feat: solve No.948

Author: BaffinLee

901-1000/948. Bag of Tokens.md

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+# 948. Bag of Tokens
+
+- Difficulty: Medium.
+- Related Topics: Array, Two Pointers, Greedy, Sorting.
+- Similar Questions: .
+
+## Problem
+
+You start with an initial **power** of `power`, an initial **score** of `0`, and a bag of tokens given as an integer array `tokens`, where each `tokens[i]` donates the value of token**i**.
+
+Your goal is to **maximize** the total **score** by strategically playing these tokens. In one move, you can play an **unplayed** token in one of the two ways (but not both for the same token):
+
+
+	
+- **Face-up**: If your current power is **at least** `tokens[i]`, you may play token**i**, losing `tokens[i]` power and gaining `1` score.
+	
+- **Face-down**: If your current score is **at least** `1`, you may play token**i**, gaining `tokens[i]` power and losing `1` score.
+
+
+Return **the **maximum** possible score you can achieve after playing **any** number of tokens**.
+
+ 
+Example 1:
+
+
+**Input:** tokens = [100], power = 50
+
+**Output:** 0
+
+**Explanation****:** Since your score is `0` initially, you cannot play the token face-down. You also cannot play it face-up since your power (`50`) is less than `tokens[0]` (`100`).
+
+
+Example 2:
+
+
+**Input:** tokens = [200,100], power = 150
+
+**Output:** 1
+
+**Explanation:** Play token**1** (`100`) face-up, reducing your power to `50` and increasing your score to `1`.
+
+There is no need to play token**0**, since you cannot play it face-up to add to your score. The maximum score achievable is `1`.
+
+
+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`:
+
+
+	
+- Play token**0** (`100`) face-up, reducing power to `100` and increasing score to `1`.
+	
+- Play token**3** (`400`) face-down, increasing power to `500` and reducing score to `0`.
+	
+- Play token**1** (`200`) face-up, reducing power to `300` and increasing score to `1`.
+	
+- Play token**2** (`300`) face-up, reducing power to `0` and increasing score to `2`.
+
+
+The maximum score achievable is 2.
+
+
+ 
+**Constraints:**
+
+
+	
+- `0 <= tokens.length <= 1000`
+	
+- `0 <= tokens[i], power < 104`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} tokens
+ * @param {number} power
+ * @return {number}
+ */
+var bagOfTokensScore = function(tokens, power) {
+    tokens.sort((a, b) => a - b);
+    var left = 0;
+    var right = tokens.length - 1;
+    var maxScore = 0;
+    var score = 0;
+    while (left <= right) {
+        if (power >= tokens[left]) {
+            power -= tokens[left++];
+            score += 1;
+            maxScore = Math.max(maxScore, score);
+        } else if (score > 0) {
+            score -= 1;
+            power += tokens[right--];
+        } else {
+            break;
+        }
+    }
+    return maxScore;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(1).