PYTHON🔥|| BEAT 100%🔥|| 100% EFFICIENT🔥|| BEGINNER'S APPROACH || EASY TO UNDERSTAND ||
if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n Describe your first thoughts on how to solve this problem. \n- My first 
# if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- My first thought was to simulate the movement of the ball among the children in the queue. Given that the ball changes direction when it reaches either end of the line, I realized that the ball\'s movement forms a repeating pattern or period. This periodic behavior can be exploited to reduce the problem size and efficiently determine the final position of the ball after k seconds.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Period Identification:\n- - The ball completes one round trip (from the start to the end and back to the start) in 2 * (n - 1) seconds, where n is the number of children.\nThis period indicates that the ball\'s position after every 2 * (n - 1) seconds will be the same.\n- Effective Steps Calculation:\n- - To handle large values of k, compute the effective number of steps the ball moves within one period using k % (2 * (n - 1)). This reduces the problem size significantly.\nSimulate the Movement:\n- Initialize the ball\'s position at 0 and set the initial direction to the right.\n- Simulate the ball\'s movement for the calculated effective steps, adjusting the direction whenever the ball reaches either end of the line.\n- This approach ensures that we only simulate the minimal necessary steps to determine the final position, making the solution efficient.\n# Complexity\n- Time complexity: $$O(n)$$ --> The time complexity is O(effective_steps), which simplifies to O(n) in the worst case. However, since we reduce the problem using the period, the average case is much smaller and runs in constant time O(1) for large k.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: The space complexity is O(1) since we only use a few variables to keep track of the position, direction, and effective steps. No additional data structures are used that scale with the input size.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution(object):\n    def numberOfChild(self, n, k):\n        """\n        :type n: int\n        :type k: int\n        :rtype: int\n        """\n        # Calculate the effective number of steps within one period\n        period = 2 * (n - 1)\n        effective_steps = k % period\n        \n        # Simulate the ball\'s movement for effective_steps\n        position = 0\n        direction = 1  # 1 means right, -1 means left\n        \n        for _ in range(effective_steps):\n            if position == 0:\n                direction = 1\n            elif position == n - 1:\n                direction = -1\n            position += direction\n        \n        return position\n```\n![7c17f983-3b8b-4738-b722-c1c1466f9510_1682940288.2823734.jpeg](https://assets.leetcode.com/users/images/991e5760-934a-4085-8be6-cfbcadf754af_1717934544.848592.jpeg)\n
['Python']