# why we do not need to differentiate two players?

• After seeing some code in discussion I came up with the following code and passed the judge. My question is: why we do not need to differentiate two players? That is, when we mark the elements in the map to true or false, how do we know it is true/false for the first or second player? Why it does not matter?

``````class Solution {
private:
bool wins(unordered_map<int, bool> &map, int used, int maxChoosableInteger, int target)
{
// if the number combination have been calculated, return the result
if (map[used]) return map[used];
// if the number combination have not been calculated
// if used is 10100, then 2, 4 have been chosen, while 1,3 are not
// so we start from the beginning, if the number has not been chosen, then we examine this number
for (int i = 1; i <= maxChoosableInteger; ++i)
{
int mask = 1 << i;
// 1. if this number >= the target, then of course wins
// 2. if this number < target, and we choose this number, then the choice is passed to the second player
//    only when second player loses can the first player wins
if ((mask & used) == 0 && (i >= target || !wins(map, mask | used, maxChoosableInteger, target - i)))
{
// for the above situations, the first player wins, and we can mark the map
// we don't need to differentiate two players
return map[used] = true;
}
}
return  map[used] = false;
}

public:
// without pruning O(n!)
// with pruning O(2^n)
bool canIWin(int maxChoosableInteger, int desiredTotal)
{
if (((1 + maxChoosableInteger) * maxChoosableInteger / 2) < desiredTotal) return false;
unordered_map<int, bool> map;
return wins(map, 0, maxChoosableInteger, desiredTotal);
}
};
``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.