3181. Maximum Total Reward Using Operations II

Description

You are given an integer array rewardValues of length n, representing the values of rewards.

Initially, your total reward x is 0, and all indices are unmarked. You are allowed to perform the following operation any number of times:

• Choose an unmarked index i from the range [0, n - 1].
• If rewardValues[i] is greater than your current total reward x, then add rewardValues[i] to x (i.e., x = x + rewardValues[i]), and mark the index i.

Return an integer denoting the maximum total reward you can collect by performing the operations optimally.

 

Example 1:

Input: rewardValues = [1,1,3,3]

Output: 4

Explanation:

During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum.

Example 2:

Input: rewardValues = [1,6,4,3,2]

Output: 11

Explanation:

Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.

 

Constraints:

• 1 <= rewardValues.length <= 5 * 104
• 1 <= rewardValues[i] <= 5 * 104

Solutions

Solution 1: Dynamic Programming + Bit Manipulation

We define \(f[i][j]\) as whether it is possible to obtain a total reward of \(j\) using the first \(i\) reward values. Initially, \(f[0][0] = \textit{True}\), and all other values are \(\textit{False}\).

We consider the \(i\)-th reward value \(v\). If we do not choose it, then \(f[i][j] = f[i - 1][j]\); if we choose it, then \(f[i][j] = f[i - 1][j - v]\), where \(0 \leq j - v < v\). Thus, the state transition equation is:

\[ f[i][j] = f[i - 1][j] \vee f[i - 1][j - v] \]

The final answer is \(\max\{j \mid f[n][j] = \textit{True}\}\).

Since \(f[i][j]\) only depends on \(f[i - 1][j]\) and \(f[i - 1][j - v]\), we can optimize away the first dimension and use only a one-dimensional array for state transitions. Additionally, due to the large data range of this problem, we need to use bit manipulation to optimize the efficiency of state transitions.

We define a binary number \(f\) to save the current state, where the \(i\)-th bit of \(f\) being \(1\) indicates that a total reward of \(i\) is reachable.

Observing the state transition equation \(f[j] = f[j] \vee f[j - v]\), this is equivalent to taking the lower \(v\) bits of \(f\), shifting them left by \(v\) bits, and then performing an OR operation with the original \(f\).

Thus, the answer is the position of the highest bit in \(f\).

The time complexity is \(O(n \times M / w)\), and the space complexity is \(O(n + M / w)\). Where \(n\) is the length of the rewardValues array, \(M\) is twice the maximum value in the rewardValues array, and the integer \(w = 32\) or \(64\).

Python3JavaC++GoTypeScript

class Solution:
    def maxTotalReward(self, rewardValues: List[int]) -> int:
        nums = sorted(set(rewardValues))
        f = 1
        for v in nums:
            f |= (f & ((1 << v) - 1)) << v
        return f.bit_length() - 1

import java.math.BigInteger;

class Solution {
    public int maxTotalReward(int[] rewardValues) {
        int[] nums = Arrays.stream(rewardValues).distinct().sorted().toArray();
        BigInteger f = BigInteger.ONE;
        for (int v : nums) {
            BigInteger mask = BigInteger.ONE.shiftLeft(v).subtract(BigInteger.ONE);
            BigInteger shifted = f.and(mask).shiftLeft(v);
            f = f.or(shifted);
        }
        return f.bitLength() - 1;
    }
}

class Solution {
public:
    int maxTotalReward(vector<int>& rewardValues) {
        sort(rewardValues.begin(), rewardValues.end());
        rewardValues.erase(unique(rewardValues.begin(), rewardValues.end()), rewardValues.end());
        bitset<100000> f{1};
        for (int v : rewardValues) {
            int shift = f.size() - v;
            f |= f << shift >> (shift - v);
        }
        for (int i = rewardValues.back() * 2 - 1;; i--) {
            if (f.test(i)) {
                return i;
            }
        }
    }
};

func maxTotalReward(rewardValues []int) int {
    slices.Sort(rewardValues)
    rewardValues = slices.Compact(rewardValues)
    one := big.NewInt(1)
    f := big.NewInt(1)
    p := new(big.Int)
    for _, v := range rewardValues {
        mask := p.Sub(p.Lsh(one, uint(v)), one)
        f.Or(f, p.Lsh(p.And(f, mask), uint(v)))
    }
    return f.BitLen() - 1
}

function maxTotalReward(rewardValues: number[]): number {
    rewardValues.sort((a, b) => a - b);
    rewardValues = [...new Set(rewardValues)];
    let f = 1n;
    for (const x of rewardValues) {
        const mask = (1n << BigInt(x)) - 1n;
        f = f | ((f & mask) << BigInt(x));
    }
    return f.toString(2).length - 1;
}

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