🌟 JavaScript || Professional Approach and In-Depth Explanation
Intuition\nTo find the largest divisible subset, we need to identify elements in the input array such that each element is divisible by the previous one. We can
# Intuition\nTo find the largest divisible subset, we need to identify elements in the input array such that each element is divisible by the previous one. We can use dynamic programming to solve this problem efficiently.\n\n# Approach\n1. Sort the input array in ascending order.\n2. Initialize two arrays:\n   - `dp`: to store the length of the largest divisible subset ending at each index.\n   - `prev`: to store the index of the previous element in the largest divisible subset for each element.\n3. Initialize `maxIndex` to keep track of the index of the element with the largest divisible subset.\n4. Iterate through each element in the input array.\n   - For each element, iterate through all previous elements.\n   - If the current element is divisible by the previous element and adding the current element to the subset would result in a larger subset, update `dp[i]` and `prev[i]`.\n5. Update `maxIndex` if necessary.\n6. Reconstruct the largest divisible subset using the `prev` array.\n7. Return the largest divisible subset in reversed order.\n\n# Complexity\n- Time complexity: O(n^2), where n is the number of elements in the input array. The nested loops iterate over each element in the worst-case scenario.\n- Space complexity: O(n) for the `dp` and `prev` arrays, and O(n) for the `result` array. Therefore, the overall space complexity is O(n).\n\n# Code\n```\nvar largestDivisibleSubset = function(nums) {\n    const n = nums.length;\n    \n    // Edge case: if there are less than or equal to one element, return the input array\n    if (n <= 1) {\n        return nums;\n    }\n\n    // Sort the input array in ascending order\n    nums.sort((a, b) => a - b);\n    \n    // Initialize arrays to store the length of the largest divisible subset ending at each index,\n    // as well as the previous index of each element in the largest divisible subset\n    const dp = new Array(n).fill(1);\n    const prev = new Array(n).fill(-1);\n    \n    // Initialize a variable to keep track of the index of the element with the largest divisible subset\n    let maxIndex = 0;\n\n    // Loop through each element in the input array\n    for (let i = 1; i < n; ++i) {\n        // Loop through each element before the current element\n        for (let j = 0; j < i; ++j) {\n            // Check if the current element is divisible by the previous element\n            // and if adding the current element to the subset would result in a larger subset\n            if (nums[i] % nums[j] === 0 && dp[i] < dp[j] + 1) {\n                // Update the length of the largest divisible subset ending at the current index\n                dp[i] = dp[j] + 1;\n                // Update the previous index of the current element in the largest divisible subset\n                prev[i] = j;\n            }\n        }\n\n        // Update the index of the element with the largest divisible subset if necessary\n        if (dp[i] > dp[maxIndex]) {\n            maxIndex = i;\n        }\n    }\n\n    // Reconstruct the largest divisible subset using the previous indices\n    const result = [];\n    while (maxIndex !== -1) {\n        result.push(nums[maxIndex]);\n        maxIndex = prev[maxIndex];\n    }\n\n    // Return the largest divisible subset in reversed order\n    return result.reverse();\n};\n\n```
['JavaScript']