Algorithm
Problem Name: 740. Delete and Earn
Problem Link: https://leetcode.com/problems/delete-and-earn/
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 earnnums[i]points. Afterwards, you must delete every element equal tonums[i] - 1and every element equal tonums[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 * 1041 <= nums[i] <= 104
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int deleteAndEarn(vector<int>& nums) {
int n = nums.size();
if(n == 0) return 0;
sort(nums.begin(), nums.end());
vector<vector < int>>dp(n, vector<int>(2));
dp[0][0] = 0;
dp[0][1] = nums[0];
for(int i = 1; i < n; i++){
if(nums[i] == nums[i - 1]){
dp[i][0] = dp[i - 1][0];
dp[i][1] = dp[i - 1][1] + nums[i];
continue;
}
dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
dp[i][1] = nums[i] == nums[i - 1] + 1 ? dp[i - 1][0] + nums[i] : dp[i][0] + nums[i];
}
return max(dp[n - 1][0], dp[n - 1][1]>;
}
};
Input
nums = [3,4,2]
Output
6
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const deleteAndEarn = function (nums) {
const n = 10001
const values = new Array(n).fill(0)
for (let num of nums) values[num] += num
let take = 0,
skip = 0
for (let i = 0; i < n; i++) {
const takei = skip + values[i]
const skipi = Math.max(skip, take)
take = takei
skip = skipi
}
return Math.max(take, skip)
}
Input
nums = [3,4,2]
Output
6
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def deleteAndEarn(self, nums):
cnt, dp, maxs = collections.Counter(nums), {}, {}
nums = sorted(set(nums))
if len(nums) < 2:
return nums and nums[0] * cnt[nums[0]] or 0
for i in range(len(nums)):
dp[i] = nums[i] * cnt[nums[i]]
if i >= 2:
if nums[i - 1] < nums[i] - 1:
dp[i] += maxs[i - 1]
else:
dp[i] += maxs[i - 2]
maxs[i] = max(dp[i], maxs[i - 1])
elif i:
if nums[i - 1] < nums[i] - 1:
dp[i] += dp[i - 1]
maxs[i] = max(dp[i], dp[i - 1])
else:
maxs[i] = dp[i]
return max(dp[len(nums) - 1], dp[len(nums) - 2])
Input
nums = [2,2,3,3,3,4]
Output
9