Leetcode-3180

Maximum Total Reward Using Operations I

Time complexity: $O(mn)$

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#include <iostream>
#include <vector>
#include <algorithm>

class Solution {
public:
    int maxTotalReward(std::vector<int>& rewardValues) {
        std::sort(rewardValues.begin(), rewardValues.end());
        int n = rewardValues.size();
        int m = rewardValues.back();
        std::vector<int> dp(2 * m);
        
        dp[0] = 1;
        for (int i : rewardValues) {
            for (int j = 2 * i - 1; j >= i; --j) {
                dp[j] |= dp[j - i];
            }
        }
        
        for (int i = 2 * m - 1; i >= 0; --i) {
            if (dp[i]) {
                return i;
            }
        }
        
        return -1;
    }
};

Leetcode-3180

Licensed under CC BY-NC-SA 4.0