O(1) | Efficiently Finding the Child Who Holds the Ball After K Seconds
Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve the problem of finding the child who has the ball after k seconds, we can leve
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve the problem of finding the child who has the ball after k seconds, we can leverage the observation that the ball\'s movement is periodic. The ball moves back and forth between the two ends of the line. Therefore, after a certain number of seconds, the ball\'s position starts to repeat in a predictable pattern. By understanding this pattern, we can determine the position of the ball without simulating each second.\n\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Identify the Periodicity:\n\nSince the ball changes direction at the ends of the line, the ball\'s position after k seconds can be found by considering the total length of movement before it starts repeating.\nThe ball travels from the start to the end and back again, forming a complete cycle. The length of this cycle is 2 * (n - 1).\n\n- Calculate Effective Position:\n\nThe effective number of moves is k % (2 * (n - 1)) because the position repeats every 2 * (n - 1) seconds.\nDepending on whether the ball is moving forward or backward within this effective number of moves, we can determine its exact position.\n\n- Determine Direction:\n\nIf the ball has moved past n - 1 moves in the effective number of moves, it means it is on its way back. Thus, we need to adjust the position accordingly.\n\n# Complexity\n- Time complexity:\nThe time complexity is O(1) since we are performing a constant number of operations regardless of the input size.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\nThe space complexity is O(1) as we are using only a fixed amount of extra space.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int numberOfChild(int n, int k) {\n        n--;\n        int op = k%(2*n);\n        if(op/n){\n            return 2*n-op;\n        }\n        return op;\n    }\n};\n```\nIn this solution, we decrement n by 1 initially to simplify the calculation since we often reference n-1 in the process. The variable op represents the effective position within the cycle. If op is greater than or equal to n, it indicates the ball is on the return trip, and we adjust accordingly to find the correct child\'s position. This approach ensures that we determine the ball\'s position efficiently without unnecessary simulations.\n
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