.

Tracing n = 5¶

i	`n - i` spaces	Row printed
1	- (border)	`*****`
2	3	`···*`
3	2	`··*`
4	1	`·*`
5	- (border)	`*****`

(dots represent spaces)

Tracing n = 3¶

i	Row printed
1	`***`
2	`·*` (1 space + star)
3	`***`

Solution approaches¶

PythonicExplanatoryUsing f-stringUsing lambda

n = int(input())
for i in range(1, n + 1):
    if i == 1 or i == n:
        print("*" * n)
    else:
        print(" " * (n - i) + "*")

n = int(input())
for i in range(1, n + 1):
    if i == 1 or i == n:
        row = "*" * n                   # full row of stars
    else:
        leading_spaces = " " * (n - i)  # decreases each row
        row = leading_spaces + "*"       # no trailing spaces
    print(row)

n = int(input())
for i in range(1, n + 1):
    if i == 1 or i == n:
        print("*" * n)
    else:
        print(f"{'*':>{n - i + 1}}")

f"{'*':>{n - i + 1}}" right-aligns * in a field of width n - i + 1. This produces n - i leading spaces followed by * - no trailing spaces.

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

The lambda encapsulates the row-building logic - pass the row number, get the row string. Clean separation of the pattern logic from the loop.

Common mistake - trailing spaces¶

A common mistake is printing each row padded to width n:

# ❌ Wrong - adds trailing spaces to diagonal rows
print(f"{'*':>{n - i + 1}:<{n}}")   # pads right side too

The problem says no trailing spaces. Always end diagonal rows right after the *.

Key takeaways¶

01

Border rows vs diagonal rows

Pattern problems almost always have special cases for the first and last row. Handle them with if i == 1 or i == n and apply the general formula for everything in between.

02

No trailing spaces - end at the star

" " * (n - i) + "*" stops right after the star. Never pad to the full row width when trailing spaces are forbidden.

03

Diagonal moves left as i increases

For Z: star starts at the right and moves left. Leading spaces = n - i - decreases as i increases. Contrast with triangles where stars grow; here only one star moves.