CF 102785J - R u really ready?

The task is to decide whether a text string can be generated by a pattern string. The pattern is made of ordinary lowercase letters and optional repetition symbols.

codeforcescompetitive-programming

CF 102785J - R u really ready?

Rating: -
Tags: -
Solve time: 48s
Verified: yes

Solution

Problem Understanding

The task is to decide whether a text string can be generated by a pattern string. The pattern is made of ordinary lowercase letters and optional repetition symbols. A + after a letter means that letter must appear at least once more after its first appearance, while a * means that letter may appear any number of times, including zero. Only the immediately preceding letter is affected by these symbols.

For example, the pattern ab*c can describe ac, abc, abbc, and so on. The pattern ab+c cannot describe ac because the b must appear at least once. The program must print Yes when the entire text can be described by the entire pattern, and No otherwise.

The strings are at most length 1000. A quadratic solution is comfortable because one million operations is small for a one second limit, while approaches that repeatedly try all possible repetitions can become much slower. A direct simulation that explores every possible way of expanding every * or + can branch heavily and approach exponential behavior, so the solution needs to remember already computed matching states.

Several details make this problem easy to get wrong. A pattern can contain a repetition symbol that matches nothing. For input:

a*
""

the answer is Yes, because * allows zero copies. A careless implementation that always consumes at least one character would output No.

A + cannot behave like *. For input:

a+
"a"

the answer is No, because + requires one or more repetitions in addition to the original symbol. Treating both symbols as identical would incorrectly accept it.

A repetition symbol only belongs to the previous character. For input:

ab*
"abb"

the answer is Yes, because only b repeats. An implementation that groups the pattern incorrectly could try to repeat ab and produce incorrect results.

Approaches

A natural first attempt is to recursively process the pattern and text together. When the current pattern character has no modifier, it either matches the current text character or fails. When it has * or +, the recursive search tries every possible number of repetitions and continues from each possible stopping position.

This approach is correct because every possible expansion of the pattern is considered. The problem is the number of repeated states. A pattern such as a*a*a*a*a* matched against a long string of a characters creates many different recursion paths that reach the same positions in the pattern and text. In the worst case, the number of explored possibilities grows exponentially, which is far beyond what length 1000 allows.

The useful observation is that the result only depends on two positions: which pattern token has already been processed and how many characters of the text have been consumed. There is no need to remember the exact history that reached a state.

We can convert the pattern into tokens, where each token contains a character and its repetition rule. Then dynamic programming stores whether the first several tokens can match the first several characters of the text. When processing a token, we compute all possible text prefixes it can consume.

The repeated-character cases can be optimized with a running scan. For *, a state is reachable either from a previous reachable state at the same position, or from the previous position if the current text character matches the token. For +, the same scan is used, but the token must consume at least one character before a state becomes valid.

Approach Time Complexity Space Complexity Verdict
Brute Force Exponential in the worst case O(pattern length + text length) Too slow
Dynamic Programming O(pattern length × text length) O(pattern length × text length) Accepted

Algorithm Walkthrough

  1. Parse the pattern into tokens. Each token stores a lowercase letter and whether it is a normal character, a + repetition, or a * repetition. This separates the syntax of the pattern from the matching logic.
  2. Create a dynamic programming array where dp[j] means that the tokens processed so far can match exactly the first j characters of the text. Initially, no tokens match an empty prefix, so only dp[0] is true.
  3. Process each token one at a time. For a normal character, move from position j to j + 1 only when the next text character is the same letter. A normal token consumes exactly one character.
  4. For a * token, scan the text from left to right while maintaining whether the current prefix can be reached. The empty usage of * keeps all previously reachable positions valid, and matching characters allow the token to extend the match by one more position.
  5. For a + token, use the same idea as *, but prevent the empty repetition case. The first matching character must be consumed before the token can contribute a valid state.
  6. After all tokens are processed, check whether the state for the complete text is reachable. If dp[len(text)] is true, the whole text matches the pattern.

Why it works: after every processed token, the dynamic programming array represents exactly all text prefixes that can be produced by those tokens. The transition rules enumerate every legal way the next token can consume characters, and they never allow an illegal repetition count. Since each token is processed using only states that already describe valid previous matches, the final state is true exactly when there exists a valid expansion of the whole pattern that equals the whole text.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    p = input().rstrip("\n")
    t = input().rstrip("\n")

    tokens = []
    i = 0
    while i < len(p):
        ch = p[i]
        kind = 0
        if i + 1 < len(p) and (p[i + 1] == '+' or p[i + 1] == '*'):
            kind = 1 if p[i + 1] == '+' else 2
            i += 1
        tokens.append((ch, kind))
        i += 1

    n = len(t)
    dp = [False] * (n + 1)
    dp[0] = True

    for ch, kind in tokens:
        ndp = [False] * (n + 1)

        if kind == 0:
            for j in range(n):
                if dp[j] and t[j] == ch:
                    ndp[j + 1] = True

        elif kind == 2:
            active = False
            for j in range(n + 1):
                if dp[j]:
                    active = True
                if active:
                    ndp[j] = True
                if j < n and t[j] != ch:
                    active = False

        else:
            active = False
            for j in range(n + 1):
                if dp[j]:
                    if j > 0:
                        active = True
                if active:
                    ndp[j] = True
                if j < n and t[j] != ch:
                    active = False

            for j in range(1, n + 1):
                if t[j - 1] == ch and dp[j - 1]:
                    ndp[j] = True

        dp = ndp

    print("Yes" if dp[n] else "No")

if __name__ == "__main__":
    solve()

The parser converts the raw pattern into a sequence of independent tokens. The variable kind distinguishes between a normal character, +, and *, which avoids repeatedly checking the original pattern while matching.

