CF 2218A - The 67th Integer Problem

Rating: 800
Tags: brute force, games, implementation, math
Solve time: 1m 27s
Verified: yes

Solution

Problem Understanding

The problem asks us to pick an integer $y$ for a given integer $x$ so that the minimum of $x$ and $y$ is as large as possible. We are given multiple test cases, each consisting of a single integer $x$ between -67 and 67. Our output must be an integer $y$ within the same bounds for each test case.

Restated simply, if we visualize $x$ and $y$ on a number line, $\min(x, y)$ is always the smaller of the two. To maximize this minimum, we need $y$ to be strictly larger than $x$. Then $\min(x, y)$ will equal $x$, which is the largest it can be. If $y \le x$, the minimum would be $y$, which is strictly smaller than $x$, so that would not maximize the minimum.

The constraints are very small. $x$ is bounded between -67 and 67, and there are at most 6767 test cases. Each test case is independent, and computing a solution per case is trivial. Even an $O(t)$ approach with simple arithmetic is easily within the 1-second time limit.

Non-obvious edge cases include the maximum and minimum values. For $x = 67$, we cannot pick $y > 67$ because the bounds restrict us. Therefore, $y = 67$ itself is valid. Similarly, for $x = -67$, any $y \ge -67$ will yield $\min(x, y) = -67$, so we can pick $y = -66$ or any larger number within bounds. The key is always choosing a number just above $x$ if possible, but respecting the -67 to 67 bounds.

Approaches

The brute-force approach is trivial. For a given $x$, we could try every $y$ from -67 to 67, compute $\min(x, y)$ for each, and pick the largest result. This works because there are only 135 possible $y$ values, but it is unnecessary because the pattern is predictable. The brute-force complexity would be $O(t \cdot 135)$, which is acceptable but cumbersome and inelegant.

The key insight is that $\min(x, y)$ is maximized whenever $y > x$, because then the minimum equals $x$. Therefore, the optimal choice for $y$ is simply $x+1$, unless that exceeds the upper bound of 67. In that case, $y = 67$ is the only valid choice. This observation reduces the problem to a constant-time calculation per test case.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(t * 135)	O(1)	Accepted but unnecessary
Optimal	O(t)	O(1)	Accepted

Algorithm Walkthrough

Read the number of test cases $t$. This determines how many times we repeat the computation.
For each test case, read $x$. We will determine $y$ for this $x$.
Compute $y = x + 1$. This guarantees that $\min(x, y) = x$, which is maximal.
If $y > 67$, adjust $y$ to be 67 to stay within bounds.
Output $y$. Repeat for all test cases.

The reason this works is that increasing $y$ above $x$ does not decrease the minimum, so the minimum cannot be larger than $x$. Choosing $y = x + 1$ keeps it simple, avoids equality issues, and respects the problem constraints.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    x = int(input())
    y = min(x + 1, 67)
    print(y)

The code first reads $t$, the number of test cases. Then, for each test case, it reads $x$, calculates $y$ using the logic described above, and prints it. Using min(x + 1, 67) ensures that we never exceed the upper bound. Fast I/O with sys.stdin.readline is used, which is standard practice for competitive programming, although here input() would also be sufficient due to the small input size.

Worked Examples

Example 1

Input: 1

Step	x	y = x + 1	min(x, y)
1	1	2	1

The minimum of 1 and 2 is 1, which is maximal because any smaller $y$ would give a smaller minimum. This confirms the algorithm works for positive numbers.

Example 2

Input: 67

Step	x	y = x + 1	Adjust y	min(x, y)
1	67	68	67	67

Here, $x + 1$ exceeds the bound, so we clamp it to 67. The minimum remains 67, which is correct. This confirms that boundary conditions are handled properly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is processed in constant time.
Space	O(1)	Only a few integers are stored per test case.

The time complexity is acceptable because $t \le 6767$. The memory footprint is minimal, well below the 256 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    
    t = int(input())
    for _ in range(t):
        x = int(input())
        y = min(x + 1, 67)
        print(y)
    
    return output.getvalue().strip()

# Provided sample
assert run("3\n1\n3\n5\n") == "2\n4\n6", "sample 1"

# Custom tests
assert run("2\n67\n-67\n") == "67\n-66", "boundary max and min"
assert run("1\n0\n") == "1", "zero input"
assert run("4\n-1\n0\n1\n2\n") == "0\n1\n2\n3", "consecutive small numbers"
assert run("1\n66\n") == "67", "near upper bound"

Test input	Expected output	What it validates
67, -67	67, -66	Upper and lower bounds handling
0	1	Zero input handled correctly
-1,0,1,2	0,1,2,3	Consecutive small numbers, normal case
66	67	Edge case near maximum bound

Edge Cases

For $x = 67$, $x + 1 = 68$ exceeds the allowed range. The algorithm clamps $y$ to 67, which keeps $\min(x, y) = 67$. For $x = -67$, adding 1 gives -66, which is within bounds, ensuring the minimum is maximized at -67. This confirms that both ends of the allowed range are correctly handled. For any intermediate $x$, $y = x + 1$ guarantees $\min(x, y) = x$, which is the largest possible minimum.