Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.
Example 1:
nums: [1,2,3] Result: [1,2] (of course, [1,3] will also be ok)
Example 2:
nums: [1,2,4,8] Result: [1,2,4,8]
Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.
Analysis:
Sort the array, then for nums[i], if nums[i] % nums[j] == 0, then nums[i] is in the largest divisible set of nums[j]. We just need to find out the largest subset of nums[i].
Solution:
public class Solution {
public List<Integer> largestDivisibleSubset(int[] nums) {
List<Integer> resList = new LinkedList<Integer>();
if (nums.length == 0) return resList;
Arrays.sort(nums);
int[] size = new int[nums.length];
int[] pre = new int[nums.length];
int maxSize = 0;
int maxInd = -1;
for (int i=0;i<nums.length;i++){
int localMax = 0;
int localInd = -1;
for (int j=0;j<i;j++)
if (nums[i] % nums[j] == 0 && localMax < size[j]+1){
localMax = size[j]+1;
localInd = j;
}
if (localInd == -1){
localMax = 1;
localInd = -1;
}
size[i] = localMax;
pre[i] = localInd;
if (maxSize < localMax){
maxSize = localMax;
maxInd = i;
}
}
resList.add(nums[maxInd]);
int preInd = pre[maxInd];
while (preInd != -1){
maxInd = preInd;
preInd = pre[maxInd];
resList.add(nums[maxInd]);
}
Collections.reverse(resList);
return resList;
}
}