368. Largest Divisible Subset - Explanation

Description

Given a set of distinct positive integers nums, return the largest subset answer such that every pair (answer[i], answer[j]) of elements in this subset satisfies:

answer[i] % answer[j] == 0, or
answer[j] % answer[i] == 0

If there are multiple solutions, return any of them.

Example 1:

Input: nums = [1,2,3]

Output: [1,2]

Explanation: [1,3] is also accepted.

Example 2:

Input: nums = [1,2,4,8]

Output: [1,2,4,8]

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 2,000,000,000
All the integers in nums are unique.

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Dynamic Programming (Memoization) - The problem requires caching recursive results to avoid recomputation
Sorting - Elements must be sorted first to establish the divisibility chain property
Divisibility and Modulo Operations - Understanding when one number divides another evenly
Recursion - Top-down approaches use recursive function calls with state tracking

1. Dynamic Programming (Top-Down)

Intuition

A divisible subset has a special property: if we sort the numbers, then for any pair in the subset, the larger number must be divisible by the smaller one. This means if we sort the array and pick elements in order, we only need to check divisibility with the most recently picked element. We can use recursion with memoization to try including or skipping each number.

Algorithm

Sort the array in ascending order.
Define a recursive function dfs(i, prevIndex) that returns the largest divisible subset starting from index i, where prevIndex is the index of the last included element (or -1 if none).
At each index:
- Option 1: Skip the current number.
- Option 2: If no previous element exists or the current number is divisible by the previous one, include it and recurse.
Memoize results based on (i, prevIndex) to avoid recomputation.
Return the larger of the two options.

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        cache = {}  # (i, prevIndex) -> List

        def dfs(i, prevIndex):
            if i == len(nums):
                return []
            if (i, prevIndex) in cache:
                return cache[(i, prevIndex)]

            res = dfs(i + 1, prevIndex)  # Skip nums[i]
            if prevIndex == -1 or nums[i] % nums[prevIndex] == 0:
                tmp = [nums[i]] + dfs(i + 1, i)  # Include nums[i]
                res = tmp if len(tmp) > len(res) else res

            cache[(i, prevIndex)] = res
            return res

        return dfs(0, -1)

public class Solution {
    private List<Integer>[][] cache;

    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        cache = new ArrayList[n][n + 1];
        return dfs(0, -1, nums);
    }

    private List<Integer> dfs(int i, int prevIndex, int[] nums) {
        if (i == nums.length) return new ArrayList<>();
        if (cache[i][prevIndex + 1] != null) return cache[i][prevIndex + 1];

        List<Integer> res = dfs(i + 1, prevIndex, nums);

        if (prevIndex == -1 || nums[i] % nums[prevIndex] == 0) {
            List<Integer> tmp = new ArrayList<>();
            tmp.add(nums[i]);
            tmp.addAll(dfs(i + 1, i, nums));
            if (tmp.size() > res.size()) res = tmp;
        }

        cache[i][prevIndex + 1] = res;
        return res;
    }
}

class Solution {
private:
    vector<vector<vector<int>>> cache;

public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        cache = vector<vector<vector<int>>>(n, vector<vector<int>>(n + 1));
        return dfs(0, -1, nums);
    }

    vector<int> dfs(int i, int prevIndex, vector<int>& nums) {
        if (i == nums.size()) return {};
        if (!cache[i][prevIndex + 1].empty()) return cache[i][prevIndex + 1];

        vector<int> res = dfs(i + 1, prevIndex, nums);

        if (prevIndex == -1 || nums[i] % nums[prevIndex] == 0) {
            vector<int> tmp = {nums[i]};
            vector<int> next = dfs(i + 1, i, nums);
            tmp.insert(tmp.end(), next.begin(), next.end());
            if (tmp.size() > res.size()) res = tmp;
        }

        return cache[i][prevIndex + 1] = res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    largestDivisibleSubset(nums) {
        nums.sort((a, b) => a - b);
        const cache = new Map();

        const dfs = (i, prevIndex) => {
            if (i === nums.length) return [];

            let key = `${i},${prevIndex}`;
            if (cache.has(key)) return cache.get(key);

            let res = dfs(i + 1, prevIndex);
            if (prevIndex === -1 || nums[i] % nums[prevIndex] === 0) {
                let tmp = [nums[i], ...dfs(i + 1, i)];
                if (tmp.length > res.length) res = tmp;
            }

            cache.set(key, res);
            return res;
        };

        return dfs(0, -1);
    }
}

public class Solution {
    private List<int>[][] cache;
    private int[] nums;

    public List<int> LargestDivisibleSubset(int[] nums) {
        Array.Sort(nums);
        this.nums = nums;
        int n = nums.Length;
        cache = new List<int>[n + 1][];
        for (int i = 0; i <= n; i++) {
            cache[i] = new List<int>[n + 1];
        }
        return Dfs(0, -1);
    }

    private List<int> Dfs(int i, int prevIndex) {
        if (i == nums.Length) return new List<int>();
        if (cache[i][prevIndex + 1] != null) return cache[i][prevIndex + 1];

