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368. Largest Divisible Subset

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 * 109
  • All the integers in nums are unique.

Solutions

Solution 1

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class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        f = [1] * n
        k = 0
        for i in range(n):
            for j in range(i):
                if nums[i] % nums[j] == 0:
                    f[i] = max(f[i], f[j] + 1)
            if f[k] < f[i]:
                k = i
        m = f[k]
        i = k
        ans = []
        while m:
            if nums[k] % nums[i] == 0 and f[i] == m:
                ans.append(nums[i])
                k, m = i, m - 1
            i -= 1
        return ans

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class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int[] f = new int[n];
        Arrays.fill(f, 1);
        int k = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] % nums[j] == 0) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
            if (f[k] < f[i]) {
                k = i;
            }
        }
        int m = f[k];
        List<Integer> ans = new ArrayList<>();
        for (int i = k; m > 0; --i) {
            if (nums[k] % nums[i] == 0 && f[i] == m) {
                ans.add(nums[i]);
                k = i;
                --m;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int f[n];
        int k = 0;
        for (int i = 0; i < n; ++i) {
            f[i] = 1;
            for (int j = 0; j < i; ++j) {
                if (nums[i] % nums[j] == 0) {
                    f[i] = max(f[i], f[j] + 1);
                }
            }
            if (f[k] < f[i]) {
                k = i;
            }
        }
        int m = f[k];
        vector<int> ans;
        for (int i = k; m > 0; --i) {
            if (nums[k] % nums[i] == 0 && f[i] == m) {
                ans.push_back(nums[i]);
                k = i;
                --m;
            }
        }
        return ans;
    }
};

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func largestDivisibleSubset(nums []int) (ans []int) {
    sort.Ints(nums)
    n := len(nums)
    f := make([]int, n)
    k := 0
    for i := 0; i < n; i++ {
        f[i] = 1
        for j := 0; j < i; j++ {
            if nums[i]%nums[j] == 0 {
                f[i] = max(f[i], f[j]+1)
            }
        }
        if f[k] < f[i] {
            k = i
        }
    }
    m := f[k]
    for i := k; m > 0; i-- {
        if nums[k]%nums[i] == 0 && f[i] == m {
            ans = append(ans, nums[i])
            k = i
            m--
        }
    }
    return
}

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