CF 102697141 - Majestic Strings

CF 102697141 - Majestic Strings

Rating: -
Tags: -
Solve time: 42s
Verified: yes

Solution

Problem Understanding

A string is called majestic when every character is immediately followed by the next character in the alphabet. The alphabet is circular, so after z comes a. For example, cdefghij is majestic because every transition advances by one letter, while acd is not because a is followed by c instead of b. Similarly, xyzab is valid because the transition from z to a is allowed.

The input consists of one string s. The output should be YES if every adjacent pair follows this rule, and NO as soon as at least one adjacent pair does not.

The official statement gives a 1 second time limit and 256 MB of memory. It does not expose a separate maximum length for the string, so the safe design is to make the running time linear in the number of characters. A single pass performs one constant-time check per adjacent pair, which is optimal because an invalid character transition can occur anywhere in the input and we may need to inspect the entire string. A quadratic or exponential method would become unnecessarily expensive as the string grows.

There are a few small cases that can make an implementation fail silently. A one-character string such as a has no adjacent pair at all, so it is majestic and the correct output is YES. Code that assumes there is always a pair to inspect can incorrectly reject it.

The wraparound from z to a is another essential case. For input

yzab

the correct output is YES. A check that simply requires the ASCII value of the next character to be exactly one larger would reject za, even though the problem explicitly treats the alphabet as circular.

A repeated character must also be rejected. For input

hellllo

the correct output is NO. In particular, the repeated l characters do not advance through the alphabet, so checking only whether characters are in alphabetical order would be insufficient.

Approaches

A completely brute-force interpretation would generate every possible string of the same length and test whether each candidate is majestic, stopping when the input itself is found. For a string of length n, there are 26^n possible lowercase strings, and checking one candidate costs n - 1 adjacent comparisons. In the worst case this gives approximately 26^n(n - 1) character checks. The method is correct because it explicitly considers every possible string, but its exponential growth makes it useless even for modest values of n.

The brute-force approach works because it eventually checks exactly the property we care about, but it spends almost all of its work considering strings that have nothing to do with the input. The key observation is that we do not need to search for a majestic string at all. The input is already fixed, and being majestic is determined independently by each adjacent pair.

For a character c, the only valid next character is (c + 1) mod 26 when characters are represented by their zero-based alphabet positions. We can consequently scan the input once. At position i, we compare s[i + 1] with the required successor of s[i]. A mismatch proves immediately that the whole string is not majestic. If the scan reaches the end without a mismatch, every adjacent transition is valid and the string is majestic.

This changes the work from exponential in the string length to linear, while using constant extra space.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(26^n · n)	O(n)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the single input string and remove the trailing newline.
2. Traverse every adjacent pair s[i], s[i + 1]. There are exactly n - 1 such pairs for a string of length n, so no transition is skipped.
3. Convert the current character into a zero-based alphabet position. Its required successor is one position later, wrapping from z back to a.
4. Compare that required successor with s[i + 1]. If they differ, print NO immediately. One invalid transition is enough to make the entire string non-majestic, so examining later characters cannot change the answer.
5. If every adjacent pair passes the check, print YES. At that point every transition in the complete string is valid.

Why it works

The invariant is that after processing the first i transitions, every transition among the first i + 1 characters is valid. When the next pair is checked, the algorithm rejects exactly when that pair violates the definition. If it does not reject, the invariant extends to the next pair. After all n - 1 transitions have been processed, every adjacent pair is valid, which is precisely the definition of a majestic string. Thus the algorithm accepts exactly the valid strings.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().strip()

    for i in range(len(s) - 1):
        current = ord(s[i]) - ord('a')
        expected = (current + 1) % 26
        actual = ord(s[i + 1]) - ord('a')

        if actual != expected:
            print("NO")
            return

    print("YES")

if __name__ == "__main__":
    solve()

The loop corresponds directly to the adjacent-pair scan in the algorithm. ord(s[i]) - ord('a') maps a through z to 0 through 25. Adding one and taking modulo 26 handles both ordinary transitions, such as c -> d, and the special transition z -> a.

The range ends at len(s) - 1 because the last character has no character after it. Using range(len(s)) would attempt to access s[i + 1] when i is the final index and cause an out-of-bounds error.

