Optimized using Circular Approach with full Explanation !!! 🔥🔥🔥
Intuition\n Describe your first thoughts on how to solve this problem. \nIn summary, the code leverages the concept of circular movement to efficiently calculat
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn summary, the code leverages the concept of circular movement to efficiently calculate the minimum time required to move elements to the same index. It does this by simulating the circular arrangement of elements and calculating the time needed for each element to reach its destination index. The result is the minimum of these calculated times for all elements, giving the minimum time required for the entire array to reach a uniform state.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Storing Positions:**\nThe code first creates an unordered map indexPos to store the positions of each element in the array. This map will have keys as elements and values as vectors containing the positions where those elements appear in the array.\n\n**Cyclic Movement:**\nThe idea is to think of the array as a circle, where elements can wrap around from the end of the array to the beginning. This means that instead of moving elements through the entire array, we can calculate the time it takes for an element to move to its destination index considering the circular arrangement.\n\n**Finding Minimum Time:**\nThe code initializes a variable res to store the minimum time required to move all elements to the same index. This value is initialized with a large value (INT_MAX) to ensure it gets updated with the smallest time found during the process.\n\n**Processing Each Element:**\nThe code uses a loop to iterate through each element in the indexPos map.\n\nFor each element, it creates a vector positions containing the positions of that element, including a virtual "wrap-around" position. This wrap-around position is calculated by adding the length of the array n to the first position of the element. This simulates the circular movement of the element.\n\nIt then initializes a variable seconds to keep track of the maximum time required for the current element to move to its destination index.\n\nIt loops through the positions of the current element. Starting from the second position, it calculates the time required to move the element from the previous position to the current position, considering the circular movement. This time is half of the difference between the current and previous positions ((positions[i] - positions[i-1]) / 2).\n\nThe seconds variable keeps track of the maximum time required among all positions of the current element.\n\n**Updating Minimum Time:**\nAfter processing all positions of the current element, the code updates the res variable with the minimum of its current value and the calculated seconds value. This ensures that the overall minimum time across all elements is recorded.\n\n**Returning Minimum Time:**\nOnce all elements have been processed, the function returns the res value, which represents the minimum time required to move all elements to the same index while considering the circular movement.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int minimumSeconds(vector<int>& nums) {\n        //Map to store the indexes of each element in the array\n        unordered_map<int,vector<int>> indexPos;\n\n        int n = nums.size(); \n        for(int i=0;i<n;i++) {\n            indexPos[nums[i]].push_back(i);\n        }\n\n\n        int res = INT_MAX;\n\n        for(auto &[element,positions] : indexPos) {\n            positions.push_back(positions[0]+n);\n\n            int seconds = 0;\n\n            for(int i = 1;i<positions.size();i++) {\n                seconds = max(seconds,(positions[i]-positions[i-1])/2);\n            }\n\n            res = min(res,seconds);\n        }\n\n        return res;\n    }\n};\n```
['Ordered Map', 'C++']