The problem at hand involves Alice and Bob playing a game with an array of piles, where each pile contains a positive integer number of stones. Alice and Bob take turns choosing a pile of stones either from the beginning or the end of the array, and the objective is to end up with the most stones. Alice always starts first, and they both play optimally. Since the total number of stones across all piles is odd, there will be no ties at the end of the game.
The provided solution checks if Alice will win based on the sum of all the stones. The key observation in the solution is that the total sum of the stones is odd, meaning there is an unequal distribution of stones. Alice, starting first, always has the advantage in such scenarios. This is because Alice can always choose the best option on her turn and will always end up with more stones than Bob, assuming both play optimally.
The code calculates the sum of all the stones in the piles and checks if the total sum is odd or even. If the sum is odd, the function returns `true`, indicating that Alice will win. If the sum were even (though in this specific problem, the sum is guaranteed to be odd), the function would return `false`, suggesting Bob would win. The logic here is simplified by the fact that the sum being odd guarantees that Alice, with her first-move advantage, will always be able to accumulate more stones than Bob, ensuring her victory.
In summary, the code essentially checks if the sum of the stones is odd, and since the total is guaranteed to be odd, it returns `true`, indicating Alice wins the game. The code does not consider the actual sequence of moves or game strategies because Alice’s first-move advantage in a game with an odd total ensures her victory.
CODE:
class Solution {
public:
bool stoneGame(vector<int>& arr) {
int i,n,sum=0;
n=arr.size();
for(i=0;i<n;i++){
sum+=arr[i];
}
if(sum%2!=0){
return true;
}
else
return false;
}
};
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