Find the Child Who Has the Ball After K Seconds

• Initialize currentChild = 0 and direction = 1 (1 for right, -1 for left).
• Loop k times to simulate k passes of the ball.
• In each iteration, update the position by one step: currentChild += direction.
• After updating, check if the new position is an endpoint (0 or n-1).
• If it is, reverse the direction for the next pass: direction *= -1.
• After the loop completes, return the final currentChild.

We can solve this problem by simulating the movement of the ball for k seconds.

• We'll use a variable currentChild to track the index of the child with the ball, starting at 0.
• Another variable, direction, will keep track of the movement direction (1 for right, -1 for left), starting at 1.
• We then loop k times. In each iteration, representing one pass:
  1. We update currentChild by adding the direction. This simulates the pass.
  2. After the pass, we check if the ball has landed on an endpoint (0 or n-1).
  3. If it has, we reverse the direction for the subsequent passes by multiplying it by -1.

After k iterations, the value of currentChild is the final answer.

1class Solution {2    public int findTheChild(int n, int k) {3        int currentChild = 0;4        int direction = 1; // 1 for right, -1 for left5 6        for (int time = 0; time < k; time++) {7            currentChild += direction;8            if (currentChild == 0 || currentChild == n - 1) {9                direction *= -1;10            }11        }12        return currentChild;13    }14}

• Calculate the time for one full round trip (from 0 to n-1 and back to 0): period = 2 * (n - 1).
• Calculate the effective time within a single cycle using the modulo operator: effectiveTime = k % period.
• Determine if the ball is in the forward or backward part of its journey:
  • If effectiveTime <= n - 1, the ball is moving from left to right. Its position is effectiveTime.
  • If effectiveTime > n - 1, the ball is moving from right to left. Its position is (n - 1) - (effectiveTime - (n - 1)).
• Return the calculated position.

The movement of the ball is cyclical. It travels from child 0 to n-1 (which takes n-1 seconds) and then back from n-1 to 0 (which also takes n-1 seconds).

• A complete round trip, therefore, takes (n-1) + (n-1) = 2 * (n-1) seconds.
• We can determine the final position by figuring out where in this cycle the k-th second falls. The length of one full cycle is period = 2 * (n - 1).
• We can find the effective time within a single cycle using the modulo operator: effectiveTime = k % period.
• Now, we analyze the position based on effectiveTime:
  • Forward Pass: If effectiveTime is less than or equal to n-1, the ball is still on its way from 0 to n-1. In this case, its position is simply effectiveTime.
  • Backward Pass: If effectiveTime is greater than n-1, the ball has reached child n-1 and is on its way back. It reached n-1 at time n-1 (within the cycle) and has been moving back for effectiveTime - (n-1) seconds. So, its position is (n-1) - (effectiveTime - (n-1)).

1class Solution {2    public int findTheChild(int n, int k) {3        // The time for a full round trip (0 -> n-1 -> 0) is 2 * (n - 1) seconds.4        int period = 2 * (n - 1);5 6        // We only need to consider the position within the last, incomplete round trip.7        int effectiveTime = k % period;8 9        // If the effective time is less than or equal to the time to reach the end (n - 1),10        // the ball is moving forward from 0. The position is equal to the time elapsed.11        if (effectiveTime <= n - 1) {12            return effectiveTime;13        } else {14            // Otherwise, the ball is moving backward from n - 1.15            // It reached n-1 at time (n-1) and is now moving back.16            // The number of steps back from n-1 is (effectiveTime - (n-1)).17            return (n - 1) - (effectiveTime - (n - 1));18        }19    }20}