You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:
- Pick any
nums[i]and delete it to earnnums[i]points. Afterwards, you must delete every element equal tonums[i] - 1and every element equal tonums[i] + 1.
Return the maximum number of points you can earn by applying the above operation some number of times.
Input: nums = [3,4,2] Output: 6 Explanation: You can perform the following operations: - Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2]. - Delete 2 to earn 2 points. nums = []. You earn a total of 6 points.
Input: nums = [2,2,3,3,3,4] Output: 9 Explanation: You can perform the following operations: - Delete a 3 to earn 3 points. All 2's and 4's are also deleted. nums = [3,3]. - Delete a 3 again to earn 3 points. nums = [3]. - Delete a 3 once more to earn 3 points. nums = []. You earn a total of 9 points.
1 <= nums.length <= 2 * 1041 <= nums[i] <= 104
implSolution{pubfndelete_and_earn(nums:Vec<i32>) -> i32{letmut sums = vec![0;*nums.iter().max().unwrap()asusize + 1];letmut dp = vec![(0,0); sums.len()];for&num in&nums { sums[num asusize] += num;} dp[1] = (0, sums[1]);for i in2..dp.len(){ dp[i].0 = dp[i - 1].0.max(dp[i - 1].1); dp[i].1 = sums[i] + dp[i - 1].0.max(dp[i - 2].1);} dp[dp.len() - 1].0.max(dp[dp.len() - 1].1)}}