# Can I Win Problem

## Description

LeetCode Problem 464.

In the “100 game” two players take turns adding, to a running total, any integer from 1 to 10. The player who first causes the running total to reach or exceed 100 wins.

What if we change the game so that players cannot re-use integers?

For example, two players might take turns drawing from a common pool of numbers from 1 to 15 without replacement until they reach a total >= 100.

Given two integers maxChoosableInteger and desiredTotal, return true if the first player to move can force a win, otherwise, return false. Assume both players play optimally.

Example 1:

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Input: maxChoosableInteger = 10, desiredTotal = 11
Output: false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.

Example 2:

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Input: maxChoosableInteger = 10, desiredTotal = 0
Output: true

Example 3:

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Input: maxChoosableInteger = 10, desiredTotal = 1
Output: true

Constraints:

- 1 <= maxChoosableInteger <= 20
- 0 <= desiredTotal <= 300

## Sample C++ Code using Depth First Search

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class Solution {
public:
bool canIWin(int maxChoosableInteger, int desiredTotal) {
if(maxChoosableInteger * (maxChoosableInteger + 1) / 2 < desiredTotal) return false;
vector<int8_t> vis(1 << maxChoosableInteger, -1);
return isWinState(0, maxChoosableInteger, desiredTotal, vis);
}
private:
bool isWinState(int used, int maxChoose, int desire, vector<int8_t>& vis){
if(~vis[used]) return vis[used];
for(int i = maxChoose; i > 0; --i){
if(used & (1 << (i - 1))) continue;
if(i >= desire || !isWinState(used | (1 << (i - 1)), maxChoose, desire - i, vis))
return vis[used] = 1;
}
return vis[used] = 0;
}
};