CF 102697116 - Diverse String

CF 102697116 - Diverse String

Rating: -
Tags: -
Solve time: 45s
Verified: yes

Solution

Problem Understanding

A string is called diverse when every character has a different character immediately after it. In other words, for every position except the last one, the character at that position must not equal the character at the next position.

The input contains one string with no spaces. The required output is YES when the string satisfies the condition everywhere, and NO as soon as there is at least one pair of equal adjacent characters. The original problem has a 1 second time limit and a 256 MB memory limit.

The key constraint for the algorithm is the length of the string. The statement does not give a useful small upper bound, so the solution should scale linearly with the input size. An O(n²) method would become impractical for a large string, while an O(n) scan only needs to inspect each character a constant number of times.

There are a few small cases where a careless implementation can fail. For input a, the correct output is YES, because there are no adjacent pairs at all. Code that assumes there is always a next character can access outside the string.

For input aa, the correct output is NO. The first character equals the second, so this is the smallest possible counterexample. A loop that accidentally stops before comparing the final adjacent pair could incorrectly print YES.

For input aba, the correct output is YES. The first pair is ab and the second is ba, so both pairs contain different characters. An implementation that checks whether all characters are globally distinct would incorrectly reject this string, even though repeated non-adjacent characters are allowed.

For input abb, the correct output is NO. The first pair is valid, but the final pair bb is not. This catches an off-by-one error where only pairs starting before the last two positions are inspected.

Approaches

A direct brute-force approach could inspect every pair of positions (i, j) and test whether it is an adjacent pair, meaning j = i + 1. If such a pair contains equal characters, the answer is NO; otherwise the answer is YES. This is correct because every potentially invalid pair is examined.

The problem is that this method examines all n(n-1)/2 pairs, even though only n-1 of them can possibly matter. In the worst case, that is roughly n²/2 pair checks. For a string of length 100,000, this is about 4,999,950,000 checks, far beyond what is reasonable for a 1 second limit.

The structure of the condition gives us a much smaller search space. Whether the string is diverse depends only on adjacent pairs, so there is no reason to inspect any non-adjacent pair. We can simply walk from left to right and compare each character with the character immediately before it.

The first equal adjacent pair proves that the answer is NO, so the scan can stop immediately. If the entire string is traversed without finding such a pair, every required comparison was successful and the answer is YES.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the string. A string with fewer than two characters is automatically diverse because it contains no adjacent pair that could violate the condition.
2. Start at the second character and compare it with the character immediately before it. These are exactly the first adjacent pair that needs to be checked.
3. If the two characters are equal, print NO and stop. One invalid adjacent pair is enough to disqualify the entire string.
4. Continue the same comparison for every remaining character. Each iteration checks exactly one new adjacent pair, so every relevant pair is examined once.
5. If the loop finishes without finding equal adjacent characters, print YES. At that point every adjacent pair in the string has been verified to contain different characters.

Why it works

After processing position i, every adjacent pair ending at or before i has been checked and found valid. This invariant starts with the first pair and is preserved because each iteration checks the next adjacent pair. If an equal pair exists, the iteration containing that pair prints NO; if no such iteration exists, every adjacent pair is different, which is exactly the definition of a diverse string.

Python Solution

import sys
input = sys.stdin.readline

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

    for i in range(1, len(s)):
        if s[i] == s[i - 1]:
            print("NO")
            return

    print("YES")

if __name__ == "__main__":
    solve()

The loop starts at index 1 because index 0 has no character before it. For each i, the expression s[i] == s[i - 1] examines exactly one adjacent pair.

Returning immediately after finding equality avoids unnecessary work. It is also logically safe because the existence of one invalid pair completely determines the answer.

The loop condition i < len(s) is handled naturally by Python's range(1, len(s)). For a one-character string, this becomes range(1, 1), which is empty, so the program correctly prints YES.

There is no need for an auxiliary array, set, or frequency table. The condition is local to neighboring characters, so constant extra space is sufficient.

Worked Examples

Sample 1

For the input coderamsclub, the scan compares each neighboring pair.

i	Previous character	Current character	Equal?	Decision
1	c	o	No	Continue
2	o	d	No	Continue
3	d	e	No	Continue
4	e	r	No	Continue
5	r	a	No	Continue
6	a	m	No	Continue
7	m	s	No	Continue
8	s	c	No	Continue
9	c	l	No	Continue
10	l	u	No	Continue
11	u	b	No	Continue

No comparison finds equal characters, so the final answer is YES.

Sample 2

For the input helloworld, the first comparison already finds the repeated l.

i	Previous character	Current character	Equal?	Decision
1	h	e	No	Continue
2	e	l	No	Continue
3	l	l	Yes	Print NO

The scan stops at the first invalid pair. There is no reason to inspect the rest of the string because the result cannot change after finding one violation.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each adjacent pair is compared once.
Space	O(1)	Only the input string and a constant number of variables are used.

The linear scan is suitable even for very large input strings because the amount of work grows directly with the input size. The algorithm also uses constant auxiliary memory, comfortably within the 256 MB memory limit.

Test Cases

import sys
import io

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

    for i in range(1, len(s)):
        if s[i] == s[i - 1]:
            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:
        from contextlib import redirect_stdout
        output = io.StringIO()
        with redirect_stdout(output):
            solve()
        return output.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# Provided samples
assert run("coderamsclub\n") == "YES\n", "sample 1"
assert run("helloworld\n") == "NO\n", "sample 2"

# Minimum-size input
assert run("a\n") == "YES\n", "single character has no adjacent pair"

# Smallest invalid string
assert run("aa\n") == "NO\n", "equal adjacent characters"

# Repeated but non-adjacent character is allowed
assert run("aba\n") == "YES\n", "non-adjacent repetition is valid"

# Invalid pair at the very end
assert run("abb\n") == "NO\n", "last adjacent pair must be checked"

# Large input, exercising linear behavior
large = ("ab" * 50000) + "\n"
assert run(large) == "YES\n", "large diverse string"

# Large input with one violation at the end
large_bad = ("ab" * 49999) + "abb\n"
assert run(large_bad) == "NO\n", "violation at the final pair"

Test input	Expected output	What it validates
a	YES	Minimum-size string and empty comparison loop
aa	NO	Smallest possible invalid string
aba	YES	Repeated non-adjacent characters are allowed
abb	NO	The final adjacent pair must not be skipped
abab...ab with 100,000 characters	YES	Linear performance on a large input
abab...abb with 100,000 characters	NO	Detecting a violation at the final comparison

Edge Cases

For the one-character input a, the algorithm executes no iterations because there is no adjacent pair. It reaches the final print("YES"), which is correct because the condition is vacuously satisfied.

For aa, the loop begins with i = 1. It compares s[1], which is a, with s[0], also a, and immediately prints NO. This confirms that the first possible adjacent violation is detected.

For aba, the comparisons are a versus b, followed by b versus a. Both differ, so the scan finishes and prints YES. The repeated a at positions 0 and 2 does not matter because those positions are not adjacent.

For abb, the first comparison, a versus b, succeeds. The second comparison, b versus b, fails, so the algorithm prints NO. This specifically verifies the upper boundary of the loop, since the pair involving the final character is included.

For a large alternating string such as abababab, every comparison succeeds. The algorithm never constructs or examines unrelated pairs, so its running time remains proportional to the string length. If the final characters are changed to bb, the algorithm still checks that final pair and reports NO, demonstrating both linear behavior and correct boundary handling.