Simple Java Code ☠️
Code\n\nclass Solution {\n    public int numberOfChild(int n, int k) {\n        if(k>=(2*(n-1))){\n            k = k%(2*(n-1));\n        }\n        if(k>=(n-1))
# Code\n```\nclass Solution {\n    public int numberOfChild(int n, int k) {\n        if(k>=(2*(n-1))){\n            k = k%(2*(n-1));\n        }\n        if(k>=(n-1)){\n            k = k%(n-1);\n            return (n-k-1);\n        }\n        else{\n            return k;\n        }\n    }\n}\n```\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nThe problem involves determining the position of a ball passed among `n` children in a line, with the direction of passing reversing when the ball reaches either end. The objective is to find out which child will have the ball after `k` seconds.\n\n#### Key Insights:\n\n1. **Cyclic Nature of the Ball Passing**:\n   - When the ball is passed from the first child (child 0) to the last child (child \\( n-1 \\)) and back to the first child, it forms a cycle.\n   - The length of one complete cycle is \\( 2 \\times (n - 1) \\) seconds (moving to the right and then back to the left).\n\n2. **Normalization of `k`**:\n   - If \\( k \\) is greater than or equal to the length of one complete cycle, we can reduce \\( k \\) using the modulo operation. This is because after every complete cycle, the position of the ball repeats.\n   - Thus, `k = k % (2 times (n - 1))` gives us the effective number of seconds within a single cycle.\n\n3. **Determining the Ball\'s Position**:\n   - After normalizing \\( k \\), we need to determine the ball\'s position within the current cycle:\n     - If \\( k \\) is less than \\( n - 1 \\), the ball is moving to the right, and the position is directly \\( k \\).\n     - If \\( k \\) is greater than or equal to \\( n - 1 \\), the ball has reached the end and is now moving back to the left. In this case, we compute the position as \\( (n - 1) - (k \\% (n - 1)) \\).\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Cycle Length Calculation**:\n   - The ball travels from child 0 to child \\( n-1 \\) and back to child 0. This round trip constitutes a complete cycle of length \\( 2 \\times (n - 1) \\) seconds.\n\n2. **Reduction of `k` within a Single Cycle**:\n   - If `k` is larger than or equal to the length of a complete cycle, it can be reduced to within the cycle using the modulo operation. This is based on the observation that the ball\'s position repeats every \\( 2 \\times (n - 1) \\) seconds.\n   - `k = k % (2 \\times (n - 1))` ensures that `k` falls within the range of one complete cycle.\n\n3. **Determining the Position within the Cycle**:\n   - After normalizing `k`, we need to find out where the ball is in its current movement:\n     - **Forward Movement**:\n       - If `k` is less than \\( n - 1 \\), the ball is still moving towards the right (from child 0 towards child \\( n-1 \\)).\n       - In this case, the position of the ball is simply `k`.\n     - **Backward Movement**:\n       - If `k` is greater than or equal to \\( n - 1 \\), the ball has reached the end (child \\( n-1 \\)) and is moving back towards the left (towards child 0).\n       - To find the exact position during the backward movement, we use the expression `n - k - 1`. This accounts for the reverse traversal from child \\( n-1 \\) back to child 0.\n\nCertainly! Let\'s go through your code step-by-step with a specific example to illustrate how it determines which child has the ball after `k` seconds.\n\n### Example:\n\nLet\'s take `n = 5` children and `k = 12` seconds.\n\n### Step-by-Step Execution:\n\n#### Step 1: Calculate the Cycle Length\n- The cycle length is \\(2 \\times (n - 1)\\):\n  \\[\n  2 \\times (5 - 1) = 8 \\text{ seconds}\n  \\]\n- This means the ball travels from child 0 to child 4 and back to child 0 in 8 seconds.\n\n#### Step 2: Normalize `k` Within a Single Cycle\n- Given `k = 12`, which is larger than the cycle length:\n  ```java\n  if (k >= 2 * (n - 1)) {\n      k = k % (2 * (n - 1));\n  }\n  ```\n- So, we reduce `k` using modulo operation:\n  \\[\n  12 \\% 8 = 4\n  \\]\n- Now, `k = 4` seconds, which is within a single cycle.\n\n#### Step 3: Determine the Ball\'s Position Within the Cycle\n- Compare `k` to `n - 1`:\n  \\[\n  n - 1 = 5 - 1 = 4\n  \\]\n- Since `k = 4` is equal to `n - 1`, it means the ball is at the last child (child 4) and will start moving back to the left in the next second.\n\nHere\u2019s the relevant code segment:\n```java\nif (k >= n - 1) {\n    k = k % (n - 1);\n    return (n - k - 1);\n} else {\n    return k;\n}\n```\n- Given `k = 4`:\n  ```java\n  k = 4 % 4 = 0;\n  return (5 - 0 - 1) = 4;\n  ```\n- The ball is with child 4 after `12` seconds.\n\n# Complexity\n\n- **Time Complexity**: \\( O(1) \\)\n  - The code performs a constant number of operations regardless of the input size `n` and `k`. This includes arithmetic operations and conditional checks, all of which execute in constant time.\n\n- **Space Complexity**: \\( O(1) \\)\n  - The code uses a constant amount of extra space, independent of the input size. It only requires space for a few variables (`n` and `k`), and does not use any additional data structures that would scale with the input.\n\n
['Java']