Problem:

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 everyelement 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:

• The length of nums is at most 20000.
• Each element nums[i] is an integer in the range [1, 10000].

Idea:

Reduce the problem to House Robber Problem

Key observations: If we take nums[i]

1. We can safely take all of its copies.
2. We can’t take any of copies of nums[i – 1] and nums[i + 1]

This problem is reduced to 198 House Robber.

Houses[i] has all the copies of num whose value is i.

[3 4 2] -> [0 2 3 4], rob([0 2 3 4]) = 6            

[2, 2, 3, 3, 3, 4] -> [0 2*2 3*3 4], rob([0 2*2 3*3 4]) = 9

Time complexity: O(n+r) reduction + O(r) solving rob = O(n + r)

Space complexity: O(r)

r = max(nums) – min(nums) + 1

Time complexity: O(n + r)

Space complexity: O(r)

Solution:

// Author: Huahua
// Runtime: 6 ms
class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        if (nums.empty()) return 0;
        const auto range = minmax_element(nums.begin(), nums.end());
        const int l = *(range.first);
        const int r = *(range.second);        
        vector<int> points(r - l + 1, 0);
        for (const int num : nums)
            points[num - l] += num;
        return rob(points);
    }
private:
    // From LeetCode 198. House Robber
    int rob(const vector<int>& nums) {
        int dp2 = 0;
        int dp1 = 0;
        for (int i = 0; i < nums.size() ; ++i) {
            int dp = max(dp2 + nums[i], dp1);
            dp2 = dp1;
            dp1 = dp;
        }
        return dp1;
    }
};

Related Problem:

• 花花酱 LeetCode 198. House Robber

The post 花花酱 LeetCode 740. Delete and Earn appeared first on Huahua's Tech Road.