S3Q2 · Horizontal Bar Chart

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⚡ Quick Reference

Type: Full I/O -formatted output

Core idea: parse all pairs first, find the longest key, then print each key right-aligned in that width followed by : and # × value.

n = int(input())
pairs = [input().split(":") for _ in range(n)]
width = max(len(k) for k, _ in pairs)
for key, val in pairs:
    print(f"{key:>{width}}:{'#' * int(val)}")

Key rules: - Must read all pairs before printing -need max key length first - f"{key:>{width}}" right-aligns key in a field of width characters - Bars are '#' * count -no spaces between : and #

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Problem Statement

Problem (I/O type)

Read n key-value pairs (separated by :). Print a horizontal bar chart where keys are right-aligned so all colons line up vertically.

Example:

Input

6
apple:10
banana:5
plum:6
cherry:3
dates:8
strawberry:15

Output

     apple:##########
    banana:#####
      plum:######
    cherry:###
     dates:########
strawberry:###############

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Understanding the problem

Two passes are needed:

1. First pass -read all pairs, find the longest key ("strawberry" = 10 chars)
2. Second pass -print each key right-aligned in a field of width 10

"strawberry" → width = 10

     apple  → 5 leading spaces (10 - 5 = 5)
    banana  → 4 leading spaces (10 - 6 = 4)
      plum  → 6 leading spaces (10 - 4 = 6)

f-string right-alignment: f\"{s:>width}\"

f"{key:>{width}}" right-aligns key in a field of width characters, padding with spaces on the left. The > means right-align; < would be left-align; ^ would be centred.

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Tracing the example

width = len("strawberry") = 10

Key	len	Padding	Bar	Printed line
apple	5	5 spaces	##########	·····apple:##########
banana	6	4 spaces	#####	····banana:#####
plum	4	6 spaces	######	······plum:######
cherry	6	4 spaces	###	····cherry:###
dates	5	5 spaces	########	·····dates:########
strawberry	10	0 spaces	###############	strawberry:###############

(dots represent spaces)

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Solution approaches

Pythonic (f-string)ExplanatoryUsing str.rjust()Using lambda + map

n = int(input())
pairs = [input().split(":") for _ in range(n)]
width = max(len(k) for k, _ in pairs)
for key, val in pairs:
    print(f"{key:>{width}}:{'#' * int(val)}")

Read all into a list, compute width once, then print with f-string alignment.

n = int(input())
pairs = []
for _ in range(n):
    line = input()
    key, val = line.split(":")
    pairs.append((key, int(val)))

width = max(len(key) for key, _ in pairs)

for key, val in pairs:
    padding = " " * (width - len(key))
    bar = "#" * val
    print(f"{padding}{key}:{bar}")

Explicit padding calculation -width - len(key) leading spaces. Every step named for clarity.

n = int(input())
pairs = [input().split(":") for _ in range(n)]
width = max(len(k) for k, _ in pairs)
for key, val in pairs:
    print(key.rjust(width) + ":" + "#" * int(val))

str.rjust(width) right-justifies a string in a field of width characters -equivalent to f"{key:>{width}}" but without an f-string.

n = int(input())
pairs = [input().split(":") for _ in range(n)]
width = max(len(k) for k, _ in pairs)
output = list(map(
    lambda kv: f"{kv[0]:>{width}}:{'#' * int(kv[1])}",
    pairs
))
print("\n".join(output))

map(lambda kv: ..., pairs) applies the formatting to every pair. "\n".join(output) prints all lines in one call.

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Key takeaways

01

Read all before printing

Right-alignment requires knowing the maximum key length first. Always collect all input before producing any output when the formatting depends on the full dataset.

02

f"{s:>{width}}" for right-alignment

The > in a format spec means right-align. < = left, ^ = centre. These work for any type inside an f-string -strings, numbers, anything.

03

"#" * n for bars

String multiplication "#" * n creates a bar of exactly n characters. The same pattern works for any character-based visualisation -stars, dashes, equals signs.