<p>In the &quot;100 game&quot; two players take turns adding, to a running total, any integer from <code>1</code> to <code>10</code>. The player who first causes the running total to <strong>reach or exceed</strong> 100 wins.</p>

<p>What if we change the game so that players <strong>cannot</strong> re-use integers?</p>

<p>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 &gt;= 100.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> maxChoosableInteger = 10, desiredTotal = 11
<strong>Output:</strong> false
<strong>Explanation:</strong>
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 &gt;= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> maxChoosableInteger = 10, desiredTotal = 0
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> maxChoosableInteger = 10, desiredTotal = 1
<strong>Output:</strong> true
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= maxChoosableInteger &lt;= 20</code></li>
	<li><code>0 &lt;= desiredTotal &lt;= 300</code></li>
</ul>

class Solution {
public:
    bool canIWin(int mx, int sum) {
        // Edge case: if the total sum of numbers [1, 2, ..., mx] is less than the target sum, nobody can win.
        if ((mx * (mx + 1)) / 2 < sum) return false;
        if(sum==0)return true;
        // Memoization to store whether the current player can win from a given mask state
        vector<int> dp(1 << mx, -1); // Only depends on mask
        
        // Recursive function with memoization
        function<bool(int, int)> rec = [&](int mask, int rem) -> bool {
            if (rem <= 0) return false;  // If remaining sum is <= 0, current player loses
            if (dp[mask] != -1) return dp[mask];  // Return memoized result if available

            for (int i = 1; i <= mx; ++i) {
                // Check if number i is available in the current mask
                if ((mask >> (i - 1)) & 1) {
                    // If the opponent cannot win after picking i, then the current player can win
                    if (!rec(mask & ~(1 << (i - 1)), rem - i)) {
                        return dp[mask] = true;
                    }
                }
            }

            return dp[mask] = false;  // If no winning move, return false
        };

        // Initial call to the recursive function with all numbers available (mask) and target sum
        return rec((1 << mx) - 1, sum);
    }
};

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