CF 58A - Chat room

CF 58A - Chat room

Rating: 1000
Tags: greedy, strings
Solve time: 1m 20s
Verified: yes

Solution

Problem Understanding

The problem asks whether Vasya, who typed a string of lowercase letters, effectively managed to say "hello". The goal is not to check if the typed string is exactly "hello", but whether we can remove some letters (possibly zero) to produce the sequence h, e, l, l, o in order.

The input is a single string of length between 1 and 100. The output is either "YES" if it is possible to extract the word "hello" from it or "NO" otherwise.

The constraints are very relaxed. With n ≤ 100 and a time limit of 1 second, even a naive solution that looks at all subsequences could theoretically work, but the problem's essence is really about scanning the string for the target sequence. Edge cases that can trick a naive approach include strings where letters are present but out of order, such as hlelo, which contains all the letters of "hello" but does not preserve their order. Another edge case is minimal input, e.g., h, which should clearly output "NO".

Approaches

The brute-force approach would try every possible subsequence of the input string to see if it matches "hello". For a string of length n, there are 2^n subsequences, which is infeasible even for n = 100, producing roughly 10^30 checks. Clearly, this approach is correct in principle but not practical.

The key insight is that we do not need to consider all subsequences explicitly. We only care about whether we can match each character of "hello" in order as we scan the string. This suggests a greedy scan: keep an index j for the target word "hello". Traverse the input string s from left to right. Each time a character matches hello[j], increment j. If j reaches the length of "hello" (5), we have successfully found the word. This is linear in the length of the string and directly encodes the problem's sequential requirement.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(1)	Too slow
Greedy Scan	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize target = "hello" and an index j = 0 to track which character of "hello" we are currently looking for.
2. Iterate through each character c in the input string s.
3. If c equals target[j], increment j. This moves us to the next character we need to match. We skip all other characters automatically because they cannot disrupt the order.
4. After scanning the string, check if j equals 5. If it does, print "YES", otherwise print "NO".

Why it works: The greedy approach relies on the invariant that j always points to the next character of "hello" that needs to be matched. Every time a character matches, it advances the index. Any characters in between are ignored because they do not prevent forming the subsequence. This guarantees that if j reaches 5, all letters have been matched in the correct order.

Python Solution

import sys
input = sys.stdin.readline

s = input().strip()
target = "hello"
j = 0

for c in s:
    if c == target[j]:
        j += 1
        if j == len(target):
            break

print("YES" if j == len(target) else "NO")

The code reads the input and strips any trailing newline. We maintain a pointer j to track our progress through "hello". As we iterate over s, we increment j whenever we match the current target character. If we reach the end of target, we stop early for efficiency. Finally, we print "YES" if all letters were matched, otherwise "NO".

Worked Examples

Example 1

Input: ahhellllloou

s[i]	c	target[j]	j
0	a	h	0
1	h	h	1
2	h	e	1
3	e	e	2
4	l	l	3
5	l	l	4
6	l	o	4
7	l	o	4
8	l	o	4
9	o	o	5

All target letters are matched in order. Output: YES.

Example 2

Input: hlelo

s[i]	c	target[j]	j
0	h	h	1
1	l	e	1
2	e	e	2
3	l	l	3
4	o	l	3

j never reaches 5. Output: NO.

These traces confirm that the algorithm correctly tracks sequential matches and rejects out-of-order letters.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We scan each character once, performing at most one comparison and pointer increment per character
Space	O(1)	Only a few variables are used regardless of input length

With n ≤ 100, this is comfortably within time and memory limits. Even if n were larger, this linear scan scales well.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    s = input().strip()
    target = "hello"
    j = 0
    for c in s:
        if c == target[j]:
            j += 1
            if j == len(target):
                break
    return "YES" if j == len(target) else "NO"

# Provided sample
assert run("ahhellllloou") == "YES", "sample 1"

# Custom cases
assert run("hlelo") == "NO", "letters out of order"
assert run("h") == "NO", "single letter"
assert run("helicopterlo") == "YES", "interleaved extra letters"
assert run("aaaaa") == "NO", "no relevant letters"
assert run("hhheeellllooo") == "YES", "duplicates before matches"

Test input	Expected output	What it validates
hlelo	NO	sequence order enforcement
h	NO	minimum input handling
helicopterlo	YES	skips irrelevant letters
aaaaa	NO	absence of target letters
hhheeellllooo	YES	multiple consecutive matches handled

Edge Cases

For s = "h", the algorithm iterates once, sees h matches target[0], increments j = 1. There are no further characters, j < 5, so output is NO. For s = "hhheeellllooo", the algorithm increments j only when the next required letter matches, skipping duplicates. After all characters, j = 5, correctly outputting YES. These confirm the greedy scan handles both insufficient and repetitive input correctly.