Sponsored
Sponsored
This approach involves using a two-pointer technique to process the tokens. By sorting the tokens array, we aim to utilize the smallest possible tokens first to maximize score when power allows, and then balance power by utilizing score if achievable.
Start by sorting the tokens array. Maintain two pointers: one at the start to play face-up when you have enough power, and the other at the end to play face-down when a score can be sacrificed to gain more power. Keep track of the current score and maximum score achieved. This greedy approach helps in maximizing score by trying to play as many tokens face-up as possible while being able to trade score for power when needed.
The time complexity is O(n log n) due to sorting the tokens array. The space complexity is O(1) as no additional space besides variables is used.
1#include <iostream>
2#include <vector>
3#include <algorithm>
4
5using namespace std;
6
7int bagOfTokensScore(vector<int>& tokens, int power) {
8 sort(tokens.begin(), tokens.end());
9 int left = 0, right = tokens.size() - 1;
10 int score = 0, maxScore = 0;
11 while (left <= right) {
12 if (power >= tokens[left]) {
13 power -= tokens[left++];
14 score++;
maxScore = max(maxScore, score);
} else if (score > 0) {
power += tokens[right--];
score--;
} else {
break;
}
}
return maxScore;
}
int main() {
vector<int> tokens = {100, 200, 300, 400};
int power = 200;
int result = bagOfTokensScore(tokens, power);
cout << "Maximum Score: " << result << endl;
return 0;
}The C++ solution uses a similar approach as the C solution. The sort function is used to sort the tokens vector. Two pointers designate the possible moves, left and right. The loop continues until no more beneficial moves are available, updating maximum score as needed.
In the dynamic programming approach, we consider decisions at each token in terms of playing face-up or face-down, and store results for subproblems to optimize the score obtained.
The dynamic programming table, dp[i][j], keeps track of the maximum score achievable with the first i tokens and with j power. For each token, either spend or gain power while adjusting the score, and calculate the best possible outcome for each state.
The time complexity is O(n * m) where n is the number of tokens and m is the initial power range explored. The space complexity is also O(n * m) due to the DP table usage.
The Python implementation makes use of a 2D list (DP table) to handle scenarios of gaining or spending power for each token processed. Each state computes a maximum score from either face-up or face-down actions, ensuring the highest score possible with the given initial power.