        List<int> res = Dfs(i + 1, prevIndex);
        if (prevIndex == -1 || nums[i] % nums[prevIndex] == 0) {
            List<int> tmp = new List<int> { nums[i] };
            tmp.AddRange(Dfs(i + 1, i));
            if (tmp.Count > res.Count) res = tmp;
        }

        cache[i][prevIndex + 1] = res;
        return res;
    }
}

func largestDivisibleSubset(nums []int) []int {
    sort.Ints(nums)
    n := len(nums)
    cache := make([][][]int, n)
    for i := range cache {
        cache[i] = make([][]int, n+1)
    }

    var dfs func(i, prevIndex int) []int
    dfs = func(i, prevIndex int) []int {
        if i == n {
            return []int{}
        }
        if cache[i][prevIndex+1] != nil {
            return cache[i][prevIndex+1]
        }

        res := dfs(i+1, prevIndex)
        if prevIndex == -1 || nums[i]%nums[prevIndex] == 0 {
            tmp := append([]int{nums[i]}, dfs(i+1, i)...)
            if len(tmp) > len(res) {
                res = tmp
            }
        }

        cache[i][prevIndex+1] = res
        return res
    }

    return dfs(0, -1)
}

class Solution {
    private lateinit var cache: Array<Array<List<Int>?>>
    private lateinit var nums: IntArray

    fun largestDivisibleSubset(nums: IntArray): List<Int> {
        nums.sort()
        this.nums = nums
        val n = nums.size
        cache = Array(n) { arrayOfNulls<List<Int>>(n + 1) }
        return dfs(0, -1)
    }

    private fun dfs(i: Int, prevIndex: Int): List<Int> {
        if (i == nums.size) return emptyList()
        cache[i][prevIndex + 1]?.let { return it }

        var res = dfs(i + 1, prevIndex)
        if (prevIndex == -1 || nums[i] % nums[prevIndex] == 0) {
            val tmp = listOf(nums[i]) + dfs(i + 1, i)
            if (tmp.size > res.size) res = tmp
        }

        cache[i][prevIndex + 1] = res
        return res
    }
}

class Solution {
    func largestDivisibleSubset(_ nums: [Int]) -> [Int] {
        var nums = nums.sorted()
        let n = nums.count
        var cache = [[Int]?](repeating: nil, count: n * (n + 1))

        func dfs(_ i: Int, _ prevIndex: Int) -> [Int] {
            if i == n { return [] }
            let key = i * (n + 1) + (prevIndex + 1)
            if let cached = cache[key] { return cached }

            var res = dfs(i + 1, prevIndex)
            if prevIndex == -1 || nums[i] % nums[prevIndex] == 0 {
                let tmp = [nums[i]] + dfs(i + 1, i)
                if tmp.count > res.count { res = tmp }
            }

            cache[key] = res
            return res
        }

        return dfs(0, -1)
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n ^ 2)$

2. Dynamic Programming (Top-Down) Space Optimized

Intuition

We can simplify the state by observing that we only need to track the starting index, not the previous index. For each starting position, we find the longest divisible subset that begins there. When building the subset from index i, we look at all later indices j where nums[j] is divisible by nums[i] and take the best result.

Algorithm

Sort the array in ascending order.
Define dfs(i) that returns the largest divisible subset starting at index i.
At each index i:
- Initialize the result with just nums[i].
- For each later index j where nums[j] % nums[i] == 0, recursively get the subset starting at j.
- Prepend nums[i] to the best result and keep the longest.
Memoize results for each starting index.
Try all starting positions and return the longest subset found.

