CF 102697105 - Planet of Visions

We are given two strings of capital letters. The first string is the text that actually appears on the eye chart, while the second is the text that was read. The strings have the same length, so every position in the first string corresponds to exactly one position in the second.

codeforcescompetitive-programming

CF 102697105 - Planet of Visions

Rating: -
Tags: -
Solve time: 2m 31s
Verified: yes

Solution

Problem Understanding

We are given two strings of capital letters. The first string is the text that actually appears on the eye chart, while the second is the text that was read. The strings have the same length, so every position in the first string corresponds to exactly one position in the second.

Some pairs of letters are declared visually similar. Similarity is transitive: if D looks similar to P, and P looks similar to G, then D, P, and G all belong to the same similarity group. A character is classified as correct when the two letters at its position are identical. If they are different but belong to the same similarity group, the character is considered close. Otherwise it is completely wrong.

The input gives the correct string, the interpreted string, and then the number of similarity relationships followed by the pairs themselves. The output contains three counts, in the order correct, close, and wrong.

The most useful constraint here is not a large numeric bound, but the fact that every letter is an uppercase English letter. The similarity graph therefore has only 26 possible vertices. Even if the strings are very long, the graph itself is tiny. A solution should consequently spend essentially linear time reading the relationships and scanning the two strings, rather than repeatedly searching the graph.

The page for this problem specifies a 1 second time limit and 256 MB memory limit, but does not publish explicit upper bounds for the string length or the number of similarity pairs. Since those values can be large enough to make repeated graph searches undesirable, the natural target is O(|s| + n) time.

A first edge case is identical letters. Consider:

A
A
1
A B

The output is:

1 0 0

The explicit similarity relation must not turn this into a close match. Equality has priority, because a letter that was read exactly correctly is counted as correct.

A second edge case is transitive similarity. Consider:

A
C
2
A B
B C

The output is:

0 1 0

A careless implementation that checks only whether A C appears directly among the input pairs would report the character as wrong. The problem defines similarity through the entire connected component, so the path A -> B -> C is sufficient.

A third edge case is a similarity relationship that has nothing to do with the letters being compared:

A
C
1
B C

The output is:

0 0 1

The fact that C resembles B does not make it resemble A. Similarity must be established between the two particular letters through the graph.

Approaches

The direct approach is to build a graph whose vertices are the 26 letters and whose edges are the declared similarity pairs. For every position in the two strings, we compare the corresponding letters. If they are equal, we increment the correct counter. Otherwise, we could run a graph search from the correct letter and check whether the interpreted letter is reachable. Reachability exactly matches the transitive definition of similarity, so this method is correct.

The problem with doing a search independently for every position is that the same tiny graph is searched again and again. If the strings have length L and there are n input pairs, one BFS or DFS can examine up to 26 vertices and 2n adjacency entries. Repeating that for every character can take up to L(26 + 2n) vertex and edge examinations. The input itself only needs to describe the graph once, so repeating this work is unnecessary.

The key observation is that transitive similarity is exactly connectedness in an undirected graph. Since the graph contains only 26 letters, we can compute its connected components once. A disjoint-set union structure, also called DSU or union-find, is particularly natural here. Whenever the input says two letters are similar, we merge their components. After all pairs have been processed, two different letters are similar precisely when their DSU representatives are equal.

The brute force works because it answers each similarity question by explicitly following paths in the graph, but fails when there are many positions in the strings. The observation that all positions query the same 26-letter graph lets us preprocess the graph once. Each subsequent comparison then becomes a pair of nearly constant-time DSU operations.

Approach Time Complexity Space Complexity Verdict
Brute Force O(L(26 + n)) O(26 + n) Too slow for large L and n
Optimal O(n α(26) + L α(26)) O(26 + n) Accepted

Here L is the string length and α is the inverse Ackermann function, which is effectively constant for any practical input size.