The dp array is updated after every token. For a plain character, only a single forward transition is possible because exactly one character must be consumed.

For *, the scan keeps an active flag. When a previous state is reachable, the token can match zero or more copies as long as consecutive text characters equal the token character. The flag is reset as soon as a different character appears.

For +, the implementation prevents the empty match. The extra loop handles the first required occurrence, while the scan allows additional copies afterward. This distinction is the most common source of mistakes in this problem.

The array indices represent positions between characters, so dp[j] refers to the prefix ending before character j. This convention avoids off-by-one errors when handling empty matches.

Worked Examples

Sample 1

Input:

pa*t+ern
pattern

The pattern becomes the tokens p, a*, t+, e, r, n.

Token processed Text position states reachable Reason
Start 0 Empty pattern prefix matches empty text prefix
p 1 Consumes p
a* 2 Consumes a once
t+ 3 Consumes required t
e 4 Consumes e
r 5 Consumes r
n 6 Consumes n

The final position is reachable, so the algorithm prints Yes. This trace shows that * can consume exactly one copy and + can consume the required copy.

Sample 2

Input:

c*cp+p
cpp

The tokens are c*, c, p+, p.

Token processed Text position states reachable Reason
Start 0 Empty prefix
c* 0, 1, 2 Zero, one, or two c characters are possible
c 3 Consumes the remaining c
p+ 4 Consumes one p
p 5 Consumes final p

The whole text is matched. This example demonstrates why the state must store all reachable positions rather than only one greedy choice.

Complexity Analysis

Measure Complexity Explanation
Time O(P × T) Each pattern token scans the text once
Space O(T) Only the current and next text-prefix state arrays are stored

Here P is the number of parsed pattern tokens and T is the text length. Both are at most 1000, so the dynamic programming approach performs about one million state operations and fits easily within the limits.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    p = input().rstrip("\n")
    t = input().rstrip("\n")

    tokens = []
    i = 0
    while i < len(p):
        ch = p[i]
        kind = 0
        if i + 1 < len(p) and p[i + 1] in "+*":
            kind = 1 if p[i + 1] == "+" else 2
            i += 1
        tokens.append((ch, kind))
        i += 1

    dp = [False] * (len(t) + 1)
    dp[0] = True

    for ch, kind in tokens:
        ndp = [False] * (len(t) + 1)

        if kind == 0:
            for j in range(len(t)):
                if dp[j] and t[j] == ch:
                    ndp[j + 1] = True
        elif kind == 2:
            active = False
            for j in range(len(t) + 1):
                if dp[j]:
                    active = True
                if active:
                    ndp[j] = True
                if j < len(t) and t[j] != ch:
                    active = False
        else:
            active = False
            for j in range(len(t) + 1):
                if dp[j] and j > 0:
                    active = True
                if active:
                    ndp[j] = True
                if j < len(t) and t[j] != ch:
                    active = False
            for j in range(1, len(t) + 1):
                if dp[j - 1] and t[j - 1] == ch:
                    ndp[j] = True

        dp = ndp

    return "Yes\n" if dp[len(t)] else "No\n"

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    return solve()

assert run("pa*t+ern\npattern\n") == "Yes\n", "sample 1"
assert run("c*cp+p\ncpp\n") == "Yes\n", "sample 2"
assert run("b+b\nb\n") == "No\n", "plus requires repetition"
assert run("a*\n\n") == "Yes\n", "star can match empty"
assert run("a+\na\n") == "No\n", "plus cannot match zero repetitions"
assert run("a*a*a*\naaaaa\n") == "Yes\n", "many repeated tokens"
assert run("ab*\na\n") == "Yes\n", "star can consume zero copies"
Test input Expected output What it validates
a*\n\n Yes Empty text matched by *
a+\na\n No Difference between + and *
a*a*a*\naaaaa\n Yes Multiple repetition tokens
ab*\na\n Yes A repetition token affects only its previous character

Edge Cases

For the empty match case:

a*

The parser creates one token a*. The initial state dp[0] is true, and the * transition keeps position zero reachable because using zero copies is legal. The final state is true, so the answer is Yes.

For the + minimum-length case:

a+
a

The token a+ must consume at least one character beyond the original symbol interpretation. The dynamic programming transition does not allow the empty position to survive, so the only text character is insufficient and the answer is No.

For the modifier scope case:

ab*
a

The tokens are a and b*. After matching a, the second token can consume zero b characters. The final text position remains reachable, giving Yes.

For consecutive repetition handling:

a*a*a*
aaaaa

Each token is processed independently. The first a* can consume some prefix, the next token continues from every reachable position, and the last token fills any remaining valid suffix. The DP states preserve all possible splits of the five a characters, so the algorithm accepts the pattern.