Overview

This is a problem where you compare two strings position by position and determine the restoration result one character at a time according to the given rules.

Analysis

In this problem, the restoration result at each position \(i\) is determined by looking at only the two characters \(S[i]\) and \(T[i]\), so each position can be processed independently.

The case analysis consists of the following 4 patterns:

For example, if \(S = \) ab?d?f and \(T = \) a?cd?g:

• Position 1: a and a → match → a
• Position 2: b and ? → adopt b
• Position 3: ? and c → adopt c
• Position 4: d and d → match → d
• Position 5: ? and ? → ?
• Position 6: f and g → contradiction → !

The restoration result is abcd?!, and since it contains !, we output Yes.

Since this problem involves simple case analysis, naively looping \(N\) times is sufficiently fast. No special algorithm is needed, but since \(N\) can be up to \(5 \times 10^5\), it is important to handle input and output efficiently.

Algorithm

1. Read strings \(S\) and \(T\).
2. For each position \(i = 0, 1, \ldots, N-1\), compare \(S[i]\) and \(T[i]\), and determine the restoration character according to the rules above.
3. Output the string formed by concatenating all restoration characters.
4. If the restoration result contains at least one !, output Yes; otherwise, output No.

Complexity

• Time complexity: \(O(N)\) — each position is examined exactly once
• Space complexity: \(O(N)\) — for storing the restoration result string

Implementation Tips

• Fast input: In Python, using sys.stdin.buffer.read() allows large inputs to be read quickly. This is far more efficient than calling input() \(N\) times.

• Append to a list, then join: Repeatedly concatenating strings can result in \(O(N^2)\) complexity, so the standard approach is to append to a list and then combine everything at once with ''.join(result).

• Flag management: Whether ! has appeared is tracked using a boolean variable has_conflict, and the final judgment is made at the end. This is more efficient than scanning the entire restoration string to search for ! (though the difference is only in constant factors).

  Source Code

import sys

def main():
    input_data = sys.stdin.buffer.read().decode()
    lines = input_data.split()
    N = int(lines[0])
    S = lines[1]
    T = lines[2]
    
    result = []
    has_conflict = False
    
    for i in range(N):
        s = S[i]
        t = T[i]
        if s == '?' and t == '?':
            result.append('?')
        elif s == '?':
            result.append(t)
        elif t == '?':
            result.append(s)
        elif s == t:
            result.append(s)
        else:
            result.append('!')
            has_conflict = True
    
    print(''.join(result))
    print('Yes' if has_conflict else 'No')

main()

This editorial was generated by claude4.6opus-thinking.