Algorithm
Problem Name: 948. Bag of Tokens
You have an initial power of power, an initial score of 0, and a bag of tokens where tokens[i] is the value of the ith token (0-indexed).
Your goal is to maximize your total score by potentially playing each token in one of two ways:
- If your current power is at least
tokens[i], you may play theithtoken face up, losingtokens[i]power and gaining1score. - If your current score is at least
1, you may play theithtoken face down, gainingtokens[i]power and losing1score.
Each token may be played at most once and in any order. You do not have to play all the tokens.
Return the largest possible score you can achieve after playing any number of tokens.
Example 1:
Input: tokens = [100], power = 50 Output: 0 Explanation: Playing the only token in the bag is impossible because you either have too little power or too little score.
Example 2:
Input: tokens = [100,200], power = 150 Output: 1 Explanation: Play the 0th token (100) face up, your power becomes 50 and score becomes 1. There is no need to play the 1st token since you cannot play it face up to add to your score.
Example 3:
Input: tokens = [100,200,300,400], power = 200 Output: 2 Explanation: Play the tokens in this order to get a score of 2: 1. Play the 0th token (100) face up, your power becomes 100 and score becomes 1. 2. Play the 3rd token (400) face down, your power becomes 500 and score becomes 0. 3. Play the 1st token (200) face up, your power becomes 300 and score becomes 1. 4. Play the 2nd token (300) face up, your power becomes 0 and score becomes 2.
Constraints:
0 <= tokens.length <= 10000 <= tokens[i], power < 104
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int bagOfTokensScore(int[] tokens, int power) {
int score = 0;
Arrays.sort(tokens);
int leftIdx = 0;
int rightIdx = tokens.length - 1;
int currScore = 0;
while (leftIdx < = rightIdx) {
if (power >= tokens[leftIdx]) {
currScore++;
power -= tokens[leftIdx++];
} else {
if (currScore == 0) {
break;
}
currScore--;
power += tokens[rightIdx--];
}
score = Math.max(score, currScore);
}
return score;
}
}
Input
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const bagOfTokensScore = function (tokens, P) {
tokens.sort((a, b) => a - b)
let res = 0,
score = 0,
i = 0,
j = tokens.length - 1
while (i <= j) {
if (P >= tokens[i]) {
P -= tokens[i++]
res = Math.max(res, ++score)
} else if (score > 0) {
score--
P += tokens[j--]
} else break
}
return res
}
Input
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def bagOfTokensScore(self, tokens, P):
"""
:type tokens: List[int]
:type P: int
:rtype: int
"""
tokens.sort()
l, r, score = 0, len(tokens) - 1, 0
while l <= r:
if P >= tokens[l]:
P -= tokens[l]
score += 1
l += 1
elif score and l != r:
P += tokens[r]
score -= 1
r -= 1
else:
break
return score
Input
Output