3178. Find the Child Who Has the Ball After K Seconds LeetCode Solution
In this guide, you will get 3178. Find the Child Who Has the Ball After K Seconds LeetCode Solution with the best time and space complexity. The solution to Find the Child Who Has the Ball After K Seconds problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Find the Child Who Has the Ball After K Seconds
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.
Note: This question is the same as 2582: Pass the Pillow.
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(1)
3178. Find the Child Who Has the Ball After K Seconds LeetCode Solution in C++
class Solution {
public:
int numberOfChild(int n, int k) {
// the time for the ball to return to 0
const int roundTime = 2 * (n - 1);
const int pos = k % roundTime;
return pos < n ? pos : roundTime - pos;
}
}; /* code provided by PROGIEZ */
3178. Find the Child Who Has the Ball After K Seconds LeetCode Solution in Java
class Solution {
public int numberOfChild(int n, int k) {
// the time for the ball to return to 0
final int roundTime = 2 * (n - 1);
final int pos = k % roundTime;
return pos < n ? pos : roundTime - pos;
}
} // code provided by PROGIEZ
3178. Find the Child Who Has the Ball After K Seconds LeetCode Solution in Python
class Solution:
def numberOfChild(self, n: int, k: int) -> int:
# the time for the ball to return to 0
roundTime = 2 * (n - 1)
pos = k % roundTime
return pos if pos < n else roundTime - pos # code by PROGIEZ
