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S3Q2 · Dual Pattern Printing

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Type: Full I/O - two patterns separated by a blank line

n = int(input())

# Pattern 1 - symmetric concatenation
for i in range(1, n + 1):
    print("".join(str(j) for j in range(1, i + 1)))
for i in range(n - 1, 0, -1):
    print("".join(str(j) for j in range(1, i + 1)))

print()   # blank line

# Pattern 2 - triangular sequential numbers
num = 1
for i in range(1, n + 1):
    print(" ".join(str(num + j) for j in range(i)))
    num += i

Key rules: - Pattern 1: rows go 1 → n → 1, each row concatenates digits 1..i with no spaces - Pattern 2: rows have 1, 2, 3, … n numbers; numbers increase sequentially across rows - Blank line between the two patterns - No trailing spaces in Pattern 2

Problem Statement

Problem (I/O type)

Read n. Print two patterns separated by a blank line: 1. Symmetric concatenation pattern (ascending then descending) 2. Triangular sequential number pattern

Example - n = 4:

Input 
4
Output 
1
12
123
1234
123
12
1

1
2 3
4 5 6
7 8 9 10

Pattern 1 - Symmetric Concatenation

Each row i concatenates digits 1, 2, …, i with no spaces. The pattern goes up to n, then back down to 1:

i=1 → "1"
i=2 → "12"
i=3 → "123"
i=4 → "1234"   ← peak
i=3 → "123"
i=2 → "12"
i=1 → "1"

for i in range(1, n + 1):
    print("".join(str(j) for j in range(1, i + 1)))
for i in range(n - 1, 0, -1):        # n-1 down to 1
    print("".join(str(j) for j in range(1, i + 1)))

Total rows: 2n - 1 (n going up, n-1 going down).

Pattern 2 - Triangular Sequential Numbers

Row i contains i numbers, continuing from where the previous row left off:

Row 1: 1           (1 number)
Row 2: 2 3         (2 numbers)
Row 3: 4 5 6       (3 numbers)
Row 4: 7 8 9 10    (4 numbers)

Track a running counter num:

num = 1
for i in range(1, n + 1):
    print(" ".join(str(num + j) for j in range(i)))
    num += i

After row i, num advances by i - next row starts at the right value.

Tracing n = 4

Pattern 1:

i Output
1 1
2 12
3 123
4 1234
3 123
2 12
1 1

Pattern 2 (num tracker):

i num before Numbers num after
1 1 1 2
2 2 2 3 4
3 4 4 5 6 7
4 7 7 8 9 10 11

Solution approaches

n = int(input())

# Pattern 1
for i in range(1, n + 1):
    print("".join(str(j) for j in range(1, i + 1)))
for i in range(n - 1, 0, -1):
    print("".join(str(j) for j in range(1, i + 1)))

print()

# Pattern 2
num = 1
for i in range(1, n + 1):
    print(" ".join(str(num + j) for j in range(i)))
    num += i

n = int(input())

# Pattern 1 - ascending
for i in range(1, n + 1):
    row = ""
    for j in range(1, i + 1):
        row += str(j)
    print(row)

# Pattern 1 - descending
for i in range(n - 1, 0, -1):
    row = ""
    for j in range(1, i + 1):
        row += str(j)
    print(row)

print()   # blank line

# Pattern 2
num = 1
for i in range(1, n + 1):
    row = []
    for _ in range(i):
        row.append(str(num))
        num += 1
    print(" ".join(row))

def concat_row(i):
    return "".join(str(j) for j in range(1, i + 1))

n = int(input())

# Pattern 1
rows = [concat_row(i) for i in range(1, n + 1)]
print("\n".join(rows + rows[-2::-1]))

print()

# Pattern 2
num = 1
for i in range(1, n + 1):
    end = num + i
    print(" ".join(map(str, range(num, end))))
    num = end

rows[-2::-1] reverses the list starting from the second-to-last element - elegant way to get the descending half without the peak row being duplicated.

n = int(input())

concat = lambda i: "".join(map(str, range(1, i + 1)))

# Pattern 1
print("\n".join(map(concat, range(1, n + 1))))
print("\n".join(map(concat, range(n - 1, 0, -1))))

print()

# Pattern 2
num = 1
for i in range(1, n + 1):
    print(" ".join(map(str, range(num, num + i))))
    num += i

Key takeaways

01

Two loops for the symmetric pattern

The ascending half uses range(1, n+1); the descending half uses range(n-1, 0, -1). The peak row (n) appears only once - the descending loop starts at n-1.

02

Running counter for sequential numbers

Track num and advance it by i after each row. range(num, num+i) generates exactly i sequential numbers starting from num.

03

print() for a blank line

print() with no arguments prints an empty line - the separator between the two patterns. Cleaner than print("").