The single-character case naturally works because the loop executes zero times and the algorithm prints YES. No special branch is necessary.

Python integers do not overflow, and the only arithmetic performed is on values from 0 through 25. The solution also stops at the first invalid transition, although its worst-case complexity remains linear.

Worked Examples

Sample 1

For the input cdefghij, the alphabet positions are 2, 3, 4, 5, 6, 7, 8, 9. Every character is exactly one position after its predecessor.

i	Current	Expected next	Actual next	Result
0	c	d	d	valid
1	d	e	e	valid
2	e	f	f	valid
3	f	g	g	valid
4	g	h	h	valid
5	h	i	i	valid
6	i	j	j	valid

The scan reaches the end without finding a bad transition, so the output is YES. This demonstrates the normal case where every pair satisfies the invariant.

Sample 2

For the input tuvwxyzab, the scan eventually reaches the circular transition from z to a.

i	Current	Expected next	Actual next	Result
0	t	u	u	valid
1	u	v	v	valid
2	v	w	w	valid
3	w	x	x	valid
4	x	y	y	valid
5	y	z	z	valid
6	z	a	a	valid
7	a	b	b	valid

The modulo operation makes z have successor a, so the entire string is accepted. The output is YES.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each of the n - 1 adjacent pairs is inspected once.
Space	O(1)	Only a constant number of character positions and integer values are stored.

The linear scan is appropriate for the problem's 1 second limit because it does only constant work per input character. The memory usage is also independent of the string length apart from the input string itself.

Test Cases

The official samples are cdefghij -> YES, tuvwxyzab -> YES, and hellllo -> NO.

# helper: run solution on input string, return output string
import sys
import io

def solve():
    s = input().strip()

    for i in range(len(s) - 1):
        current = ord(s[i]) - ord('a')
        expected = (current + 1) % 26
        actual = ord(s[i + 1]) - ord('a')

        if actual != expected:
            print("NO")
            return

    print("YES")

def run(inp: str) -> str:
    global input
    old_stdin = sys.stdin
    old_input = input

    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    try:
        output = io.StringIO()
        old_stdout = sys.stdout
        sys.stdout = output
        try:
            solve()
        finally:
            sys.stdout = old_stdout
        return output.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# provided samples
assert run("cdefghij\n") == "YES\n", "sample 1"
assert run("tuvwxyzab\n") == "YES\n", "sample 2"
assert run("hellllo\n") == "NO\n", "sample 3"

# minimum-size input
assert run("a\n") == "YES\n", "single character has no invalid transition"

# all-equal values
assert run("aaaaa\n") == "NO\n", "equal consecutive characters are invalid"

# wraparound at z -> a
assert run("xyzab\n") == "YES\n", "alphabet must wrap around"

# invalid transition at the final pair
assert run("abcdefgq\n") == "NO\n", "invalid transition at the end"

# large input
large = "a" + "bcdefghijklmnopqrstuvwxyz" * 4000
assert run(large + "\n") == "NO\n", "large input with a boundary mismatch"

Test input	Expected output	What it validates
a	YES	Minimum-size input and zero adjacent pairs
aaaaa	NO	Repeated characters are rejected
xyzab	YES	Correct z -> a wraparound
abcdefgq	NO	A mismatch at the final comparison is detected
Large constructed string	NO	Linear scanning remains practical on large input

Edge Cases

The one-character input a is handled by the exact input a\n. The loop has no iterations because there is no adjacent pair. The algorithm prints YES, which follows directly from the definition: there is no transition that violates the rule.

The wraparound case xyzab exercises the transition that a plain character-value comparison often mishandles. The relevant part of the execution is x -> y, y -> z, z -> a, and a -> b. For z, the zero-based value is 25, so (25 + 1) % 26 becomes 0, which corresponds to a. Every pair passes and the result is YES.

The repeated-character case aaaaa fails at the first comparison. The current character is a, whose required successor is b, while the actual next character is a. The algorithm immediately prints NO, rather than incorrectly treating the string as alphabetically nondecreasing.

Finally, abcdefgq demonstrates that the implementation does not merely check a prefix. The first six transitions are valid, but g -> q is invalid. The scan reaches that final pair, detects the mismatch, and prints NO. This is why checking every adjacent pair, rather than stopping after confirming an initial increasing run, is necessary.