Given an array nums of integers, you can perform operations on the array.
In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete every element equal to nums[i] - 1 or nums[i] + 1.
You start with 0 points. Return the maximum number of points you can earn by applying such operations.
Example 1:
Input: nums = [3, 4, 2]
Output: 6
Explanation:
Delete 4 to earn 4 points, consequently 3 is also deleted.
Then, delete 2 to earn 2 points. 6 total points are earned.
Example 2:
Input: nums = [2, 2, 3, 3, 3, 4]
Output: 9
Explanation:
Delete 3 to earn 3 points, deleting both 2's and the 4.
Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
9 total points are earned.
Note:
The length of nums is at most 20000.
Each element nums[i] is an integer in the range [1, 10000].
/*
If we sort all the numbers into buckets indexed by these numbers, this is essentially asking you to repetitively
take an bucket while giving up the 2 buckets next to it. (the range of these numbers is [1, 10000])
The optimal final result can be derived by keep updating 2 variables skip_i, take_i, which stands for:
skip_i : the best result for sub-problem of first (i+1) buckets from 0 to i, while you skip the ith bucket.
take_i : the best result for sub-problem of first (i+1) buckets from 0 to i, while you take the ith bucket.
DP formula:
take[i] = skip[i-1] + values[i];
skip[i] = Math.max(skip[i-1], take[i-1]);
take[i] can only be derived from: if you skipped the [i-1]th bucket, and you take bucket[i].
skip[i] through, can be derived from either take[i-1] or skip[i-1], whatever the bigger;
/
for numbers from [1 - 10000], each has a total sum sums[i]; if you earn sums[i], you cannot earn sums[i-1] and sums[i+1]
kind of like house robbing. you cannot rob 2 connected houses.
/
*/
class Solution {
public int deleteAndEarn(int[] nums) {
int n = 10001;
int[] values = new int[n];
for (int num : nums)
values[num] += num;
int take = 0, skip = 0;
for (int i = 0; i < n; i++) {
int takei = skip + values[i];
int skipi = Math.max(skip, take);
take = takei;
skip = skipi;
}
return Math.max(take, skip);
}
}