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        cache = {}

        def dfs(i):
            if i in cache:
                return cache[i]

            res = [nums[i]]
            for j in range(i + 1, len(nums)):
                if nums[j] % nums[i] == 0:
                    tmp = [nums[i]] + dfs(j)
                    if len(tmp) > len(res):
                        res = tmp

            cache[i] = res
            return res

        res = []
        for i in range(len(nums)):
            tmp = dfs(i)
            if len(tmp) > len(res):
                res = tmp
        return res

public class Solution {
    private List<Integer>[] cache;

    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        cache = new ArrayList[n];

        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            List<Integer> tmp = dfs(i, nums);
            if (tmp.size() > res.size()) {
                res = tmp;
            }
        }
        return res;
    }

    private List<Integer> dfs(int i, int[] nums) {
        if (cache[i] != null) return cache[i];

        List<Integer> res = new ArrayList<>();
        res.add(nums[i]);
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] % nums[i] == 0) {
                List<Integer> tmp = new ArrayList<>();
                tmp.add(nums[i]);
                tmp.addAll(dfs(j, nums));

                if (tmp.size() > res.size()) {
                    res = tmp;
                }
            }
        }
        return cache[i] = res;
    }
}

class Solution {
private:
    vector<vector<int>> cache;

public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        cache.resize(n, vector<int>());

        vector<int> res;
        for (int i = 0; i < n; i++) {
            vector<int> tmp = dfs(i, nums);
            if (tmp.size() > res.size()) {
                res = tmp;
            }
        }
        return res;
    }

    vector<int> dfs(int i, vector<int>& nums) {
        if (!cache[i].empty()) return cache[i];

        vector<int> res = {nums[i]};
        for (int j = i + 1; j < nums.size(); j++) {
            if (nums[j] % nums[i] == 0) {
                vector<int> tmp = {nums[i]};
                vector<int> next = dfs(j, nums);
                tmp.insert(tmp.end(), next.begin(), next.end());

                if (tmp.size() > res.size()) {
                    res = tmp;
                }
            }
        }
        return cache[i] = res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    largestDivisibleSubset(nums) {
        nums.sort((a, b) => a - b);
        const n = nums.length;
        const cache = new Array(n).fill(null);

        const dfs = (i) => {
            if (cache[i] !== null) return cache[i];

            let res = [nums[i]];
            for (let j = i + 1; j < n; j++) {
                if (nums[j] % nums[i] === 0) {
                    let tmp = [nums[i], ...dfs(j)];
                    if (tmp.length > res.length) {
                        res = tmp;
                    }
                }
            }
            return (cache[i] = res);
        };

        let res = [];
        for (let i = 0; i < n; i++) {
            let tmp = dfs(i);
            if (tmp.length > res.length) {
                res = tmp;
            }
        }
        return res;
    }
}

public class Solution {
    private Dictionary<int, List<int>> cache;
    private int[] nums;

    public List<int> LargestDivisibleSubset(int[] nums) {
        Array.Sort(nums);
        this.nums = nums;
        cache = new Dictionary<int, List<int>>();
        List<int> res = new List<int>();

        for (int i = 0; i < nums.Length; i++) {
            List<int> tmp = Dfs(i);
            if (tmp.Count > res.Count) {
                res = tmp;
            }
        }
        return res;
    }

    private List<int> Dfs(int i) {
        if (cache.ContainsKey(i)) return cache[i];

        List<int> res = new List<int> { nums[i] };
        for (int j = i + 1; j < nums.Length; j++) {
            if (nums[j] % nums[i] == 0) {
                List<int> tmp = new List<int> { nums[i] };
                tmp.AddRange(Dfs(j));
                if (tmp.Count > res.Count) {
                    res = tmp;
                }
            }
        }

        cache[i] = res;
        return res;
    }
}

func largestDivisibleSubset(nums []int) []int {
    sort.Ints(nums)
    n := len(nums)
    cache := make([][]int, n)

    var dfs func(i int) []int
    dfs = func(i int) []int {
        if cache[i] != nil {
            return cache[i]
        }

        res := []int{nums[i]}
        for j := i + 1; j < n; j++ {
            if nums[j]%nums[i] == 0 {
                tmp := append([]int{nums[i]}, dfs(j)...)
                if len(tmp) > len(res) {
                    res = tmp
                }
            }
        }

        cache[i] = res
        return res
    }

    res := []int{}
    for i := 0; i < n; i++ {
        tmp := dfs(i)
        if len(tmp) > len(res) {
            res = tmp
        }
    }
    return res
}

class Solution {
    private lateinit var cache: Array<List<Int>?>
    private lateinit var nums: IntArray

    fun largestDivisibleSubset(nums: IntArray): List<Int> {
        nums.sort()
        this.nums = nums
        cache = arrayOfNulls(nums.size)

