solution with comments|| using map
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\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        // Create an unordered_map to store positions of each distinct value\n        unordered_map<int, vector<int>> mp;\n\n        int n = nums.size(); // Get the total number of elements in nums\n\n        // Step 1: Populate the \'mp\' map with positions of each distinct value\n        for (int i = 0; i < n; i++) {\n            mp[nums[i]].push_back(i);\n        }\n\n        int ans = n - 1; // Initialize the answer to the maximum possible value\n\n        // Step 2: Iterate through the \'mp\' map for each distinct value\n        for (auto it : mp) {\n            vector<int> pos = it.second; // Get the positions of the current value\n            int len = pos.size(); // Get the number of occurrences of the current value\n\n            int temp = (n - 1 - pos[len - 1]) + pos[0]; // Initialize \'temp\' with time to reach the last occurrence and back to 0\n            for (int j = 1; j < len; j++) {\n                // Calculate the time to reach each consecutive occurrence\n                temp = max(temp, pos[j] - pos[j - 1] - 1);\n            }\n\n            // Update \'ans\' with the minimum time required for the current value\n            ans = min(ans, temp);\n        }\n\n        // Return the final answer (add 1 and divide by 2 as required)\n        return (ans + 1) / 2;\n    }\n};\n\n```
['C++']