CF 2044B - Normal Problem

CF 2044B - Normal Problem

Rating: 800
Tags: implementation, strings
Solve time: 1m 30s
Verified: yes

Solution

Problem Understanding

We are given a string consisting only of the letters p, q, and w. Ship observes this string from outside a glass window, giving us string a. When Ship moves inside the store and looks through the same window, he sees string b. Our task is to determine what b looks like given a.

The key observation is that the window effectively mirrors certain characters. From the sample, we see p and q swap roles depending on the viewing direction, while w remains unchanged. For example, qwq observed outside becomes pwp inside. Therefore, there is a fixed mapping: p <-> q, and w -> w.

The constraints are small: each string has length at most 100, and there are at most 100 test cases. This means the total number of characters processed is at most 10,000. A simple linear scan per string is acceptable.

Non-obvious edge cases include strings with only w or strings where all characters are identical. For instance, wwww should remain wwww because w maps to itself. Another edge case is alternating characters, like pqpq, which requires us to apply the mapping consistently at every position.

Approaches

The brute-force approach is straightforward: for each character in the string a, check its value and replace it according to the mirror mapping. This requires examining every character and performing a simple substitution, which works because the total number of operations is only about 10,000 in the worst case.

The key insight for the optimal solution is realizing that we do not need any complicated data structures or algorithms. The mapping is constant and small (p, q, w). Using a dictionary or direct conditional statements allows us to translate each character in O(1) time. Because the string length is limited and the number of test cases is moderate, this linear-time approach is already optimal.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force / Direct Mapping	O(n) per string	O(n) per string	Accepted

There is no faster asymptotic solution because we must inspect every character to apply the mapping.

Algorithm Walkthrough

1. Define a mapping for the mirrored characters: p becomes q, q becomes p, and w stays w.
2. Read the number of test cases t.
3. For each test case, read the input string a.
4. Initialize an empty string or list to hold the result b.
5. Iterate through each character c in a:

• If c is p, append q to b.
• If c is q, append p to b.
• If c is w, append w to b.

6. After processing all characters, output b.

Why it works: The mapping preserves the exact behavior observed in the examples. Each character is independently transformed, so the final string b is exactly what Ship sees inside the store. The invariant is that after processing the first k characters, the partial string corresponds correctly to what Ship would see from inside.

Python Solution

import sys
input = sys.stdin.readline

# mapping function
mirror = {'p': 'q', 'q': 'p', 'w': 'w'}

t = int(input())
for _ in range(t):
    a = input().strip()
    b = ''.join(mirror[c] for c in a)
    print(b)

The solution reads all input efficiently using sys.stdin.readline. The mirror dictionary encodes the character transformation in a single place, reducing the chance of mistakes. Using ''.join() with a generator avoids unnecessary string concatenation in loops, which is more efficient in Python.

Worked Examples

Example 1: qwq

Step	Character c	Transformed	Partial b
1	q	p	p
2	w	w	pw
3	q	p	pwp

Output: pwp

This trace confirms that the mapping is applied consistently for each character.

Example 2: pppwwwqqq

Step	Character c	Transformed	Partial b
1	p	q	q
2	p	q	qq
3	p	q	qqq
4	w	w	qqqw
5	w	w	qqqww
6	w	w	qqqwww
7	q	p	qqqwwwp
8	q	p	qqqwwwpp
9	q	p	qqqwwwppp

Output: qqqwwwppp

This confirms the correct application over longer strings and shows that blocks of identical characters are handled correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per string	Each character is visited once and replaced in O(1)
Space	O(n) per string	Output string b stores transformed characters

Given the constraints (t <= 100 and |a| <= 100), the total operations are well under 10,000, which fits comfortably within the 1-second time limit. Memory usage is minimal since only the mapping and the output string are stored.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # solution
    mirror = {'p': 'q', 'q': 'p', 'w': 'w'}
    t = int(input())
    for _ in range(t):
        a = input().strip()
        b = ''.join(mirror[c] for c in a)
        print(b)
    sys.stdout = sys.__stdout__
    return output.getvalue().strip()

# provided samples
assert run("5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\n") == "pwp\nqqqqq\nqqqwwwppp\nqpwwpqwpqpw\npqpqpqpq"

# custom cases
assert run("1\nw\n") == "w", "single w"
assert run("1\np\n") == "q", "single p"
assert run("1\nq\n") == "p", "single q"
assert run("1\npppppppppp\n") == "qqqqqqqqqq", "all p's"
assert run("1\npqwqpwq\n") == "qwpwqpq", "mixed pattern"

Test input	Expected output	What it validates
w	w	single character mapping, self-mapped
p	q	single character mapping, swaps to q
q	p	single character mapping, swaps to p
pppppppppp	qqqqqqqqqq	long string of identical letters
pqwqpwq	qwpwqpq	alternating pattern, consistency

Edge Cases

If the input string contains only w, the algorithm leaves it unchanged. For example, input wwww produces wwww. The loop correctly maps each w to itself without error. For a string like pqpq, the mapping alternates between q and p, producing qpqp, which demonstrates correct handling of repeated swaps.