CF 1326D1 - Prefix-Suffix Palindrome (Easy version)

CF 1326D1 - Prefix-Suffix Palindrome (Easy version)

Rating: 1500
Tags: hashing, string suffix structures, strings
Solve time: 1m 10s
Verified: yes

Solution

Problem Understanding

We are given a string consisting of lowercase English letters, and the task is to construct the longest palindrome by combining a prefix and a suffix of this string. Formally, we need a string t such that it can be split into two parts a and b with t = a + b, where a is a prefix of the original string and b is a suffix. The string t must itself be a palindrome and cannot exceed the original string in length. For each input string, we need to produce such a t of maximal length.

The constraints are relatively small. The number of test cases can be up to 1000, and the total sum of the lengths of all strings does not exceed 5000. This means that even an algorithm with quadratic complexity in the length of a single string is feasible, since the total number of character operations remains manageable. However, anything significantly worse than O(n^2) per string would be unnecessary.

Non-obvious edge cases appear when the longest palindrome is not strictly contained in the middle of the string or when the entire string is already a palindrome. For example, in the string acbba, the longest prefix-suffix palindrome is abba rather than acbba because the first and last characters do not match. A naive approach that only checks palindromes starting at the beginning or ending at the end would miss this. Another edge case is a string like codeforces, where the longest prefix-suffix palindrome reduces to a single character, either c or s, depending on which side we check first.

Approaches

The brute-force approach iterates over every possible split of the string into a prefix and a suffix, then concatenates them and checks whether the resulting string is a palindrome. For a string of length n, there are O(n^2) possible concatenations, and each palindrome check takes O(n). This leads to an O(n^3) time complexity per string, which is unnecessarily slow even for our constraints.

The key insight is that the maximal palindrome we can form has a specific structure: if the original string has a matching prefix and suffix, those characters can be included in the palindrome, and the remainder in the middle can be any palindrome. Thus, we can first find the longest matching prefix and suffix, strip them, and focus on the remaining middle substring. Within this middle substring, the longest palindrome is either its longest prefix palindrome or its longest suffix palindrome. By appending this inner palindrome between the matching prefix and suffix, we construct the maximal prefix-suffix palindrome efficiently.

This reduces the problem to three main operations: finding the longest prefix-suffix match, finding the longest palindromic prefix of a substring, and finding the longest palindromic suffix of a substring. Each operation can be performed in linear time relative to the substring length, giving an overall O(n) solution per string.

Algorithm Walkthrough

1. Initialize two pointers, l at the start of the string and r at the end, and move them inward as long as s[l] == s[r]. This identifies the maximal matching prefix and suffix.
2. Extract the middle substring mid that lies between l and r. This is the portion of the string that is not part of the matching prefix or suffix.
3. Find the longest palindromic prefix of mid. Start from the first character of mid and expand to the right until a palindrome fails. Keep track of the longest successful prefix.
4. Find the longest palindromic suffix of mid. Start from the last character of mid and expand to the left until a palindrome fails. Keep track of the longest successful suffix.
5. Compare the lengths of the palindromic prefix and suffix of mid and choose the longer one.
6. Concatenate the matching prefix, the chosen palindrome from mid, and the matching suffix to form the final result.

The algorithm works because the prefix and suffix that match at the ends are always part of any maximal prefix-suffix palindrome. Within the middle substring, the longest palindrome must lie either at the start or the end because including any non-edge palindrome would reduce the length of the total concatenated palindrome.

Python Solution

import sys
input = sys.stdin.readline

def longest_prefix_suffix_palindrome(s):
    n = len(s)
    l, r = 0, n - 1
    while l < r and s[l] == s[r]:
        l += 1
        r -= 1

    if l >= r:
        return s

    mid = s[l:r+1]

    # Longest palindromic prefix
    pref_len = 0
    for i in range(len(mid)):
        if mid[:i+1] == mid[:i+1][::-1]:
            pref_len = i+1

    # Longest palindromic suffix
    suff_len = 0
    for i in range(len(mid)):
        if mid[len(mid)-i-1:] == mid[len(mid)-i-1:][::-1]:
            suff_len = i+1

    middle = mid[:pref_len] if pref_len >= suff_len else mid[len(mid)-suff_len:]
    return s[:l] + middle + s[r+1:]

t = int(input())
for _ in range(t):
    s = input().strip()
    print(longest_prefix_suffix_palindrome(s))

The solution first determines the maximal matching prefix and suffix using two pointers. If the entire string matches in this manner, it is already a palindrome. Otherwise, the remaining middle substring is analyzed for its longest palindromic prefix and suffix. Care is taken to correctly slice substrings and avoid off-by-one errors, particularly in computing suffix lengths.

Worked Examples

Input: abcdfdcecba