Algorithm Walkthrough

  1. Create a DSU containing one separate component for each of the 26 uppercase letters. Initially, no two distinct letters are known to be similar.
  2. Read every similarity pair and union the two corresponding letters. If several pairs connect through intermediate letters, repeated unions naturally merge the whole connected component. For example, processing A B and then B C makes A, B, and C share one representative.
  3. Scan the two strings position by position. If the two letters are equal, increment the correct counter immediately. This check must happen before the similarity check because identical letters are classified as correct, not merely close.
  4. For different letters, compare their DSU representatives. Equal representatives mean the letters are in the same similarity component, so increment the close counter. Different representatives mean there is no chain of similarity relationships between them, so increment the wrong counter.
  5. Print the three counters in the required order.

The invariant is that after all similarity pairs have been processed, two letters have the same DSU representative if and only if there is a chain of declared similarity relationships connecting them. Every union adds exactly one declared relationship and therefore cannot merge unrelated components. Conversely, every path of relationships is processed edge by edge, so all letters on that path eventually receive the same representative. Each string position is then classified according to exactly the definition of correct, close, or wrong.

Python Solution

import sys
input = sys.stdin.readline

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n

    def find(self, x):
        while self.parent[x] != x:
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x

    def union(self, a, b):
        a = self.find(a)
        b = self.find(b)

        if a == b:
            return

        if self.size[a] < self.size[b]:
            a, b = b, a

        self.parent[b] = a
        self.size[a] += self.size[b]

def solve():
    s = input().strip()
    t = input().strip()
    n = int(input())

    dsu = DSU(26)

    for _ in range(n):
        a, b = input().split()
        x = ord(a) - ord('A')
        y = ord(b) - ord('A')
        dsu.union(x, y)

    correct = 0
    close = 0
    wrong = 0

    for a, b in zip(s, t):
        if a == b:
            correct += 1
        else:
            x = ord(a) - ord('A')
            y = ord(b) - ord('A')

            if dsu.find(x) == dsu.find(y):
                close += 1
            else:
                wrong += 1

    print(correct, close, wrong)

if __name__ == "__main__":
    solve()

The DSU class stores one parent for each letter. find returns the representative of a component and performs path compression, while union combines two components and attaches the smaller one below the larger one.

The conversion ord(ch) - ord('A') maps A through Z to integers 0 through 25. This keeps the graph representation compact and avoids dictionaries or string-based graph nodes.

The similarity pairs are processed before the strings are examined. That ordering matters because a pair such as A B, followed later by B C, must already have merged all three letters when the comparison A versus C is made.

The comparison loop explicitly tests a == b first. Without that branch, an identical pair whose letters happen to lie in the same DSU component would incorrectly be counted as close.

Python integers have arbitrary precision, and all counters are at most the string length, so there is no overflow concern.

Worked Examples

For Sample 1, the input is:

DEFPOTEC
PEFDOTED
3
D P
C D
A Z

The two similarity edges involving the relevant letters merge D, P, and C into one component.

Position Correct Read Classification Correct Close Wrong
0 D P close 0 1 0
1 E E correct 1 1 0
2 F F correct 2 1 0
3 P D close 2 2 0
4 O O correct 3 2 0
5 T T correct 4 2 0
6 E E correct 5 2 0
7 C D close 5 3 0

The final output is 5 3 0. This demonstrates both the equality check and the fact that several different letters can belong to one similarity component.

For Sample 2, the relationships form several chains:

A - H - D
B - Z - Y - E
F - C - T

The comparison trace is:

Position Correct Read Classification Correct Close Wrong
0 D A wrong 0 0 1
1 E B close 0 1 1
2 F C close 0 2 1
3 P D wrong 0 2 2
4 O E wrong 0 2 3
5 T F close 0 3 3
6 E G wrong 0 3 4
7 C H close 0 4 4

The output is 0 4 4. The example exercises transitive similarity because B and E are not directly paired, but the chain B -> Z -> Y -> E puts them in the same component.

