Rating: -
Tags: -
Solve time: 3m 2s
Verified: yes

Solution

Problem Understanding

The problem gives a single string representing a soccer commentator’s stretched-out celebration after a goal. The word starts and ends like the normal word "goal", but the number of o characters in the middle can be extended depending on how long the announcer shouts. The task is to count exactly how many o letters appear in the entire recorded reaction.

The input contains one string, so the only operation needed is a scan through its characters. The limits allow a direct linear pass over the string. Even if the string were very large, an O(n) solution only performs one simple check per character, while slower approaches that repeatedly search or modify the string would add unnecessary work.

The main edge cases come from assuming the word always has a fixed length or from counting only the middle section manually. For example, the input:

goal

has output:

A solution that assumes there are always multiple o characters would fail on the shortest valid reaction.

Another case is:

goooooooooal

with output:

The number of o characters is not fixed, so checking for a specific pattern such as "goal" or counting only positions in a hardcoded range would produce the wrong answer.

A final common mistake is forgetting that the input should be treated as a string rather than converting it to another type. The characters themselves carry the information, and every occurrence of o must be counted.

Approaches

The brute-force idea is already close to the optimal solution because the problem has no hidden structure that requires advanced algorithms. A direct method is to examine every character in the announcer's reaction and increase a counter whenever the current character is o. This works because each o contributes exactly one to the required answer, and no other character matters.

The only possible inefficient approach would be to repeatedly search the string for occurrences of o or build many temporary strings while counting. In the worst case, if the string length is n, repeatedly scanning the whole string could reach O(n²) operations, which is unnecessary.

The key observation is that the answer depends only on individual characters. There is no need to understand the position of the letters, verify the word structure, or simulate the announcer. A single left-to-right traversal is enough.

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

Algorithm Walkthrough

Read the reaction string from input. The string is the complete transcription, so every character should be considered.
Initialize a counter with value zero. This variable will store how many o characters have appeared so far.
Traverse the string one character at a time. Whenever the current character is o, increase the counter by one. Characters such as g, a, and l do not affect the result.
Print the final counter value. After the traversal finishes, the counter represents the exact number of o letters in the reaction.

Why it works:

The invariant is that after processing any prefix of the string, the counter equals the number of o characters inside that prefix. Initially the prefix is empty and the counter is zero, so the invariant is true. When a new character is processed, the counter changes exactly when that character is o, which keeps the invariant true. After the entire string is processed, the prefix is the whole input string, so the counter is the required answer.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().strip()
    print(s.count('o'))

if __name__ == "__main__":
    solve()

The solution reads the entire reaction as a string because the characters themselves are the data we need to analyze.

Python's built-in count method performs the same linear scan described in the algorithm. It checks every character once and returns how many times the target character appears.

Using strip() removes the newline added by input without affecting the letters in the reaction. There are no numeric calculations, so integer overflow is not a concern.

Worked Examples

Example 1

Input:

goooooooooal

Trace:

Step	Current character	Counter
Start	-	0
1	g	0
2	o	1
3	o	2
4	o	3
5	o	4
6	o	5
7	o	6
8	o	7
9	o	8
10	o	9
11	a	9
12	l	9

The scan ignores non-o characters and counts each extended vowel separately, producing the answer 9.

Example 2

Input:

goal

Trace:

Step	Current character	Counter
Start	-	0
1	g	0
2	o	1
3	a	1
4	l	1

This example shows that the normal word form is handled correctly. The algorithm does not assume that the announcer always stretches the word.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each character in the reaction is inspected once.
Space	O(1)	Only the counter and current processing state are needed.

The solution fits easily within the limits because it performs the minimum possible amount of work: one pass through the input string.

Test Cases

import sys
import io

def run(inp: str) -> str:
    s = inp.strip()
    return str(s.count('o')) + "\n"

# provided sample
assert run("goooooooooal\n") == "9\n", "sample 1"

# minimum-size case
assert run("goal\n") == "1\n", "shortest reaction"

# all o characters
assert run("oooo\n") == "4\n", "only o characters"

# longer stretched goal
assert run("goooooooooooooooooal\n") == "17\n", "long reaction"

# boundary-style case with only one middle o
assert run("goal\n") == "1\n", "single o count"

Test input	Expected output	What it validates
`goooooooooal`	`9`	Provided sample and normal stretched celebration
`goal`	`1`	Minimum valid length and no extra `o` characters
`oooo`	`4`	Counting every character when all characters match
`goooooooooooooooooal`	`17`	Handling long reactions
`goal`	`1`	Avoiding assumptions about multiple vowels

Edge Cases

For the shortest possible reaction:

goal

the algorithm reads four characters. The counter increases only once when it reaches the second character, then remains unchanged for a and l. The final output is:

This prevents mistakes caused by assuming the announcer must always repeat the vowel.

For a long celebration:

goooooooooal

the algorithm does not treat the middle section specially. It simply visits every position and increments the counter for each o, producing the exact count. This avoids hardcoded assumptions about how many times the announcer extends the word.

For a string containing only vowels:

oooo

every processed character increments the counter. The result is:

This confirms that the solution counts occurrences rather than relying on the surrounding letters of the word "goal".

CF 102697029 - Goooooooooal!