        var res = emptyList<Int>()
        for (i in nums.indices) {
            val tmp = dfs(i)
            if (tmp.size > res.size) {
                res = tmp
            }
        }
        return res
    }

    private fun dfs(i: Int): List<Int> {
        cache[i]?.let { return it }

        var res = listOf(nums[i])
        for (j in i + 1 until nums.size) {
            if (nums[j] % nums[i] == 0) {
                val tmp = listOf(nums[i]) + dfs(j)
                if (tmp.size > res.size) {
                    res = tmp
                }
            }
        }

        cache[i] = res
        return res
    }
}

class Solution {
    func largestDivisibleSubset(_ nums: [Int]) -> [Int] {
        var nums = nums.sorted()
        let n = nums.count
        var cache = [[Int]?](repeating: nil, count: n)

        func dfs(_ i: Int) -> [Int] {
            if let cached = cache[i] { return cached }

            var res = [nums[i]]
            for j in (i + 1)..<n {
                if nums[j] % nums[i] == 0 {
                    let tmp = [nums[i]] + dfs(j)
                    if tmp.count > res.count {
                        res = tmp
                    }
                }
            }

            cache[i] = res
            return res
        }

        var res = [Int]()
        for i in 0..<n {
            let tmp = dfs(i)
            if tmp.count > res.count {
                res = tmp
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

3. Dynamic Programming (Bottom-Up)

Intuition

We can convert the top-down approach to bottom-up. Processing from right to left, for each index we compute the longest divisible subset starting from that position. At each step, we check all later indices for valid extensions and build upon the precomputed results.

Algorithm

Sort the array in ascending order.
Create a DP array where dp[i] stores the longest divisible subset starting at index i.
Initialize each dp[i] with just the element at that index.
Iterate from right to left:
- For each later index j, if nums[j] % nums[i] == 0, check if prepending nums[i] to dp[j] gives a longer subset.
- Update dp[i] with the longest result.
- Track the overall longest subset found.
Return the longest subset.

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        dp = [[num] for num in nums]  # dp[i] = longest start at i
        res = []
        for i in range(len(nums) - 1, -1, -1):
            for j in range(i + 1, len(nums)):
                if nums[j] % nums[i] == 0:
                    tmp = [nums[i]] + dp[j]
                    dp[i] = tmp if len(tmp) > len(dp[i]) else dp[i]
            res = dp[i] if len(dp[i]) > len(res) else res
        return res

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        List<Integer>[] dp = new ArrayList[n];
        List<Integer> res = new ArrayList<>();

        for (int i = n - 1; i >= 0; i--) {
            dp[i] = new ArrayList<>();
            dp[i].add(nums[i]);

            for (int j = i + 1; j < n; j++) {
                if (nums[j] % nums[i] == 0) {
                    List<Integer> tmp = new ArrayList<>();
                    tmp.add(nums[i]);
                    tmp.addAll(dp[j]);

                    if (tmp.size() > dp[i].size()) {
                        dp[i] = tmp;
                    }
                }
            }
            if (dp[i].size() > res.size()) {
                res = dp[i];
            }
        }
        return res;
    }
}

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> dp(n);
        vector<int> res;

        for (int i = n - 1; i >= 0; i--) {
            dp[i].push_back(nums[i]);

            for (int j = i + 1; j < n; j++) {
                if (nums[j] % nums[i] == 0) {
                    vector<int> tmp = dp[j];
                    tmp.insert(tmp.begin(), nums[i]);

                    if (tmp.size() > dp[i].size()) {
                        dp[i] = tmp;
                    }
                }
            }
            if (dp[i].size() > res.size()) {
                res = dp[i];
            }
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    largestDivisibleSubset(nums) {
        nums.sort((a, b) => a - b);
        const n = nums.length;
        const dp = new Array(n).fill(0).map(() => []);
        let res = [];

        for (let i = n - 1; i >= 0; i--) {
            dp[i] = [nums[i]];

            for (let j = i + 1; j < n; j++) {
                if (nums[j] % nums[i] === 0) {
                    let tmp = [nums[i], ...dp[j]];

                    if (tmp.length > dp[i].length) {
                        dp[i] = tmp;
                    }
                }
            }
            if (dp[i].length > res.length) {
                res = dp[i];
            }
        }
        return res;
    }
}

public class Solution {
    public List<int> LargestDivisibleSubset(int[] nums) {
        Array.Sort(nums);
        List<List<int>> dp = new List<List<int>>();
        foreach (int num in nums) {
            dp.Add(new List<int> { num });
        }

