CF 14B - Young Photographer

CF 14B - Young Photographer

Rating: 1000
Tags: implementation
Solve time: 1m 20s
Verified: yes

Solution

Problem Understanding

The problem asks us to determine how far Bob, a photographer, must move along a straight racetrack to take pictures of every sportsman. Each sportsman runs back and forth along a fixed segment of the racetrack, defined by two positions ai and bi. Bob starts at position x0, and he can photograph a sportsman only if he stands somewhere inside that sportsman’s segment. We are asked to find the minimum distance Bob has to travel to reach a position that allows him to photograph all sportsmen. If no such position exists, we must return -1.

The input consists of the number of sportsmen n (up to 100), Bob’s starting position x0 (0 to 1000), and n segments defined by pairs of integers ai and bi (0 to 1000). Each segment may be given in either order, so the smaller number is not guaranteed to be first. The output is a single integer: the minimum distance Bob must move.

With n at most 100, we can process every sportsman individually without hitting performance limits. The problem is largely about correctly handling ranges and determining overlap. Edge cases include situations where all segments do not overlap at all, in which case no valid position exists. For example, if segments are [0, 2], [5, 6], and [8, 10], there is no single position Bob can stand to photograph all runners, so the correct output is -1. A careless approach might try to pick the first segment as reference without computing intersections, leading to a wrong answer.

Approaches

The naive approach is to examine every position from 0 to 1000, check for each position whether it lies inside all segments, and track the minimum distance from Bob’s starting point. This brute-force method works because the position range is small (maximum 1001 positions) and n is also small. In the worst case, it would perform around 100,000 checks, which is feasible under a 2-second limit. The downside is that it is unnecessarily slow and does not leverage the structure of the problem.

The key insight to optimize is that Bob only needs to stand within the intersection of all segments. If we compute the intersection by taking the maximum of all left endpoints and the minimum of all right endpoints, we immediately find the range of positions that satisfy all sportsmen. If this intersection is empty (the maximum left endpoint is greater than the minimum right endpoint), no position works and the answer is -1. Otherwise, the minimum distance Bob must move is the distance from x0 to the nearest point in this intersection.

The brute-force works because checking every integer is guaranteed to find the minimal distance, but it fails to exploit the simple property that all segments’ intersection defines the feasible region. The observation that the intersection alone determines the valid positions reduces the problem from O(n * 1000) to O(n) with trivial arithmetic.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n * 1000)	O(n)	Accepted but unnecessary
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Normalize each segment so that the smaller value is first. For a segment [ai, bi], assign left = min(ai, bi) and right = max(ai, bi). This ensures consistent comparison when finding intersections.
2. Initialize two variables, max_left to 0 and min_right to 1000. These will store the endpoints of the intersection of all segments.
3. Iterate through each segment. Update max_left as the maximum of the current max_left and the segment’s left. Update min_right as the minimum of the current min_right and the segment’s right. After this loop, [max_left, min_right] represents the intersection of all segments.
4. Check if the intersection is empty by comparing max_left and min_right. If max_left > min_right, output -1. This indicates no single position satisfies all sportsmen.
5. Otherwise, compute the distance from Bob’s starting position x0 to the intersection. If x0 is within [max_left, min_right], the distance is 0. If x0 < max_left, the distance is max_left - x0. If x0 > min_right, the distance is x0 - min_right. Return this distance.

Why it works: The intersection step guarantees that any position within [max_left, min_right] allows Bob to photograph every sportsman. By taking the nearest endpoint relative to x0, we ensure the minimal distance. No other position can yield a smaller distance since any movement outside the intersection either increases distance or violates the constraints.

Python Solution

import sys
input = sys.stdin.readline

n, x0 = map(int, input().split())

max_left = 0
min_right = 1000

for _ in range(n):
    a, b = map(int, input().split())
    left = min(a, b)
    right = max(a, b)
    max_left = max(max_left, left)
    min_right = min(min_right, right)

if max_left > min_right:
    print(-1)
else:
    if x0 < max_left:
        print(max_left - x0)
    elif x0 > min_right:
        print(x0 - min_right)
    else:
        print(0)

The first section reads input efficiently using sys.stdin.readline. Each segment is normalized using min and max to ensure proper intersection computation. The variables max_left and min_right accumulate the intersection endpoints. The final conditional handles three cases: Bob is already inside the intersection, Bob is left of the intersection, or Bob is right of it. This avoids off-by-one errors or unnecessary looping.

Worked Examples

Sample Input 1:

3 3
0 7
14 2
4 6

Step	Segment	left	right	max_left	min_right
1	0 7	0	7	0	7
2	14 2	2	14	2	7
3	4 6	4	6	4	6

x0 = 3 is left of intersection [4,6], distance = 1.

Sample Input 2:

2 5
0 4
6 10

Step	Segment	left	right	max_left	min_right
1	0 4	0	4	0	4
2	6 10	6	10	6	4

Intersection empty (6 > 4), output = -1.

These traces demonstrate the intersection calculation and the distance logic relative to x0.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single pass through n segments for intersection
Space	O(1)	Only a few integer variables are maintained

With n ≤ 100 and positions ≤ 1000, the solution executes quickly and uses minimal memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, x0 = map(int, input().split())
    max_left = 0
    min_right = 1000
    for _ in range(n):
        a, b = map(int, input().split())
        left = min(a, b)
        right = max(a, b)
        max_left = max(max_left, left)
        min_right = min(min_right, right)
    if max_left > min_right:
        return str(-1)
    if x0 < max_left:
        return str(max_left - x0)
    elif x0 > min_right:
        return str(x0 - min_right)
    return str(0)

# Provided sample
assert run("3 3\n0 7\n14 2\n4 6\n") == "1", "sample 1"

# Custom cases
assert run("2 5\n0 4\n6 10\n") == "-1", "no intersection"
assert run("1 0\n0 0\n") == "0", "single segment, already there"
assert run("2 0\n0 5\n2 4\n") == "0", "x0 inside intersection"
assert run("3 1000\n0 1000\n500 1000\n600 900\n") == "100", "x0 beyond intersection"

Test input	Expected output	What it validates
2 5\n0 4\n6 10	-1	Intersection empty
1 0\n0 0	0	Single segment, x0 at correct position
2 0\n0 5\n2 4	0	x0 already inside intersection
3 1000\n0 1000\n500 1000\n600 900	100	x0 right of intersection