CF 105902A - One Must Imagine Time Tight,

CF 105902A - One Must Imagine Time Tight,

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Solve time: 47s
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Solution

Problem Understanding

We are given three candidate routes or plans for completing the same task, each associated with a time cost. The goal is simply to choose the fastest option among the three and output its time.

Each input consists of exactly three integers. You can think of them as three competing strategies for solving the same subproblem, where each strategy has a known fixed cost. The task reduces to selecting the smallest of these three values and printing it.

The constraints are extremely small, with each value between 1 and 60. This immediately tells us that any solution, even something unnecessarily heavy, would still run in constant time. There is no need for preprocessing, sorting large structures, or repeated computation. Any algorithm that even inspects each value once is already optimal.

There are no tricky structural edge cases involving sequences or ranges, but there is one subtle correctness concern that often appears in problems of this type: ensuring that equality does not affect correctness. If two or more values are equal and minimal, any of them is valid, so the implementation must not incorrectly discard valid minimum candidates due to strict comparisons or early exits that assume uniqueness.

For example, if the input is 1 1 45, both of the first two options are optimal. A correct solution must still output 1. A buggy approach might mistakenly assume a unique minimum and skip equal values, but here equality is part of correctness.

Approaches

A direct approach is to compare the three values manually. We can check which of the three is smallest by a sequence of comparisons. For instance, we could start by assuming the first value is the answer, then compare it with the second, updating if needed, and then compare again with the third. This is already sufficient because the input size is fixed.

This works because the problem is essentially computing the minimum over a set of three elements. There is no dependency between values and no transformation required, only selection.

One might consider a more general approach such as sorting the three values and taking the first element. That also works, but introduces unnecessary overhead in both reasoning and implementation for such a small fixed-size input. A loop over three elements is also valid, but again unnecessary complexity for a constant-sized structure.

The key observation is that the structure is fixed and tiny. The brute force and optimal solutions are effectively identical in complexity; the only difference is clarity and directness.

Approach	Time Complexity	Space Complexity	Verdict
Pairwise comparisons	O(1)	O(1)	Accepted
Sorting three values	O(1)	O(1)	Accepted but unnecessary

Algorithm Walkthrough

1. Read the three integers from input. These represent the costs of the three candidate plans.
2. Initialize a variable ans with the first value. This gives us a baseline candidate that is valid by default.
3. Compare ans with the second value. If the second value is smaller, update ans to the second value. This ensures we always keep the best option seen so far.
4. Compare ans with the third value. If the third value is smaller, update ans again. This guarantees that after processing all three values, ans holds the minimum of all candidates.
5. Output ans.

Why it works

At every step, the variable ans stores the minimum value among all elements processed so far. After the first step, it is the minimum of the first element. After the second comparison, it becomes the minimum of the first two elements. After the third comparison, it becomes the minimum of all three elements. This invariant ensures that no smaller value can be missed, because every candidate is explicitly compared against the current best.

Python Solution

import sys
input = sys.stdin.readline

a, b, c = map(int, input().split())

ans = a
if b < ans:
    ans = b
if c < ans:
    ans = c

print(ans)

The solution reads the three integers in one line and stores them in variables. The variable ans is initialized with the first value to establish a valid starting point. Each subsequent comparison updates ans only when a strictly smaller value is found, preserving correctness even when values are equal.

No special handling is required for equal values because the condition uses < rather than <=, which naturally keeps the first encountered minimum without affecting correctness.

Worked Examples

Example 1

Input:

14 12 18

Step	a	b	c	ans
init	14	12	18	14
compare b	14	12	18	12
compare c	14	12	18	12

The second value becomes the answer because it is smaller than the initial candidate, and the third value does not change it since it is larger.

Example 2

Input:

1 1 45

Step	a	b	c	ans
init	1	1	45	1
compare b	1	1	45	1
compare c	1	1	45	1

This confirms that equal values do not break the logic. The algorithm correctly retains 1 even when duplicates exist.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only three constant-time comparisons are performed regardless of input values
Space	O(1)	Only a fixed number of variables are used

The constraints restrict inputs to three integers with small magnitude, so constant-time processing is trivially within limits. The solution runs instantly under any reasonable environment.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    a, b, c = map(int, input().split())
    ans = a
    if b < ans:
        ans = b
    if c < ans:
        ans = c
    return str(ans)

# provided samples
assert run("14 12 18") == "12", "sample 1"
assert run("1 1 45") == "1", "sample 2"

# custom cases
assert run("1 2 3") == "1", "increasing order"
assert run("3 2 1") == "1", "decreasing order"
assert run("5 5 5") == "5", "all equal"
assert run("60 1 60") == "1", "minimum in middle"

Test input	Expected output	What it validates
1 2 3	1	increasing order correctness
3 2 1	1	reverse order correctness
5 5 5	5	equality handling
60 1 60	1	middle-position minimum

Edge Cases

All values equal

Input: 7 7 7

The algorithm starts with ans = 7, then compares against the second and third values. Neither update triggers because neither is strictly smaller. The final output remains 7, which is correct since all options are equivalent.

Minimum value in different positions

Input: 9 2 8

The algorithm first sets ans = 9. It then updates to 2 after comparing with the second value. The third comparison does not change it. The output 2 correctly reflects the global minimum regardless of position.

Strictly decreasing order

Input: 10 6 1

The algorithm progressively updates ans at each step, demonstrating that repeated improvements are correctly handled. The final result 1 confirms that no early decision prevents later corrections.