464. Can I Win

Description

In the "100 game," two players take turns adding, to a running total, any integer from 1..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 of 1..15 without replacement until they reach a total >= 100.

Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.

You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.

Example

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.

Solution

[TLE] DFS, time extremely high, space O(n)

Brute-force解法，注意不要忘记reset used array！由于会重复计算子问题，这个solution会TLE。

class Solution {
    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal <= 0) {
            return true;
        }
        
        int sum = (1 + maxChoosableInteger) * maxChoosableInteger / 2;
        if (sum < desiredTotal) {
            return false;
        }
        
        boolean[] used = new boolean[maxChoosableInteger + 1];
        return canIWin(desiredTotal, used);
    }
    
    private boolean canIWin(int desiredTotal, boolean[] used) {
        if (desiredTotal <= 0) {
            return false;
        }
        
        for (int i = 1; i < used.length; ++i) {
            if (used[i]) {
                continue;
            }
            
            used[i] = true;
            if (!canIWin(desiredTotal - i, used)) {
                used[i] = false;    // don't forget to reset used[]!
                return true;
            }
            used[i] = false;
        }
        
        return false;
    }
}

DFS with note

由于上面的solution会有很多子问题的重复计算，时间消耗太大，所以需要尝试使用备忘。首先需要定义这里的子问题，即当前还没有被选过的数字们，可以用used[]的一个状态去表示。

注意题目中指出，“maxChoosableInteger will not be larger than 20”，所以used[]最大的size是21，可以将其用二进制转换成一个int，作为HashMap的key（这种做法还是蛮巧妙的）。

class Solution {
    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal <= 0) {
            return true;
        }
        
        int sum = (1 + maxChoosableInteger) * maxChoosableInteger / 2;
        if (sum < desiredTotal) {
            return false;
        }
        
        boolean[] used = new boolean[maxChoosableInteger + 1];
        return canIWin(desiredTotal, used, new HashMap<>());
    }
    
    private boolean canIWin(int desiredTotal, boolean[] used, Map note) {
        if (desiredTotal <= 0) {    // because "reach or exceed"
            return false;
        }
        
        int index = getIndex(used);
        if (note.containsKey(index)) {
            return note.get(index);
        }
        
        for (int i = 1; i < used.length; ++i) {
            if (used[i]) {
                continue;
            }
            
            used[i] = true;
            if (!canIWin(desiredTotal - i, used, note)) {
                used[i] = false;    // don't forget to reset used[]!
                note.put(index, true);
                return true;
            }
            used[i] = false;
        }
        
        note.put(index, false);
        return false;
    }
    
    // transfer boolean[] to an Integer 
    private int getIndex(boolean[] used) {
        int index = 0;
        for (boolean b : used) {
            index <<= 1;
            index |= b ? 1 : 0;
        }
        return index;
    }
}