CF 104396I - Elevator

Rating: -
Tags: -
Solve time: 54s
Verified: yes

Solution

Problem Understanding

There are n people inside an elevator, including yourself. Every person has selected a destination floor. Among all selected destinations, exactly m distinct floors appear.

You may choose your own destination floor freely, and your goal is to maximize how many people, including yourself, leave the elevator together on your floor. Since exactly m different floors must be represented, the destinations assigned to the other n - 1 people can be distributed in any valid way that satisfies this requirement.

For each test case, we are given only the total number of people and the number of distinct destination floors. We must compute the largest possible group that can leave on the same floor as us.

The largest input values are only around 10^4, and every test case contains just two integers. Even if there are 10^4 test cases, an O(1) solution per case performs only a few arithmetic operations overall. Any more complicated algorithm is unnecessary because the answer depends only on a simple counting argument.

One easy mistake is forgetting that every distinct floor must have at least one passenger.

For example,

n = 5
m = 5

The correct answer is 1. Every passenger must choose a different floor, so nobody can share your destination.

Another common mistake is assuming all remaining passengers can always choose your floor.

For example,

n = 6
m = 3

The correct answer is 4, not 6. Besides your own floor, there must still be two other distinct floors. At least one passenger must be assigned to each of those floors, leaving only four people on your floor.

The smallest input also deserves attention.

n = 1
m = 1

The answer is 1. You are the only passenger, so everyone leaves together.

Approaches

A brute-force viewpoint is to imagine distributing all n passengers among exactly m different floors and checking every valid distribution. For each distribution, we record the largest group assigned to a single floor and keep the maximum over all possibilities.

This approach is correct because it examines every legal assignment. Unfortunately, the number of such distributions grows exponentially with n, making it completely impractical even for moderate input sizes.

The key observation is that we never need to know the exact distribution. To maximize the number of people leaving with us, every passenger who is not forced onto another floor should choose our floor instead.

Since there must be exactly m distinct floors, one of them is ours, while the remaining m - 1 floors each require at least one passenger. Assign exactly one passenger to each of those other floors. This uses m - 1 people, leaving

n - (m - 1) = n - m + 1

people on our floor, including ourselves.

No larger answer is possible because every additional person moved onto our floor would leave one of the required distinct floors empty.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	Exponential	Too slow
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the values n and m.
Reserve one passenger for each of the other m - 1 distinct floors. This is the minimum possible assignment that still keeps all required floors represented.
Place every remaining passenger on your floor. Since no other constraint exists, this produces the largest possible group leaving together.
Compute the answer as n - m + 1.
Output the result for the current test case.

Why it works

Every valid assignment must contain at least one passenger on each of the m - 1 floors different from yours. Those passengers cannot also leave with you. This means at least m - 1 people are excluded from your group, so your group size is at most n - (m - 1). The algorithm reaches exactly this value by assigning only one passenger to every other floor and placing everyone else on your floor. Since the upper bound is achieved, the result is optimal.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    out = []

    for _ in range(t):
        n, m = map(int, input().split())
        out.append(str(n - m + 1))

    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()

The program first reads the number of test cases. Each test case is handled independently, so no information needs to be stored between cases.

The entire solution is the arithmetic expression n - m + 1. This follows directly from the counting argument developed in the algorithm. No loops over passengers or floors are needed.

Using sys.stdin.readline keeps the input fast even when the number of test cases reaches its maximum. The answers are accumulated in a list and printed together at the end, avoiding many individual output operations.

There are no overflow concerns because Python integers have arbitrary precision, and the problem constraints are already small. The only boundary condition is when m = n, in which case the formula correctly evaluates to 1.

Worked Examples

Example 1

Input:

n = 6
m = 3

Step	n	m	Other floors reserved	Answer
Read input	6	3	-	-
Reserve one passenger for each other floor	6	3	2	-
Remaining passengers stay with us	6	3	2	4

The two required extra floors each receive one passenger. The remaining four passengers, including ourselves, all leave together.

Example 2

Input:

n = 5
m = 5

Step	n	m	Other floors reserved	Answer
Read input	5	5	-	-
Reserve one passenger for each other floor	5	5	4	-
Remaining passengers stay with us	5	5	4	1

Every passenger must occupy a different floor, so nobody can share our destination.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1) per test case	Only one subtraction and one addition are performed.
Space	O(1)	Only a few integer variables are used.

Since every test case requires only constant time and constant memory, the solution easily satisfies the constraints even when the number of test cases is as large as possible.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline
    t = int(input())
    ans = []
    for _ in range(t):
        n, m = map(int, input().split())
        ans.append(str(n - m + 1))
    sys.stdout.write("\n".join(ans))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    solve()
    out = sys.stdout.getvalue()
    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return out

# custom cases
assert run("1\n1 1\n") == "1", "minimum case"
assert run("1\n6 3\n") == "4", "general case"
assert run("1\n5 5\n") == "1", "all floors distinct"
assert run("1\n10 1\n") == "10", "everyone can leave together"
assert run("1\n10000 9999\n") == "2", "boundary values"

Test input	Expected output	What it validates
`1 1`	`1`	Minimum input size
`6 3`	`4`	General counting case
`5 5`	`1`	Every floor must be different
`10 1`	`10`	Only one distinct floor exists
`10000 9999`	`2`	Large boundary values

Edge Cases

Consider the smallest possible input.

1
1 1

The algorithm computes 1 - 1 + 1 = 1. There are no other required floors, so the single passenger leaves alone. The output is correctly 1.

Now consider the case where every passenger must choose a different floor.

1
5 5

The algorithm computes 5 - 5 + 1 = 1. Since four other distinct floors are required besides ours, every other passenger must choose one of them. No one can share our destination, so the answer is correct.

Finally, consider a case with only one distinct floor.

1
8 1

The algorithm computes 8 - 1 + 1 = 8. No additional floors are required, so every passenger can choose our floor. All eight passengers leave together, which is clearly optimal.