        List<int> res = new List<int>();
        for (int i = nums.Length - 1; i >= 0; i--) {
            for (int j = i + 1; j < nums.Length; j++) {
                if (nums[j] % nums[i] == 0) {
                    List<int> tmp = new List<int> { nums[i] };
                    tmp.AddRange(dp[j]);
                    if (tmp.Count > dp[i].Count) dp[i] = tmp;
                }
            }
            if (dp[i].Count > res.Count) res = dp[i];
        }

        return res;
    }
}

func largestDivisibleSubset(nums []int) []int {
    sort.Ints(nums)
    n := len(nums)
    dp := make([][]int, n)
    for i := range dp {
        dp[i] = []int{nums[i]}
    }

    res := []int{}
    for i := n - 1; i >= 0; i-- {
        for j := i + 1; j < n; j++ {
            if nums[j]%nums[i] == 0 {
                tmp := append([]int{nums[i]}, dp[j]...)
                if len(tmp) > len(dp[i]) {
                    dp[i] = tmp
                }
            }
        }
        if len(dp[i]) > len(res) {
            res = dp[i]
        }
    }

    return res
}

class Solution {
    fun largestDivisibleSubset(nums: IntArray): List<Int> {
        nums.sort()
        val n = nums.size
        val dp = Array(n) { mutableListOf(nums[it]) }

        var res = emptyList<Int>()
        for (i in n - 1 downTo 0) {
            for (j in i + 1 until n) {
                if (nums[j] % nums[i] == 0) {
                    val tmp = listOf(nums[i]) + dp[j]
                    if (tmp.size > dp[i].size) {
                        dp[i] = tmp.toMutableList()
                    }
                }
            }
            if (dp[i].size > res.size) {
                res = dp[i]
            }
        }

        return res
    }
}

class Solution {
    func largestDivisibleSubset(_ nums: [Int]) -> [Int] {
        var nums = nums.sorted()
        let n = nums.count
        var dp = nums.map { [$0] }

        var res = [Int]()
        for i in stride(from: n - 1, through: 0, by: -1) {
            for j in (i + 1)..<n {
                if nums[j] % nums[i] == 0 {
                    let tmp = [nums[i]] + dp[j]
                    if tmp.count > dp[i].count {
                        dp[i] = tmp
                    }
                }
            }
            if dp[i].count > res.count {
                res = dp[i]
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

4. Dynamic Programming (Top-Down) + Tracing

Intuition

Instead of storing entire subsets in the DP table (which uses extra memory), we can store just two values per index: the length of the longest subset starting there, and the next index in that subset. After computing all lengths, we trace through the indices to reconstruct the actual subset.

Algorithm

Sort the array in ascending order.
Create a DP array where dp[i] = [maxLen, nextIndex].
Define dfs(i) that returns the length of the longest subset starting at index i:
- For each later index j where nums[j] % nums[i] == 0, compute the length via dfs(j) + 1.
- Track which j gave the best length for tracing.
Find the starting index with the maximum length.
Reconstruct the subset by following the nextIndex pointers.

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        dp = [[-1, -1] for _ in range(n)]  # dp[i] = [maxLen, prevIdx]

        def dfs(i):
            if dp[i][0] != -1:
                return dp[i][0]

            dp[i][0] = 1
            for j in range(i + 1, n):
                if nums[j] % nums[i] == 0:
                    length = dfs(j) + 1
                    if length > dp[i][0]:
                        dp[i][0] = length
                        dp[i][1] = j

            return dp[i][0]

        max_len, start_index = 1, 0
        for i in range(n):
            if dfs(i) > max_len:
                max_len = dfs(i)
                start_index = i

        subset = []
        while start_index != -1:
            subset.append(nums[start_index])
            start_index = dp[start_index][1]

        return subset

public class Solution {
    private int[][] dp;

