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.
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Dynamic Programming Math
/*
1. Sort
2. FInd the length of longest subset
3. Record the largest element of it.
4. DO a loop from the largest element to num[0], add every element belong to the logest subset.
*/
public List<Integer> largestDivisibleSubset(int[] nums) {
List<Integer> res = new ArrayList<Integer>();
if (nums == null || nums.length == 0) return res;
Arrays.sort(nums);
int[] dp = new int[nums.length];
dp[0] = 1;
// for each element in nums; find the length of largest subset it has.
for (int i = 1; i < nums.length; i++) {
for (int j = i - 1; j >= 0; j--) {
if (nums[i] % nums[j] == 0) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
}
// pick the index of the largest element in dp.
int maxIndex = 0;
for (int i = 1; i < nums.length; i++) {
maxIndex = dp[i] > dp[maxIndex] ? i : maxIndex;
}
// from nums[maxIndex] to 0, add every element belongs to the largest subset
int temp = nums[maxIndex];
int curDp = dp[maxIndex];
for (int i = maxIndex; i >= 0; i--) {
if (temp % nums[i] == 0 && dp[i] == curDp) {
res.add(nums[i]);
temp = nums[i];
curDp--;
}
}
return res;
}
