740. Delete and Earn

Problem

Tags: Array, Hash Table, Dynamic Programming

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

Code

C

// 740. Delete and Earn (7/16/54144)
// Runtime: 6 ms (91.55%) Memory: 5.94 MB (94.37%) 

int deleteAndEarn (int nums[], int size) {
    int vals[10002] = { 0 };
    for (int i = 0; i < size; i++) {
        vals[nums[i]] += nums[i];
    }
    
    int prev = 0, curr = 0;
    for (int i = 1; i <= 10000; i++) {
        if (vals[i - 1]) {
            int temp = prev;
            prev = curr;
            curr = temp + vals[i] > prev ? temp + vals[i] : prev;
        } else {
            prev = curr;
            curr += vals[i];
        }
    }
    
    return curr > prev ? curr : prev;
}