S3Q2 · Visualize Pattern Lock

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Type: Full I/O -grid visualization

Core idea: build a lookup label[dot_position] = visit_order, then render the 3×3 grid row by row.

seq = list(map(int, input().split()))
label = [''] * 10                    # 1-indexed, label[i] = display for dot i
for order, dot in enumerate(seq, 1):
    label[dot] = str(order)

def cell(dot):
    return f"[{label[dot] if label[dot] else ' '}]"

for row in range(3):
    dots = [row*3+1, row*3+2, row*3+3]
    print(cell(dots[0]) + "---" + cell(dots[1]) + "---" + cell(dots[2]))
    if row < 2:
        print(" |     |     |")

Key rules: - label[dot] = visit order if visited, else ' ' (space) - Cell format: [n] for visited, [ ] for unvisited - Row separator: " | | |" between rows (not after last) - No trailing spaces

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Problem Statement

Problem (I/O type)

Given a sequence of dot numbers visited in order, print a 3×3 pattern lock grid. Each visited dot shows its visit order; unvisited dots show [ ].

Example:

Input

2 5 7 4 1

Output

[5]---[1]---[ ]
 |     |     |
[4]---[2]---[ ]
 |     |     |
[3]---[ ]---[ ]

---

Understanding the structure

Dot numbering:

1  2  3     ← row 0: dots 1,2,3
4  5  6     ← row 1: dots 4,5,6
7  8  9     ← row 2: dots 7,8,9

Input 2 5 7 4 1 means:

Visit order	Dot visited
1st	Dot 2
2nd	Dot 5
3rd	Dot 7
4th	Dot 4
5th	Dot 1

So the label at each dot position:

Dot	Visited?	Label
1	Yes -5th	[5]
2	Yes -1st	[1]
3	No	[ ]
4	Yes -4th	[4]
5	Yes -2nd	[2]
6	No	[ ]
7	Yes -3rd	[3]
8	No	[ ]
9	No	[ ]

Row 0: [5]---[1]---[ ]

Building the label lookup

enumerate(seq, 1) gives (1, dot2), (2, dot5), (3, dot7), ... -the visit order and the dot number simultaneously. Storing label[dot] = order gives O(1) lookup when rendering each cell.

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Rendering the grid

Each content row: cell(d1) + "---" + cell(d2) + "---" + cell(d3)

Separator row: " | | |" -printed between content rows only (not after the last).

The separator spacing matches the 3-character cell width: [ ] is 3 chars, --- is 3 chars, | with spaces aligns under the cell centres.

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Solution approaches

PythonicExplanatoryUsing lambda + map

seq = list(map(int, input().split()))

label = [''] * 10    # 1-indexed; empty string = unvisited
for order, dot in enumerate(seq, 1):
    label[dot] = str(order)

def cell(dot):
    v = label[dot]
    return f"[{v if v else ' '}]"

for row in range(3):
    d1, d2, d3 = row*3+1, row*3+2, row*3+3
    print(f"{cell(d1)}---{cell(d2)}---{cell(d3)}")
    if row < 2:
        print(" |     |     |")

seq = list(map(int, input().split()))

# Build label lookup: label[dot_number] = visit order string
label = {i: ' ' for i in range(1, 10)}     # default: space (unvisited)
for order, dot in enumerate(seq, 1):
    label[dot] = str(order)

# Render row by row
for row in range(3):
    cells = []
    for col in range(3):
        dot = row * 3 + col + 1
        cells.append(f"[{label[dot]}]")
    print("---".join(cells))
    if row < 2:
        print(" |     |     |")

Uses a dict for the label lookup. "---".join(cells) assembles the three cells cleanly.

seq = list(map(int, input().split()))

label = {i: ' ' for i in range(1, 10)}
for order, dot in enumerate(seq, 1):
    label[dot] = str(order)

make_row = lambda r: "---".join(
    map(lambda c: f"[{label[r*3+c+1]}]", range(3))
)

rows = list(map(make_row, range(3)))
sep = " |     |     |"
print(("\n" + sep + "\n").join(rows))

make_row lambda builds one content row. ("\n" + sep + "\n").join(rows) weaves separators between the three rows in one expression.

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Key takeaways

01

enumerate(seq, 1) for 1-based ordering

enumerate(seq, 1) gives (1, first_item), (2, second_item), .... Use the start argument to avoid adding 1 manually throughout the code.

02

dot = row*3 + col + 1

Maps a (row, col) pair to the dot number 1–9. Row 0 col 0 → dot 1; row 2 col 2 → dot 9. This avoids hardcoding the grid layout.

03

Separator only between rows

Print the separator " | | |" only when row < 2. A common mistake is printing it after every row including the last -use the condition to suppress the trailing separator.