740. Delete and Earn

Description

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 * 10^4
• 1 <= nums[i] <= 10^4

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Solutions

Solution 1

Intuition

1. store number’s frequent, cnt[i] means number i ‘s frequenct.
2. f[i][0]means the max value without picking nums[i], which from 0 to i
   1. f[i][0] = max(f[i-1][0], f[i-1][1])
3. f[i][1]means the max value with nums[i], which from 0 to i
   1. f[i][1] = f[i - 1][0] + i * cnt[i];
4. return max result

Code

const int N = 10010;
int cnt[N], f[N][2];
class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        memset(cnt, 0, sizeof cnt);
        memset(f, 0, sizeof f);
        int res = 0;
        for (int num : nums) cnt[num]++;
        for (int i = 1; i < N; i++) {
            f[i][0] = max(f[i - 1][0], f[i - 1][1]);
            f[i][1] = f[i - 1][0] + i * cnt[i];
            res = max(res, max(f[i][0], f[i][1]));
        }
        return res;
    }
};

Complexity

• Time: O(n)
• Space: O(n)