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 * 104
• 1 <= nums[i] <= 104

Code

DP

同 House Robber。

Denote 10004 as $N$. Time Complexity: $O(N)$, Space Complexity: $O(N)$

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int n = 10001;
        int values[n];
        memset(values, 0, sizeof(values));
        for(int i = 0; i < nums.size(); i++) {
            values[nums[i]] += nums[i];
        }
 
        int dp[n];
        memset(dp, 0, sizeof(dp));
        dp[0] = values[0];
        dp[1] = values[1];
        for(int i = 2; i < n; i++) {
            dp[i] = max(dp[i - 2] + values[i], dp[i-1]);
        }
 
        return dp[n - 1];
    }
};

DP-Optimized

Time Complexity: $O(N)$, Space Complexity: $O(N)$

dp[i] only depends on dp[i-1] & dp[i-2], so we can use only three variables a, b, c, corresponds to dp[i-2], dp[i-1], dp[i].

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int n = 10001;
        int values[n];
        memset(values, 0, sizeof(values));
        for(int i = 0; i < nums.size(); i++) {
            values[nums[i]] += nums[i];
        }
 
        int a = values[0];
        int b = values[1];
        for(int i = 2; i < n; i++) {
            int c = max(a + values[i], b);
            a = b;
            b = c;
        }
 
        return b;
    }
};
 

Source

• Delete and Earn - LeetCode