CF 2182A - New Year String

Rating: 800
Tags: constructive algorithms, greedy, implementation, strings
Solve time: 1m 57s
Verified: yes

Solution

Problem Understanding

We are given a string composed of the characters 0, 2, 5, and 6. A string is considered a New Year string if it either contains the substring 2026 somewhere or does not contain the substring 2025 at all. For each string, we are allowed to change any character to one of the four allowed characters any number of times. Our task is to compute the minimum number of such character replacements needed to make the string a New Year string.

Each test case provides the string length n and the string s. The constraints 4 ≤ n ≤ 20 and up to 10^4 test cases imply that the strings are short enough for simple brute-force or combinatorial checks, but the number of test cases is high, so we need a solution that is efficient for each individual string.

Non-obvious edge cases arise when the string already partially contains 2025 or 2026. For example, the string 2025 is not a New Year string and requires at least one change. Replacing the last 5 with 6 makes it 2026, which satisfies the first condition. Another tricky case is 20252025, which contains 2025 twice. Simply changing the last character is insufficient; we need the minimal change that either introduces 2026 or removes 2025.

Approaches

The naive brute-force approach is to try all possible combinations of character replacements. For each character, we could attempt all four possible replacements, generating 4^n possible strings. For n = 20, this produces over a trillion possibilities, which is clearly infeasible.

The key observation is that the problem is constrained to short strings. The only substring that directly triggers a requirement is 2026 or 2025. Therefore, we do not need to generate all strings; we only need to check every substring of length 4 in the input. For each 4-length window, we compute how many character changes are needed to make it exactly 2026. The minimum of these values across all windows is the minimal number of changes needed to satisfy the first condition.

If the string contains 2025, we can either modify some characters to break 2025 or directly convert one occurrence into 2026 (which simultaneously breaks 2025 and satisfies the first condition). This makes the solution much simpler: iterate all length-4 substrings, compute mismatches with 2026, and pick the minimum. Because n ≤ 20, this check requires at most 17 * 4 = 68 comparisons per string, which is fast enough for 10^4 test cases.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (all replacements)	O(4^n)	O(n)	Too slow
Optimal (check length-4 substrings)	O(n) per test case	O(1)	Accepted

Algorithm Walkthrough

Read the number of test cases t.
For each test case, read the string s and its length n.
Initialize a variable min_changes with a large value. This will store the minimal operations needed.
Iterate over all substrings of length 4 in s. For each substring, count how many characters differ from the target string 2026.
Update min_changes with the minimum mismatch found among all substrings.
If the string does not contain 2025 at all, then zero changes are needed according to the second New Year string condition.
Output the minimal number of operations for each test case.

Why it works: by checking all length-4 substrings against 2026, we guarantee that if it is possible to create 2026 in the string with minimal changes, we will find the optimal number. For strings not containing 2025, the algorithm implicitly returns zero because no changes are required to satisfy the second condition.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    s = input().strip()
    
    if "2025" not in s:
        print(0)
        continue
    
    min_changes = float('inf')
    for i in range(n - 3):
        substring = s[i:i+4]
        changes = sum(substring[j] != target for j, target in enumerate("2026"))
        if changes < min_changes:
            min_changes = changes
    print(min_changes)

This code first checks if 2025 exists. If not, zero changes are required. Otherwise, it iterates through each 4-character substring and counts mismatches with 2026. min_changes ensures we find the minimal operation count.

Worked Examples

Example 1

Input:

4
2025

i	substring	mismatch vs 2026	min_changes
0	2025	1	1

Output: 1

The substring 2025 requires only the last character to change to 6 to become 2026.

Example 2

Input:

8
20252025

i	substring	mismatch vs 2026	min_changes
0	2025	1	1
1	0252	3	1
2	2520	4	1
3	5202	4	1
4	2025	1	1

Output: 1

Even though there are two occurrences of 2025, changing a single character in the first occurrence suffices to satisfy the first condition.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t * n)	For each of the `t` test cases, we iterate through up to `n-3` substrings of length 4.
Space	O(1)	Only a few integer variables are used; we do not store intermediate arrays.

Given t ≤ 10^4 and n ≤ 20, the total number of operations is at most 2 * 10^5, which fits comfortably within the 2-second time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # call the solution
    t = int(input())
    for _ in range(t):
        n = int(input())
        s = input().strip()
        
        if "2025" not in s:
            print(0)
            continue
        
        min_changes = float('inf')
        for i in range(n - 3):
            substring = s[i:i+4]
            changes = sum(substring[j] != target for j, target in enumerate("2026"))
            if changes < min_changes:
                min_changes = changes
        print(min_changes)
    return output.getvalue().strip()

# Provided samples
assert run("7\n4\n0000\n4\n2025\n4\n2026\n8\n20252026\n8\n20252025\n9\n202520256\n9\n202520265\n") == "0\n1\n0\n0\n1\n1\n0"

# Custom cases
assert run("2\n4\n2025\n4\n0002\n") == "1\n0", "handles single 2025 and no 2025"
assert run("1\n20\n20252025202520252025\n") == "1", "long repeated 2025"
assert run("1\n4\n2026\n") == "0", "already New Year string"
assert run("1\n5\n22222\n") == "0", "no 2025, no change needed"

Test input	Expected output	What it validates
`2025`	1	Correctly converts a single occurrence of 2025
`0002`	0	Does not contain 2025, so zero changes
`20252025202520252025`	1	Multiple consecutive 2025, minimal change applied
`2026`	0	Already contains 2026
`22222`	0	No forbidden substring, nothing to change

Edge Cases

For a string that is already valid by the second condition, such as 0000, the algorithm checks for 2025. Since it does not exist, it outputs zero immediately. For strings with multiple overlapping 2025, the algorithm examines all substrings of length 4 and finds the minimal number of character changes to create 2026, which simultaneously removes all instances of 2025 within that substring. For strings that already contain 2026, mismatches with 2026 will be zero, so the output is zero, confirming correctness.