CF 1829A - Love Story

CF 1829A - Love Story

Rating: 800
Tags: implementation, strings
Solve time: 1m
Verified: yes

Solution

Problem Understanding

We are given a fixed reference string, “codeforces”, and for each test case we receive another string of the same length, exactly 10 lowercase letters. The task is to compare these two strings position by position and count how many positions contain different characters.

In other words, imagine lining up two rows of 10 characters: the reference word on top and the given string below it. We scan from left to right and count mismatches. Each mismatch contributes 1 to the final answer, and matching positions contribute 0. The output per test case is just this mismatch count.

The constraints are small enough that each test case can be processed independently in constant time. With at most 1000 strings and each string length fixed at 10, even a direct character-by-character comparison is trivial under the time limit. This immediately rules out any need for preprocessing, hashing, or advanced string techniques.

There are no subtle edge cases in terms of structure because the length is fixed and equality is well-defined character-by-character. The only potential mistake comes from off-by-one indexing or accidentally comparing against a modified reference string.

A typical incorrect approach would be to count matches instead of mismatches and forget to convert, or to compare sorted versions of strings, which destroys positional information. For example, comparing “forcescode” to “codeforces” after sorting would incorrectly suggest full similarity, but the problem depends entirely on positions.

Approaches

The brute-force approach is exactly what the problem suggests: for each test case, compare each of the 10 positions against the corresponding character in “codeforces” and increment a counter whenever they differ. Since each string has constant length, this runs in constant time per test case.

The total work is at most 1000 × 10 character comparisons, which is negligible. There is no asymptotic bottleneck to improve upon, but the conceptual simplification is still useful: we are not solving a general string problem, only a fixed alignment comparison against a constant template.

One could imagine more complex solutions using hashing or precomputed mismatch tables, but they would only add overhead without benefit.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(10·t)	O(1)	Accepted
Optimal	O(10·t)	O(1)	Accepted

Algorithm Walkthrough

1. Store the reference string “codeforces” once before processing test cases. This avoids reconstructing or retyping it repeatedly and ensures all comparisons are consistent.
2. For each test case, read the input string of length 10.
3. Initialize a counter to zero for that test case.
4. Iterate over all indices from 0 to 9. At each index, compare the character in the input string with the character in the reference string. If they differ, increment the counter.
5. After finishing all 10 positions, output the counter for that test case.

The key idea is that every position contributes independently to the final answer. There is no interaction between indices, so a single pass is sufficient.

Why it works

Each position in the strings is an independent binary condition: either the characters match or they do not. The final answer is simply the sum of these independent mismatch indicators. Since the reference string is fixed and the input string is not modified, the comparison is deterministic and exhaustive across all indices. No alternative arrangement or hidden transformation exists, so counting mismatches directly produces the correct result.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    target = "codeforces"
    t = int(input())
    
    for _ in range(t):
        s = input().strip()
        diff = 0
        
        for i in range(10):
            if s[i] != target[i]:
                diff += 1
        
        print(diff)

if __name__ == "__main__":
    solve()

The solution fixes the reference string once and reuses it for all test cases, avoiding repeated allocations inside the loop. Each test case is handled independently, and we strictly compare characters by index, ensuring positional correctness. The .strip() is important to avoid counting newline characters as part of the string, which is a common subtle bug in competitive programming I/O.

Worked Examples

Let’s trace two cases.

First example: coolforsez compared to codeforces.

i	s[i]	target[i]	match?	diff
0	c	c	yes	0
1	o	o	yes	0
2	o	d	no	1
3	l	e	no	2
4	f	f	yes	2
5	o	o	yes	2
6	r	r	yes	2
7	s	c	no	3
8	e	e	yes	3
9	z	s	no	4

Final output is 4.

This confirms that mismatches are counted purely by position and not by character frequency.

Second example: codeforces compared to itself.

i	s[i]	target[i]	match?	diff
0-9	all equal	all equal	yes	0

This demonstrates the identity case where no mismatches occur, verifying correctness in the base case.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(10·t)	Each test case compares exactly 10 characters once
Space	O(1)	Only a fixed reference string and a counter are used

The computation is linear in the number of test cases with a tiny constant factor. With at most 1000 inputs and fixed-length strings, the solution comfortably runs within both time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

def solve():
    target = "codeforces"
    t = int(input())
    for _ in range(t):
        s = input().strip()
        diff = 0
        for i in range(10):
            if s[i] != target[i]:
                diff += 1
        print(diff)

# provided samples
assert run("""5
coolforsez
cadafurcie
codeforces
paiuforces
forcescode
""") == "4\n5\n0\n4\n9"

# custom cases
assert run("""1
codeforces
""") == "0", "all match"

assert run("""1
aaaaaaaaaa
""") == "10", "all mismatch"

assert run("""1
codeforces
""") == "0", "identity check"

assert run("""2
codeforces
codeforces
""") == "0\n0", "repeated identity"

assert run("""1
codeforcez
""") == "1", "single mismatch"

Test input	Expected output	What it validates
codeforces	0	perfect match base case
aaaaaaaaaa	10	all positions differ
repeated cases	0 0	multiple test handling
codeforcez	1	single-position mismatch

Edge Cases

One edge case is when the input string is identical to the reference. For input codeforces, every index matches, so the loop never increments the counter and the output is 0. The algorithm handles this naturally because no conditional triggers.

Another case is when every character differs, such as aaaaaaaaaa. Here, each of the 10 comparisons fails, and the counter reaches 10. Since we explicitly iterate over all indices, we do not miss any position.

A subtle case is repeated identical inputs across multiple test cases. Since we reset the counter inside the loop for each test case, previous results do not leak into the next computation, ensuring correctness across batches.