368 Largest Divisible Subset
Problem:
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]
Solutions:
public class Solution {
public List<Integer> largestDivisibleSubset(int[] nums) {
if (nums == null || nums.length == 0) {
return new LinkedList<Integer>();
}
int[] dp = new int[nums.length];
int[] index = new int[nums.length];
dp[0] = 1;
index[0] = -1;
Arrays.sort(nums);
int max = 0;
int start = 0;
for (int i = 1; i < nums.length; i ++) {
index[i] = -1;
dp[i] = 1;
for (int j = 0; j < i; j ++) {
if (nums[i] % nums[j] == 0 && dp[j] + 1 > dp[i]) {
dp[i] = dp[j] + 1;
index[i] = j;
}
}
if (dp[i] > max) {
max = dp[i];
start = i;
}
}
int i = start;
List<Integer> result = new LinkedList<Integer>();
while (i != -1) {
result.add(0, nums[i]);
i = index[i];
}
return result;
}
}
Can be resolved in DFS as well
public class Solution {
List<Integer> answer;
public List<Integer> largestDivisibleSubset(int[] nums) {
if(nums==null || nums.length==0)
return new ArrayList<Integer>();
Arrays.sort(nums);
int[] max = new int[1];
List<Integer> result = new ArrayList<Integer>();
helper(nums, 0, result, max);
return answer;
}
public void helper(int[] nums, int start, List<Integer> result, int[] max){
if(result.size()>max[0]){
max[0]=result.size();
answer=new ArrayList<Integer>(result);
}
if(start==nums.length)
return;
for(int i=start; i<nums.length; i++){
if(result.size()==0){
result.add(nums[i]);
helper(nums, i+1, result, max);
result.remove(result.size()-1);
}else{
int top = result.get(result.size()-1);
if(nums[i]%top==0){
result.add(nums[i]);
helper(nums, i+1, result, max);
result.remove(result.size()-1);
}
}
}
}
}