    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        dp = new int[n][2];
        for (int i = 0; i < n; i++) {
            dp[i][0] = -1;
            dp[i][1] = -1;
        }

        int maxLen = 1, startIndex = 0;
        for (int i = 0; i < n; i++) {
            if (dfs(i, nums) > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        List<Integer> subset = new ArrayList<>();
        while (startIndex != -1) {
            subset.add(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }

    private int dfs(int i, int[] nums) {
        if (dp[i][0] != -1) return dp[i][0];

        dp[i][0] = 1;
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] % nums[i] == 0) {
                int length = dfs(j, nums) + 1;
                if (length > dp[i][0]) {
                    dp[i][0] = length;
                    dp[i][1] = j;
                }
            }
        }
        return dp[i][0];
    }
}

class Solution {
    vector<vector<int>> dp;

public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        dp.assign(n, vector<int>(2, -1));

        int maxLen = 1, startIndex = 0;
        for (int i = 0; i < n; i++) {
            if (dfs(i, nums) > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        vector<int> subset;
        while (startIndex != -1) {
            subset.push_back(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }

private:
    int dfs(int i, vector<int>& nums) {
        if (dp[i][0] != -1) return dp[i][0];

        dp[i][0] = 1;
        for (int j = i + 1; j < nums.size(); j++) {
            if (nums[j] % nums[i] == 0) {
                int length = dfs(j, nums) + 1;
                if (length > dp[i][0]) {
                    dp[i][0] = length;
                    dp[i][1] = j;
                }
            }
        }
        return dp[i][0];
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    largestDivisibleSubset(nums) {
        nums.sort((a, b) => a - b);
        let n = nums.length;
        let dp = Array.from({ length: n }, () => [-1, -1]); // dp[i] = [maxLen, prevIdx]

        const dfs = (i) => {
            if (dp[i][0] !== -1) return dp[i][0];

            dp[i][0] = 1;
            for (let j = i + 1; j < n; j++) {
                if (nums[j] % nums[i] === 0) {
                    let length = dfs(j) + 1;
                    if (length > dp[i][0]) {
                        dp[i][0] = length;
                        dp[i][1] = j;
                    }
                }
            }
            return dp[i][0];
        };

        let maxLen = 1,
            startIndex = 0;
        for (let i = 0; i < n; i++) {
            if (dfs(i) > maxLen) {
                maxLen = dfs(i);
                startIndex = i;
            }
        }

        let subset = [];
        while (startIndex !== -1) {
            subset.push(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }
}

public class Solution {
    private int[][] dp;
    private int[] nums;
    private int n;

    public List<int> LargestDivisibleSubset(int[] nums) {
        Array.Sort(nums);
        this.nums = nums;
        n = nums.Length;
        dp = new int[n][];
        for (int i = 0; i < n; i++) {
            dp[i] = new int[] { -1, -1 };
        }

        int maxLen = 1, startIndex = 0;
        for (int i = 0; i < n; i++) {
            if (Dfs(i) > maxLen) {
                maxLen = Dfs(i);
                startIndex = i;
            }
        }

        List<int> subset = new List<int>();
        while (startIndex != -1) {
            subset.Add(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }

        return subset;
    }

    private int Dfs(int i) {
        if (dp[i][0] != -1) return dp[i][0];

        dp[i][0] = 1;
        for (int j = i + 1; j < n; j++) {
            if (nums[j] % nums[i] == 0) {
                int length = Dfs(j) + 1;
                if (length > dp[i][0]) {
                    dp[i][0] = length;
                    dp[i][1] = j;
                }
            }
        }

        return dp[i][0];
    }
}

func largestDivisibleSubset(nums []int) []int {
    sort.Ints(nums)
    n := len(nums)
    dp := make([][2]int, n)
    for i := range dp {
        dp[i] = [2]int{-1, -1}
    }

    var dfs func(i int) int
    dfs = func(i int) int {
        if dp[i][0] != -1 {
            return dp[i][0]
        }

        dp[i][0] = 1
        for j := i + 1; j < n; j++ {
            if nums[j]%nums[i] == 0 {
                length := dfs(j) + 1
                if length > dp[i][0] {
                    dp[i][0] = length
                    dp[i][1] = j
                }
            }
        }
        return dp[i][0]
    }

    maxLen, startIndex := 1, 0
    for i := 0; i < n; i++ {
        if dfs(i) > maxLen {
            maxLen = dfs(i)
            startIndex = i
        }
    }

    subset := []int{}
    for startIndex != -1 {
        subset = append(subset, nums[startIndex])
        startIndex = dp[startIndex][1]
    }

    return subset
}

class Solution {
    private lateinit var dp: Array<IntArray>
    private lateinit var nums: IntArray
    private var n = 0

    fun largestDivisibleSubset(nums: IntArray): List<Int> {
        nums.sort()
        this.nums = nums
        n = nums.size
        dp = Array(n) { intArrayOf(-1, -1) }

