Java's Tiny Treasure Hunt: Min Element Mastery! 🕵️‍♂️💎
 # Intuition  \n\n\n# Approach\n\n- Initialize res: Start with the maximum possible integer value.\n\n- Loop Through Array: Iterate through each number in the a
<!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- **Initialize res:** Start with the maximum possible integer value.\n\n- **Loop Through Array:** Iterate through each number in the array.\n\n- **Sum Digits for Numbers > 9:** For each number greater than 9, sum its digits.\n\n- **Update Element:** Replace the original number with its digit sum.\n\n- **Update Minimum Value:** Use Math.min to keep track of the smallest value.\n\n- **Return Result:** Return the minimum value found in the array.\n\n# Complexity\n- Time complexity: *O(n * d)*\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:   *O(1)*\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n    public int minElement(int[] nums) {\n        // Use the maximum possible value for initialization\n        int res = Integer.MAX_VALUE; \n\n        for (int i = 0; i < nums.length; i++) {\n            if (nums[i] > 9) {\n                int k = nums[i];\n                int sum = 0;\n                \n                // Correct the logic for reducing the number\n                while (k > 0) {\n                    // Sum the digits\n                    sum += k % 10; \n                    // Move to the next digit\n                    k /= 10;\n                }\n\n                nums[i] = sum;\n            }\n            // Update the result with the current minimum\n            res = Math.min(res, nums[i]); \n        }\n\n        return res;\n    }\n}\n\n```
['Java']