464. Can I Win

Description

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:

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:

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

Example 3:

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

 

Constraints:

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

Solutions

Solution 1: State Compression + Memoization

First, we check if the sum of all selectable integers is less than the target value. If so, it means that we cannot win no matter what, so we directly return false.

Then, we design a function dfs(mask, s), where mask represents the current state of the selected integers, and s represents the current cumulative sum. The return value of the function is whether the current player can win.

The execution process of the function dfs(mask, s) is as follows:

We iterate through each integer i from 1 to maxChoosableInteger. If i has not been selected, we can choose i. If the cumulative sum s + i after choosing i is greater than or equal to the target value desiredTotal, or if the result of the opponent choosing i is losing, then the current player is winning, return true.

If no choice can make the current player win, then the current player is losing, return false.

To avoid repeated calculations, we use a hash table f to record the calculated states, where the key is mask, and the value is whether the current player can win.

The time complexity is $O(2^n)$, and the space complexity is $O(2^n)$. Where $n$ is maxChoosableInteger.

Python3JavaC++GoTypeScript

class Solution:
    def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:
        @cache
        def dfs(mask: int, s: int) -> bool:
            for i in range(1, maxChoosableInteger + 1):
                if mask >> i & 1 ^ 1:
                    if s + i >= desiredTotal or not dfs(mask | 1 << i, s + i):
                        return True
            return False

        if (1 + maxChoosableInteger) * maxChoosableInteger // 2 < desiredTotal:
            return False
        return dfs(0, 0)

class Solution {
    private Map<Integer, Boolean> f = new HashMap<>();
    private int maxChoosableInteger;
    private int desiredTotal;

    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if ((1 + maxChoosableInteger) * maxChoosableInteger / 2 < desiredTotal) {
            return false;
        }
        this.maxChoosableInteger = maxChoosableInteger;
        this.desiredTotal = desiredTotal;
        return dfs(0, 0);
    }

    private boolean dfs(int mask, int s) {
        if (f.containsKey(mask)) {
            return f.get(mask);
        }
        for (int i = 0; i < maxChoosableInteger; ++i) {
            if ((mask >> i & 1) == 0) {
                if (s + i + 1 >= desiredTotal || !dfs(mask | 1 << i, s + i + 1)) {
                    f.put(mask, true);
                    return true;
                }
            }
        }
        f.put(mask, false);
        return false;
    }
}

class Solution {
public:
    bool canIWin(int maxChoosableInteger, int desiredTotal) {
        if ((1 + maxChoosableInteger) * maxChoosableInteger / 2 < desiredTotal) {
            return false;
        }
        unordered_map<int, int> f;
        function<bool(int, int)> dfs = [&](int mask, int s) {
            if (f.contains(mask)) {
                return f[mask];
            }
            for (int i = 0; i < maxChoosableInteger; ++i) {
                if (mask >> i & 1 ^ 1) {
                    if (s + i + 1 >= desiredTotal || !dfs(mask | 1 << i, s + i + 1)) {
                        return f[mask] = true;
                    }
                }
            }
            return f[mask] = false;
        };
        return dfs(0, 0);
    }
};

func canIWin(maxChoosableInteger int, desiredTotal int) bool {
    if (1+maxChoosableInteger)*maxChoosableInteger/2 < desiredTotal {
        return false
    }
    f := map[int]bool{}
    var dfs func(int, int) bool
    dfs = func(mask, s int) bool {
        if v, ok := f[mask]; ok {
            return v
        }
        for i := 1; i <= maxChoosableInteger; i++ {
            if mask>>i&1 == 0 {
                if s+i >= desiredTotal || !dfs(mask|1<<i, s+i) {
                    f[mask] = true
                    return true
                }
            }
        }
        f[mask] = false
        return false
    }
    return dfs(0, 0)
}

function canIWin(maxChoosableInteger: number, desiredTotal: number): boolean {
    if (((1 + maxChoosableInteger) * maxChoosableInteger) / 2 < desiredTotal) {
        return false;
    }
    const f: Record<string, boolean> = {};
    const dfs = (mask: number, s: number): boolean => {
        if (f.hasOwnProperty(mask)) {
            return f[mask];
        }
        for (let i = 1; i <= maxChoosableInteger; ++i) {
            if (((mask >> i) & 1) ^ 1) {
                if (s + i >= desiredTotal || !dfs(mask ^ (1 << i), s + i)) {
                    return (f[mask] = true);
                }
            }
        }
        return (f[mask] = false);
    };
    return dfs(0, 0);
}

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