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| 1 | +/* |
| 2 | + Alice and Bob play a game with piles of stones. There are an even number of piles arranged |
| 3 | + in a row, and each pile has a positive integer number of stones piles[i]. |
| 4 | +
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| 5 | + The objective of the game is to end with the most stones. The total number of stones across |
| 6 | + all the piles is odd, so there are no ties. |
| 7 | +
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| 8 | + Alice and Bob take turns, with Alice starting first. Each turn, a player takes the entire pile |
| 9 | + of stones either from the beginning or from the end of the row. This continues until there are |
| 10 | + no more piles left, at which point the person with the most stones wins. |
| 11 | +
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| 12 | + Assuming Alice and Bob play optimally, return true if Alice wins the game, or false if Bob wins. |
| 13 | + |
| 14 | + Ex: Input: piles = [5,3,4,5] |
| 15 | + Output: true |
| 16 | + Explanation: |
| 17 | + Alice starts first, and can only take the first 5 or the last 5. |
| 18 | + Say she takes the first 5, so that the row becomes [3, 4, 5]. |
| 19 | + If Bob takes 3, then the board is [4, 5], and Alice takes 5 to win with 10 points. |
| 20 | + If Bob takes the last 5, then the board is [3, 4], and Alice takes 4 to win with 9 points. |
| 21 | + This demonstrated that taking the first 5 was a winning move for Alice, so we return true. |
| 22 | +
|
| 23 | + Space: O(n/2) |
| 24 | + Time : O(1) |
| 25 | +*/ |
| 26 | + |
| 27 | + |
| 28 | +class Solution { |
| 29 | +public: |
| 30 | + bool stoneGame(vector<int>& piles) { |
| 31 | + int alice = 0, bob = 0, size = piles.size(); |
| 32 | + for(int i = 0 ; i < size/2; i++) { |
| 33 | + alice = alice + max( piles[i], piles[size - 1- i] ); |
| 34 | + bob = bob + min( piles[i], piles[size - 1- i] ); |
| 35 | + } |
| 36 | + if(alice > bob) |
| 37 | + return true; |
| 38 | + return false; |
| 39 | + } |
| 40 | +}; |
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