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] - 1
and 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 * 104
1 <= nums[i] <= 104
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class Solution { | |
public: | |
// Point: DP on input integer range! | |
int deleteAndEarn(vector<int>& nums) { | |
int nmax = 1e4 + 1; | |
map<int, int> cnt; | |
for(auto x: nums) cnt[x]++; | |
vector<int> dp(nmax + 1); | |
dp[0] = 0; | |
dp[1] = cnt[1]; | |
for(int i = 2; i < nmax; i++){ | |
dp[i] = max(dp[i-1], dp[i-2] + cnt[i] * i); | |
} | |
return dp[nmax - 1]; | |
} | |
}; |
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