Delete And Earn
// 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 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.
Ex1:
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.
Ex2:
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.
*/
#include <unordered_map>
#include <vector>
#include <iostream>
#include <cassert>
using namespace std;
int dfs(unordered_map<int, pair<int, int>>& m, string key, unordered_map<string, int>& cache) {
if (cache.count(key)) {
return cache[key];
}
int sum = 0;
for (auto i : m) {
int num = i.first;
int s = i.second.first;
int idx = i.second.second;
if (s == 0) {
continue;
}
int plusOne = m[num+1].first;
int minusOne = m[num-1].first;
m[num].first = 0;
m[num+1].first = 0;
m[num-1].first = 0;
char c = key[idx];
key[idx] = '#';
sum = max(sum, s + dfs(m, key, cache));
m[num+1].first = plusOne;
m[num-1].first = minusOne;
m[num].first = s;
key[idx] = c;
}
cache[key] = sum;
return sum;
}
int deleteAndEarn1(vector<int>& nums) {
unordered_map<int, pair<int, int>> m;
unordered_map<string, int> cache;
int n = nums.size();
string key = "";
for (int i = 0; i < n; i++) {
if (!m.count(nums[i])) {
m[nums[i]].second = key.size();
key += to_string(nums[i]);
}
m[nums[i]].first += nums[i];
}
return dfs(m, key, cache);
}
// Time: O(n + mx), Space: O(n + mx)
int topDownDP(int num, unordered_map<int, int>& cache,
unordered_map<int, int>& m) {
if (num == 0) {
return num;
}
if (num == 1) {
return m[1];
}
if (cache.count(num)) {
return cache[num];
}
int res = max(m[num] + topDownDP(num-2, cache, m),
topDownDP(num-1, cache, m));
cache[num] = res;
return res;
}
int deleteAndEarn(vector<int>& nums) {
int n = nums.size();
int mx = 0;
unordered_map<int, int> m;
for (int i = 0; i < n; i++) {
mx = max(nums[i], mx);
m[nums[i]] += nums[i];
}
unordered_map<int, int> cache;
return topDownDP(mx, cache, m);
}
int main() {
vector<int> nums = {3,4,2};
assert(deleteAndEarn(nums) == 6);
nums = {2,2,3,3,3,4};
assert(deleteAndEarn(nums) == 9);
return 0;
}```
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