Is Direction Assignments with Exactly K Visible People easy or hard?

The problem is generally rated Medium because the brute force idea is simple but inefficient. Recognizing that the task can be converted into a combinatorics counting problem requires deeper algorithmic insight.

Direction Assignments with Exactly K Visible People Python/Java solution

The implementation typically precomputes combinations using factorials or dynamic programming, then iterates over possible splits of visible people. The same logic works across Python, Java, C++, Go, and TypeScript because the core idea is combinatorial counting rather than language-specific data structures.

What is the best approach for Direction Assignments with Exactly K Visible People?

The combinatorics plus enumeration approach is the most efficient. Instead of testing all 2^n direction assignments, it counts how many arrangements produce exactly K visible people by splitting visibility between the left and right sides. Using precomputed combinations allows the algorithm to run in about O(nk) time with O(n) space.

Is Direction Assignments with Exactly K Visible People asked at Google/Amazon/Meta?

Problems combining visibility rules with combinatorial counting appear in interviews at large tech companies such as Google, Amazon, and Meta. They test understanding of counting techniques, permutations, and the ability to reduce exponential search spaces into mathematical counting problems.

What data structure is used in Direction Assignments with Exactly K Visible People?

The solution primarily relies on mathematical structures such as factorial arrays and combination tables for efficient nCr computation. Some implementations also use simple arrays or dynamic programming tables to store intermediate combinatorial values.

What is the time complexity of Direction Assignments with Exactly K Visible People?

The optimal solution runs in O(nk) time by enumerating how many people are visible from each side and using combinatorial counting to compute the number of valid assignments. Space complexity is typically O(n) if factorials or dynamic programming tables are used for combinations.

Description Hints 5 Solution Video 14:58

3881. Direction Assignments with Exactly K Visible People

Q: How to solve Direction Assignments with Exactly K Visible People in O(nk)?

Enumerate the number of people visible from the left and right such that their sum equals K. For each split, use combinatorics to count how many configurations produce that visibility pattern. Precomputing factorials or combination values allows each count to be computed in constant time during the enumeration loop.

Medium35.7% acceptance

You are given three integers n, pos, and k.

There are n people standing in a line indexed from 0 to n - 1. Each person independently chooses a direction:

'L': visible only to people on their right
'R': visible only to people on their left

A person at index pos sees others as follows:

A person i < pos is visible if and only if they choose 'L'.
A person i > pos is visible if and only if they choose 'R'.

Return the number of possible direction assignments such that the person at index pos sees exactly k people.

Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 3, pos = 1, k = 0

Output: 2

Explanation:

Index 0 is to the left of pos = 1, and index 2 is to the right of pos = 1.
To see k = 0 people, index 0 must choose 'R' and index 2 must choose 'L', keeping both invisible.
The person at index 1 can choose 'L' or 'R' since it does not affect the count. Thus, the answer is 2.

Example 2:

Input: n = 3, pos = 2, k = 1

Output: 4

Explanation:

Index 0 and index 1 are left of pos = 2, and there is no index to the right.
To see k = 1 person, exactly one of index 0 or index 1 must choose 'L', and the other must choose 'R'.
There are 2 ways to choose which index is visible from the left.
The person at index 2 can choose 'L' or 'R' since it does not affect the count. Thus, the answer is 2 + 2 = 4.

Example 3:

Input: n = 1, pos = 0, k = 0

Output: 2

Explanation:

There are no indices to the left or right of pos = 0.
To see k = 0 people, no additional condition is required.
The person at index 0 can choose 'L' or 'R'. Thus, the answer is 2.

Constraints:

1 <= n <= 10⁵
0 <= pos, k <= n - 1

Topics

Combinatorics Math

Hints open one at a time

Hint 1

Use combinatorics

Hint 2

Count the visible people on the left and right independently. Let a = pos and b = n - pos - 1.

Hint 3

To see exactly k people, choose i visible people from the left side and k - i from the right side.

Hint 4

Each chosen left person must pick L, each chosen right person must pick R, and every other person can pick either direction.

Hint 5

Sum over all valid i, multiplying the two binomial choices and the free choices for the remaining people.

14:58

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def countVisiblePeople(self, n: int, pos: int, k: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

3
1
0

Input

3
2
1

Input

1
0
0

Your code is saved on this device.Solution