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:
nums is at most 20000.nums[i] is an integer in the range [1, 10000].Intuition
Because all numbers are positive, if we "take" a number (use it to score points), we might as well take all copies of it, since we've already erased all its neighbors. We could keep a count of each number so we know how many points taking a number is worth total.
Now let's investigate what happens when we add a new number X (plus copies) that is larger than all previous numbers. Naively, our answer would be the previous answer, plus the value of X - which can be solved with dynamic programming. However, this fails if our previous answer had a number taken that was adjacent to X.
Luckily, we can remedy this. Let's say we knew using, the value of our previous answer, and avoid, the value of our previous answer that doesn't use the previously largest value prev. Then we could compute new values of using and avoid appropriately.
Algorithm
For each unique value k of nums in increasing order, let's maintain the correct values of avoid and using, which represent the answer if we don't take or take k respectively.
If the new value k is adjacent to the previously largest value prev, then the answer if we must take k is (the point value of k) + avoid, while the answer if we must not take k is max(avoid, using). Similarly, if k is not adjacent to prev, the answer if we must take k is (the point value of k) + max(avoid, using), and the answer if we must not take k is max(avoid, using).
At the end, the best answer may or may not use the largest value in nums, so we return max(avoid, using).
Our demonstrated solutions showcase two different kinds of sorts: a library one, and a radix sort. For each language, the other kind of solution can be done without much difficulty, by using an array (Python) or HashMap (Java) respectively.
Complexity Analysis
Time Complexity (Python): , where is the length of nums. We make a single pass through the sorted keys of , and the complexity is dominated by the sorting step.
Space Complexity (Python): , the size of our count.
Time Complexity (Java): We performed a radix sort instead, so our complexity is where is the range of allowable values for nums[i].
Space Complexity (Java): , the size of our count.
Analysis written by: @awice.