Problem Description

You are given two positive integers n and k. There are n children numbered from 0 to n - 1 standing in a queue in order from left to right.

Initially, child 0 holds a ball and the direction of passing the ball is towards the right direction. After each second, the child holding the ball passes it to the child next to them. Once the ball reaches either end of the line, i.e., child 0 or child n - 1, the direction of passing is reversed.

Return the number of the child who receives the ball after k seconds.

Intuition

The problem involves simulating the passing of a ball between children, where the direction of the pass changes upon reaching either end of the line. To solve this, we first observe a pattern in the passing sequence. The key idea is to recognize the periodic nature of how the ball passes back and forth.

Each complete back-and-forth pass among the children takes n - 1 seconds. This is because the ball moves from child 0 to n-1 and then back to 0, taking n - 1 steps in each direction.

By using modular arithmetic, we can find the position of the ball without simulating each pass individually:

1. Calculate k mod (n - 1) to determine the position of the ball after k moves within the current back-and-forth pass.
2. Determine the direction using the quotient of k divided by n - 1. If this quotient is even, the ball's final position is moving in the initial direction (from the start to the end). If the quotient is odd, the ball is moving in the opposite direction (from the end to the start).

This allows us to efficiently compute the position of the child holding the ball after k seconds.

Learn more about Math patterns.

Solution Approach

The solution uses a mathematical approach to efficiently compute the child holding the ball after k seconds, using the periodic nature of the passing and modular arithmetic. Here's a step-by-step walk-through of the solution:

1. Compute Modulo:

   • First, we compute k % (n-1), which determines how far into the current forward or backward pass the ball would be after k seconds.

2. Compute Division:

   • Divide k by n-1 to get the number of complete forward and backward passes. This quotient (k // (n-1)) will determine the current direction of the pass:
     • If the quotient is even, the direction is from the starting child 0 towards the last child n-1.
     • If the quotient is odd, the direction is from the last child n-1 back towards the starting child 0.

3. Determine the Final Position:

   • If the current direction determined by the quotient is even, the ball is moving towards the right, and you can find the ball's position directly using the modulo result.
   • If the direction is odd, reverse the movement description: the ball is passing left, so the final position is n - mod - 1.

By acting on the periodic behavior and reversing the direction based on whether the full pass count (k // (n-1)) is even or odd, the solution quickly arrives at the correct child's index.

Here's the implementation of this approach:

class Solution:
    def numberOfChild(self, n: int, k: int) -> int:
        k, mod = divmod(k, n - 1)
        return n - mod - 1 if k & 1 else mod

This code effectively calculates which child will end up with the ball after k seconds, using efficient mathematical operations without the need to simulate each move individually.

Example Walkthrough

Let's consider an example with n = 5 children and k = 10 seconds:

1. Initial Setup:

   • We have 5 children: numbered 0 to 4.
   • Initially, child 0 holds the ball and the direction is towards the right.

2. Compute Modulo:

   • Calculate 10 % (5 - 1) = 10 % 4 = 2.
   • This means that, within its current back-and-forth sequence, the ball would be 2 steps into the pass.

3. Compute Division:

   • Calculate 10 // (5 - 1) = 10 // 4 = 2.
   • This quotient tells us that there have been 2 complete back-and-forth passes.

4. Determine Direction:

   • Since the quotient 2 is even, the final direction of the ball is in its initial direction (towards the right).

5. Determine the Final Position:

   • Since the direction is towards the right, the ball ends up at position 2, which is child 2.

By following the calculation steps, child 2 will be holding the ball after 10 seconds have passed.

class Solution:
    def numberOfChild(self, n: int, k: int) -> int:
        k, mod = divmod(k, n - 1)
        return n - mod - 1 if k & 1 else mod

Through this walkthrough, we observe that the mathematical approach efficiently determines the position of the ball after k seconds by leveraging the periodic nature of the passing.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(1) because the operations (division, modulus, bitwise AND, subtraction, and conditional expression) are all constant time operations, independent of the input size.

The space complexity is O(1) as well, as the number of variables used does not increase with the input size; they remain fixed regardless of n or k.

Learn more about how to find time and space complexity quickly.