Find the Child Who Has the Ball After K Seconds

Easy

Agoda

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Topics:

Arrays

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.

Example 1:

Input: n = 3, k = 5

Output: 1

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2]`
`1`	`[0, 1, 2]`
`2`	`[0, 1, 2]`
`3`	`[0, 1, 2]`
`4`	`[0, 1, 2]`
`5`	`[0, 1, 2]`

Example 2:

Input: n = 5, k = 6

Output: 2

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2, 3, 4]`
`1`	`[0, 1, 2, 3, 4]`
`2`	`[0, 1, 2, 3, 4]`
`3`	`[0, 1, 2, 3, 4]`
`4`	`[0, 1, 2, 3, 4]`
`5`	`[0, 1, 2, 3, 4]`
`6`	`[0, 1, 2, 3, 4]`

Example 3:

Input: n = 4, k = 2

Output: 2

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2, 3]`
`1`	`[0, 1, 2, 3]`
`2`	`[0, 1, 2, 3]`

Constraints:

2 <= n <= 50
1 <= k <= 50

Note: This question is the same as 2582: Pass the Pillow.

Try coding it in LeetCode

Find the Child Who Has the Ball After K Seconds

Easy

Agoda

8 views

Topics:

Arrays

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.

Example 1:

Input: n = 3, k = 5

Output: 1

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2]`
`1`	`[0, 1, 2]`
`2`	`[0, 1, 2]`
`3`	`[0, 1, 2]`
`4`	`[0, 1, 2]`
`5`	`[0, 1, 2]`

Example 2:

Input: n = 5, k = 6

Output: 2

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2, 3, 4]`
`1`	`[0, 1, 2, 3, 4]`
`2`	`[0, 1, 2, 3, 4]`
`3`	`[0, 1, 2, 3, 4]`
`4`	`[0, 1, 2, 3, 4]`
`5`	`[0, 1, 2, 3, 4]`
`6`	`[0, 1, 2, 3, 4]`

Example 3:

Input: n = 4, k = 2

Output: 2

Explanation:

Time elapsed	Children
`0`	`[0, 1, 2, 3]`
`1`	`[0, 1, 2, 3]`
`2`	`[0, 1, 2, 3]`

Constraints:

2 <= n <= 50
1 <= k <= 50

Note: This question is the same as 2582: Pass the Pillow.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the data type of `k` and what are the possible range of values for `k`? Can `k` be negative?
What is the data type of the array representing the circle of children, and what are the possible values it can hold? Are the children represented by integers?
Is the circle represented by a circular array or a linear array, and if linear, how do we handle going past the end of the array?
What happens if the number of seconds `k` is larger than the number of children in the circle?
Is the starting position (where the ball begins) always the first child (index 0), or is it provided as input? If it's provided, what are the possible values?

Brute Force Solution

Approach

We're tracking a ball being passed between children. The brute force method simulates the ball's journey step-by-step, mimicking each pass until the desired time is reached. We'll follow the ball as it moves, noting who has it at each second.

Here's how the algorithm would work step-by-step:

Start with the child who initially has the ball.
Simulate the first pass: determine which child receives the ball from the current child.
Record that the new child now has the ball after one second.
Simulate the next pass from this new child, determining who they pass it to.
Record that the new child now has the ball after two seconds.
Continue simulating the passes, one second at a time, and updating who has the ball.
Keep doing this until we've simulated the correct number of seconds.
The child holding the ball at the very end is the answer.

Code Implementation

def find_child_with_ball_brute_force(
        number_of_children,
        initial_child,
        passing_time
):

    current_child = initial_child

    # Simulate each second of passing
    for current_second in range(passing_time):

        # Determine the next child based on who has the ball now
        if current_child % 2 == 0:
            next_child = (current_child + 1) % number_of_children

        # Odd numbered child passes to the previous child
        else:
            next_child = (current_child - 1) % number_of_children

        current_child = next_child

    return current_child

Big(O) Analysis

Time Complexity

O(k) – The provided solution simulates the passing of the ball for k seconds. Each second, the algorithm determines the next child who will have the ball based on the current child. The primary loop iterates k times, representing the k seconds of ball passing. Therefore, the time complexity is directly proportional to k, the number of seconds.

Space Complexity

O(1) – The provided plain English solution describes a step-by-step simulation that only requires tracking the current child holding the ball. This implies using a single variable to store the index of the current child. No additional data structures, like arrays or hash maps, are mentioned or needed to store intermediate results. The space used remains constant regardless of the number of children or the number of seconds. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The problem involves figuring out where a ball ends up after being passed around a circle of children for a certain amount of time. The most efficient way to solve it involves using the remainder after dividing the total time by the number of children to quickly find the final child.

Here's how the algorithm would work step-by-step:

Figure out how many kids are in the circle.
Take the total time the ball is passed and find the remainder when you divide it by the number of kids.
Add this remainder to the starting child's number.
If the result is larger than the number of kids, subtract the number of kids to find the actual final child.
The final result is the child that holds the ball after that many passes.

Code Implementation

def find_the_child_with_ball(number_of_children, start_child, time_passed):
    # Calculate the remainder to find the effective number of passes.
    remainder = time_passed % number_of_children

    final_child = start_child + remainder

    # Adjust if the final child exceeds the number of children.
    if final_child > number_of_children:

        final_child = final_child - number_of_children

    # Return the child number holding the ball.
    return final_child

Big(O) Analysis

Time Complexity

O(1) – The algorithm performs a fixed number of arithmetic operations regardless of the input size. We calculate the remainder of the time divided by the number of children, add it to the starting child, and potentially subtract the number of children once. These are all constant-time operations. Therefore, the time complexity is O(1).

Space Complexity

O(1) – The algorithm calculates the final child by using only a few variables: the number of kids, the remainder, and the initial child. These variables require constant space regardless of the number of children (N) or the number of passes (K). No additional data structures like arrays or hash maps are created to store intermediate results. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty array for child positions	Return -1 immediately, indicating no child has the ball.
K is zero	Return the initial position, as no time has passed.
Array with a single child	The ball remains with this single child regardless of K, so return that position.
Ball returns to the starting position before K seconds elapse.	Handle using the modulo operator (%) to account for cyclical movement.
K is a very large number (potential overflow if not handled correctly)	Use modulo operator to reduce K to its effective range based on array length to prevent integer overflow.
Children positions are not sequential (e.g., 1, 3, 5)	The algorithm should correctly handle jumps between children, not assuming sequential numbering.
Negative K value	Return the initial position or treat K as its absolute value, documenting chosen behavior clearly.
Array contains duplicate child indices.	Return -1 or throw an exception as this is invalid input.