Solving the 'Consecutive Game Winner' Problem: Intuitive and Efficient Approaches
Intuition\n Describe your first thoughts on how to solve this problem. \nInefficient Approach :\nIn the inefficient approach, I simulated the game as described 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n***Inefficient Approach :***\nIn the inefficient approach, I simulated the game as described in the problem statement. I maintained a count of consecutive wins for each element and rotate the array after each round until one player wins **k** consecutive rounds. This approach is straightforward but not efficient for large inputs because it involves simulating all rounds.\nSo the solution was right but the time limit gets exceeded.\n\n***Efficient Approach :***\nIn the efficient approach, I optimized the solution by directly calculating the winner without simulating each round. just iterate through the array, keeping track of the current winner and the consecutive wins. This approach avoids the overhead of simulating the game and is more suitable for large inputs.\n\n\n***The efficient approach avoids the need to simulate all rounds and provides a faster solution to the problem.***\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n***Inefficient Approach :***\n1. Initialize a count for consecutive wins for each player and the current winner.\n2. While the current winner hasn\'t won **k** consecutive rounds:\n    -> Compare the two players (arr[0] and arr[1]) in the current round.\n    -> Update the consecutive win count for the winner of the round.\n    -> If the winner has won **k** consecutive rounds, return the winner.\n    -> Rotate the array to prepare for the next round.\n3. Return the current winner as the result.\n\n***Efficient Approach :***\n1. Initialize variables to keep track of the current winner and the number of consecutive wins.\n2. Iterate through the array starting from the second element (arr[1]) since the first element is the initial winner.\n3. Compare the current player with the current winner.\n4. If the current player is larger, update the current winner and reset the consecutive win count to 1. Otherwise, increment the consecutive win count.\n5. Check if the current winner has won **k** consecutive rounds during each iteration.\n6. When the current winner has won **k** consecutive rounds, return the winner as the result.\n7. If no winner is found during the iteration, return the initial winner as the result.\n\n# Complexity\n\n***Inefficient Approach:***\n- **Time complexity:** O(n * k), where **n** is the number of elements in the input array **arr**.\n- **Space complexity:** O(1) \n\n***efficient Approach:***\n- **Time complexity:** O(n), where **n** is the number of elements in the input array **arr**.\n- **Space complexity:** O(1) \n\n# Code\n```\nclass Solution {\npublic:\n/******** Inefficient Solution********\n    int getWinner(std::vector<int>& arr, int k) {\n        map<int, int> result;\n\n        int win_count = 0;\n        int winner = arr[0];\n\n        while (win_count < k) {\n            int a = arr[0];\n            int b = arr[1];\n\n            if (a > b) {\n                result[a]++;\n                win_count = std::max(win_count, result[a]);\n                if (result[a] == k) {\n                    return a;\n                }\n                rotate(arr.begin() + 1, arr.begin() + 2, arr.end());\n            } else {\n                result[b]++;\n                win_count = max(win_count, result[b]);\n                if (result[b] == k) {\n                    return b;\n                }\n                rotate(arr.begin(), arr.begin() + 1, arr.end());\n            }\n        }\n        return winner; \n    }\n    */\n    \n    int getWinner(vector<int>& arr, int k) {\n        int first=0; int second=1; int count=0;\n        while(first < arr.size() && second< arr.size()){\n            if(count==k) return arr[first];\n            \n            if(arr[first] > arr[second]){\n                count++;\n                second++;\n            }\n            else{\n                count=1;\n                while(first!=second){\n                    first++;\n                }\n                first++;\n                int temp=second;\n                second= first;\n                first= temp;\n            }\n        }\n        return arr[first];\n    }\n};\n```
['C++']