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Formatted question description: https://leetcode.ca/all/740.html

740. Delete and Earn (Medium)

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:

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

 

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

Firstly, to avoid duplicate, store the data in a map from the number to its count.

Let dp[i] be the max point you can get at point i.

If num != prevNum + 1, we can freely pick num, then dp[i] = dp[i-1] + num * count.

Otherwise, if we don’t pick num, dp[i] = dp[i-1].

Otherwise, we pick num, dp[i] = dp[i-2] + num * count.

So in sum:

dp[i] = num == prevNum ? max(dp[i-1], dp[i-2] + num * count) : (dp[i-1] + num * count)
// OJ: https://leetcode.com/problems/delete-and-earn/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        map<int, int> m;
        for (int n : nums) m[n]++;
        int prev = 0, prev2 = 0, num = INT_MIN;
        for (auto &p : m) {
            int cur = p.first == num + 1 ? max(prev, prev2 + p.first * p.second) : (prev + p.first * p.second);
            prev2 = prev;
            prev = cur;
            num = p.first;
        }
        return prev;
    }
};
  • class Solution {
        public int deleteAndEarn(int[] nums) {
            if (nums == null || nums.length == 0)
                return 0;
            if (nums.length == 1)
                return nums[0];
            int max = 0;
            for (int num : nums)
                max = Math.max(max, num);
            int[] counts = new int[max + 1];
            for (int num : nums)
                counts[num]++;
            int[] dp = new int[max + 1];
            dp[1] = counts[1];
            dp[2] = Math.max(dp[1], 2 * counts[2]);
            for (int i = 3; i <= max; i++)
                dp[i] = Math.max(dp[i - 1], dp[i - 2] + i * counts[i]);
            return dp[max];
        }
    }
    
    ############
    
    public class Solution {
        public int deleteAndEarn(int[] nums) {
            if (nums.length == 0) {
                return 0;
            }
    
            int[] sums = new int[10010];
            int[] select = new int[10010];
            int[] nonSelect = new int[10010];
    
            int maxV = 0;
            for (int x : nums) {
                sums[x] += x;
                maxV = Math.max(maxV, x);
            }
    
            for (int i = 1; i <= maxV; i++) {
                select[i] = nonSelect[i - 1] + sums[i];
                nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
            }
            return Math.max(select[maxV], nonSelect[maxV]);
        }
    }
    
    
  • // OJ: https://leetcode.com/problems/delete-and-earn/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        int deleteAndEarn(vector<int>& A) {
            map<int, int> m;
            for (int n : A) m[n]++;
            int prev = 0, prev2 = 0, num = INT_MIN;
            for (auto &[n, cnt] : m) {
                int cur = n == num + 1 ? max(prev, prev2 + n * cnt) : (prev + n * cnt);
                prev2 = prev;
                prev = cur;
                num = n;
            }
            return prev;
        }
    };
    
  • class Solution:
        def deleteAndEarn(self, nums: List[int]) -> int:
            mx = -inf
            for num in nums:
                mx = max(mx, num)
            total = [0] * (mx + 1)
            for num in nums:
                total[num] += num
            first = total[0]
            second = max(total[0], total[1])
            for i in range(2, mx + 1):
                cur = max(first + total[i], second)
                first = second
                second = cur
            return second
    
    
    
  • func deleteAndEarn(nums []int) int {
    
    	max := func(x, y int) int {
    		if x > y {
    			return x
    		}
    		return y
    	}
    
    	mx := math.MinInt32
    	for _, num := range nums {
    		mx = max(mx, num)
    	}
    	total := make([]int, mx+1)
    	for _, num := range nums {
    		total[num] += num
    	}
    	first := total[0]
    	second := max(total[0], total[1])
    	for i := 2; i <= mx; i++ {
    		cur := max(first+total[i], second)
    		first = second
    		second = cur
    	}
    	return second
    }
    
    

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