README.md

740. Delete and Earn

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 earn nums[i] points. Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.

Return the maximum number of points you can earn by applying the above operation some number of times.

Example 1:

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.

Example 2:

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.

Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i] <= 104

Solutions (Rust)

1. Solution

impl Solution {
    pub fn delete_and_earn(nums: Vec<i32>) -> i32 {
        let mut sums = vec![0; *nums.iter().max().unwrap() as usize + 1];
        let mut dp = vec![(0, 0); sums.len()];

        for &num in &nums {
            sums[num as usize] += num;
        }

        dp[1] = (0, sums[1]);

        for i in 2..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)
    }
}