Complexity Analysis

Measure Complexity Explanation
Time O(n α(26) + L α(26)) Each similarity pair performs one union, and each string position performs at most two finds
Space O(n + 26) The DSU uses 26 nodes, while the input itself requires storage for the strings and pairs read by the program

Since the letter universe has only 26 vertices, α(26) is effectively a constant. The practical running time is thus linear in the number of similarity relationships and the length of the strings. This is comfortably suited to the 1 second and 256 MB limits specified by the problem.

Test Cases

The official statement provides two samples. The custom cases below additionally cover a one-character input, transitive similarity, all equal characters, and a large string stress test.

import sys
import io

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n

    def find(self, x):
        while self.parent[x] != x:
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x

    def union(self, a, b):
        a = self.find(a)
        b = self.find(b)

        if a == b:
            return

        if self.size[a] < self.size[b]:
            a, b = b, a

        self.parent[b] = a
        self.size[a] += self.size[b]

def solve():
    input = sys.stdin.readline

    s = input().strip()
    t = input().strip()
    n = int(input())

    dsu = DSU(26)

    for _ in range(n):
        a, b = input().split()
        dsu.union(ord(a) - 65, ord(b) - 65)

    correct = close = wrong = 0

    for a, b in zip(s, t):
        if a == b:
            correct += 1
        elif dsu.find(ord(a) - 65) == dsu.find(ord(b) - 65):
            close += 1
        else:
            wrong += 1

    return f"{correct} {close} {wrong}\n"

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()
    result = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout

    return result

# Official sample 1
assert run(
    """DEFPOTEC
PEFDOTED
3
D P
C D
A Z
"""
) == "5 3 0\n"

# Official sample 2
assert run(
    """DEFPOTEC
ABCDEFGH
7
A H
D H
B Z
Z Y
Y E
F C
C T
"""
) == "0 4 4\n"

# Minimum-size strings, with no useful similarity relation.
assert run(
    """A
B
1
C D
"""
) == "0 0 1\n"

# Transitive similarity: A and C are close through B.
assert run(
    """A
C
2
A B
B C
"""
) == "0 1 0\n"

# All positions are already correct, even though every letter is connected.
assert run(
    """AAAAAA
AAAAAA
2
A B
B C
"""
) == "6 0 0\n"

# Large input stress test.
large_s = "A" * 100000
large_t = "C" * 100000
assert run(
    large_s + "\n" +
    large_t + "\n" +
    "2\n" +
    "A B\n" +
    "B C\n"
) == "0 100000 0\n"
Test input Expected output What it validates
A / B / C D 0 0 1 Minimum-size input and unrelated similarity
A / C / A B, B C 0 1 0 Transitive connectivity
AAAAAA / AAAAAA / A B, B C 6 0 0 Equality taking priority over similarity
100000 A characters versus 100000 C characters with A-B-C 0 100000 0 Linear processing and large string handling

Edge Cases

For identical letters, the input

A
A
1
A B

starts with A == A, so the algorithm increments correct and never checks the DSU. The result is 1 0 0. This prevents equality from being counted as a weaker form of similarity.

For transitive similarity, consider:

A
C
2
A B
B C

The first union creates the component {A, B}. The second union merges C into that component, giving {A, B, C}. When the strings are compared, A != C, but both letters have the same representative, so the result is 0 1 0.

For unrelated letters, consider:

A
C
1
B C

The only component containing more than one letter is {B, C}. A remains isolated, so find(A) != find(C). The result is 0 0 1.

A final boundary case is a long string in which every position requires a transitive lookup. With 100000 copies of A in the correct string, 100000 copies of C in the interpreted string, and relationships A B and B C, every position is close. The DSU is built once, and each position performs only constant-time component checks, producing 0 100000 0. This is exactly the situation where preprocessing the 26-letter graph is preferable to running a separate graph search for every character.