        var maxLen = 1
        var startIndex = 0
        for (i in 0 until n) {
            if (dfs(i) > maxLen) {
                maxLen = dfs(i)
                startIndex = i
            }
        }

        val subset = mutableListOf<Int>()
        var idx = startIndex
        while (idx != -1) {
            subset.add(nums[idx])
            idx = dp[idx][1]
        }

        return subset
    }

    private fun dfs(i: Int): Int {
        if (dp[i][0] != -1) return dp[i][0]

        dp[i][0] = 1
        for (j in i + 1 until n) {
            if (nums[j] % nums[i] == 0) {
                val length = dfs(j) + 1
                if (length > dp[i][0]) {
                    dp[i][0] = length
                    dp[i][1] = j
                }
            }
        }

        return dp[i][0]
    }
}

class Solution {
    func largestDivisibleSubset(_ nums: [Int]) -> [Int] {
        var nums = nums.sorted()
        let n = nums.count
        var dp = [[Int]](repeating: [-1, -1], count: n)

        func dfs(_ i: Int) -> Int {
            if dp[i][0] != -1 { return dp[i][0] }

            dp[i][0] = 1
            for j in (i + 1)..<n {
                if nums[j] % nums[i] == 0 {
                    let length = dfs(j) + 1
                    if length > dp[i][0] {
                        dp[i][0] = length
                        dp[i][1] = j
                    }
                }
            }
            return dp[i][0]
        }

        var maxLen = 1
        var startIndex = 0
        for i in 0..<n {
            if dfs(i) > maxLen {
                maxLen = dfs(i)
                startIndex = i
            }
        }

        var subset = [Int]()
        var idx = startIndex
        while idx != -1 {
            subset.append(nums[idx])
            idx = dp[idx][1]
        }

        return subset
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

5. Dynamic Programming (Bottom-Up) + Tracing

Intuition

This is the iterative version of the tracing approach. We process indices from left to right, looking backward for valid predecessors. For each position, we store the length of the longest subset ending there and a pointer to the previous index. This is similar to the classic Longest Increasing Subsequence pattern.

Algorithm

Sort the array in ascending order.
Create a DP array where dp[i] = [maxLen, prevIndex], initialized with length 1 and no predecessor.
For each index i, check all earlier indices j:
- If nums[i] % nums[j] == 0 and extending from j gives a longer subset, update dp[i].
- Track the index with the overall maximum length.
Reconstruct the subset by following the prevIndex pointers backward from the best ending index.

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        dp = [[1, -1] for _ in range(n)]  # dp[i] = [maxLen, prevIdx]

        max_len, start_index = 1, 0

        for i in range(n):
            for j in range(i):
                if nums[i] % nums[j] == 0 and dp[j][0] + 1 > dp[i][0]:
                    dp[i][0] = dp[j][0] + 1
                    dp[i][1] = j

            if dp[i][0] > max_len:
                max_len = dp[i][0]
                start_index = i

        subset = []
        while start_index != -1:
            subset.append(nums[start_index])
            start_index = dp[start_index][1]
        return subset

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int[][] dp = new int[n][2]; // dp[i] = {maxLen, prevIdx}

        int maxLen = 1, startIndex = 0;
        for (int i = 0; i < n; i++) {
            dp[i][0] = 1;
            dp[i][1] = -1;
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 && dp[j][0] + 1 > dp[i][0]) {
                    dp[i][0] = dp[j][0] + 1;
                    dp[i][1] = j;
                }
            }

            if (dp[i][0] > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        List<Integer> subset = new ArrayList<>();
        while (startIndex != -1) {
            subset.add(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }
}

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> dp(n, vector<int>(2, -1)); // dp[i] = {maxLen, prevIdx}

        int maxLen = 1, startIndex = 0;
        for (int i = 0; i < n; i++) {
            dp[i][0] = 1;
            dp[i][1] = -1;
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 && dp[j][0] + 1 > dp[i][0]) {
                    dp[i][0] = dp[j][0] + 1;
                    dp[i][1] = j;
                }
            }

            if (dp[i][0] > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        vector<int> subset;
        while (startIndex != -1) {
            subset.push_back(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    largestDivisibleSubset(nums) {
        nums.sort((a, b) => a - b);
        let n = nums.length;
        let dp = Array.from({ length: n }, () => [1, -1]); // dp[i] = [maxLen, prevIdx]

        let maxLen = 1,
            startIndex = 0;
        for (let i = 0; i < n; i++) {
            for (let j = 0; j < i; j++) {
                if (nums[i] % nums[j] === 0 && dp[j][0] + 1 > dp[i][0]) {
                    dp[i][0] = dp[j][0] + 1;
                    dp[i][1] = j;
                }
            }

            if (dp[i][0] > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        let subset = [];
        while (startIndex !== -1) {
            subset.push(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }
        return subset;
    }
}

public class Solution {
    public List<int> LargestDivisibleSubset(int[] nums) {
        Array.Sort(nums);
        int n = nums.Length;
        int[][] dp = new int[n][];
        for (int i = 0; i < n; i++) {
            dp[i] = new int[] { 1, -1 };
        }

        int maxLen = 1, startIndex = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 && dp[j][0] + 1 > dp[i][0]) {
                    dp[i][0] = dp[j][0] + 1;
                    dp[i][1] = j;
                }
            }

            if (dp[i][0] > maxLen) {
                maxLen = dp[i][0];
                startIndex = i;
            }
        }

        List<int> subset = new List<int>();
        while (startIndex != -1) {
            subset.Add(nums[startIndex]);
            startIndex = dp[startIndex][1];
        }

        return subset;
    }
}

func largestDivisibleSubset(nums []int) []int {
    sort.Ints(nums)
    n := len(nums)
    dp := make([][2]int, n)
    for i := range dp {
        dp[i] = [2]int{1, -1}
    }

    maxLen, startIndex := 1, 0

    for i := 0; i < n; i++ {
        for j := 0; j < i; j++ {
            if nums[i]%nums[j] == 0 && dp[j][0]+1 > dp[i][0] {
                dp[i][0] = dp[j][0] + 1
                dp[i][1] = j
            }
        }

        if dp[i][0] > maxLen {
            maxLen = dp[i][0]
            startIndex = i
        }
    }

    subset := []int{}
    for startIndex != -1 {
        subset = append(subset, nums[startIndex])
        startIndex = dp[startIndex][1]
    }

    return subset
}

class Solution {
    fun largestDivisibleSubset(nums: IntArray): List<Int> {
        nums.sort()
        val n = nums.size
        val dp = Array(n) { intArrayOf(1, -1) }

        var maxLen = 1
        var startIndex = 0

        for (i in 0 until n) {
            for (j in 0 until i) {
                if (nums[i] % nums[j] == 0 && dp[j][0] + 1 > dp[i][0]) {
                    dp[i][0] = dp[j][0] + 1
                    dp[i][1] = j
                }
            }

            if (dp[i][0] > maxLen) {
                maxLen = dp[i][0]
                startIndex = i
            }
        }

        val subset = mutableListOf<Int>()
        var idx = startIndex
        while (idx != -1) {
            subset.add(nums[idx])
            idx = dp[idx][1]
        }

        return subset
    }
}

class Solution {
    func largestDivisibleSubset(_ nums: [Int]) -> [Int] {
        var nums = nums.sorted()
        let n = nums.count
        var dp = [[Int]](repeating: [1, -1], count: n)

        var maxLen = 1
        var startIndex = 0

        for i in 0..<n {
            for j in 0..<i {
                if nums[i] % nums[j] == 0 && dp[j][0] + 1 > dp[i][0] {
                    dp[i][0] = dp[j][0] + 1
                    dp[i][1] = j
                }
            }

            if dp[i][0] > maxLen {
                maxLen = dp[i][0]
                startIndex = i
            }
        }

        var subset = [Int]()
        var idx = startIndex
        while idx != -1 {
            subset.append(nums[idx])
            idx = dp[idx][1]
        }

        return subset
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

Common Pitfalls

Forgetting to Sort the Array First

The divisibility property only guarantees transitivity when elements are sorted. If a divides b and b divides c, then a divides c, but this chain only works reliably when processing elements in ascending order. Skipping the sort step means you might miss valid subset extensions or incorrectly reject valid ones.

Checking Divisibility in the Wrong Direction

After sorting in ascending order, when comparing nums[i] and nums[j] where i < j, you must check if nums[j] % nums[i] == 0 (larger divisible by smaller). A common mistake is checking nums[i] % nums[j] == 0, which will always be false for distinct positive integers when nums[i] < nums[j].

Only Returning the Length Instead of the Actual Subset

The problem asks for the subset itself, not just its size. When using DP with length tracking only, you must also maintain a way to reconstruct the path, typically through parent pointers or by storing the actual subsets. Forgetting this reconstruction step means you can compute the correct length but cannot output the required elements.