Sunday, November 20, 2016

LeetCode 464 - Can I Win


https://leetcode.com/problems/can-i-win/
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.
https://discuss.leetcode.com/topic/68896/java-solution-using-hashmap-with-detailed-explanation
After solving several "Game Playing" questions in leetcode, I find them to be pretty similar. Most of them can be solved using the top-down DP approach, which "brute-forcely" simulates every possible state of the game.
The key part for the top-down dp strategy is that we need to avoid repeatedly solving sub-problems. Instead, we should use some strategy to "remember" the outcome of sub-problems. Then when we see them again, we instantly know their result. By doing this, we can always reduce time complexity from exponential to polynomial.
(EDIT: Thanks for @billbirdh for pointing out the mistake here. For this problem, by applying the memo, we at most compute for every subproblem once, and there are O(2^n) subproblems, so the complexity is O(2^n) after memorization. (Without memo, time complexity should be like O(n!))
For this question, the key part is: what is the state of the game? Intuitively, to uniquely determine the result of any state, we need to know:
  1. The unchosen numbers
  2. The remaining desiredTotal to reach
A second thought reveals that 1) and 2) are actually related because we can always get the 2) by deducting the sum of chosen numbers from original desiredTotal.
Then the problem becomes how to describe the state using 1).
In my solution, I use a boolean array to denote which numbers have been chosen, and then a question comes to mind, if we want to use a Hashmap to remember the outcome of sub-problems, can we just use Map<boolean[], Boolean> ? Obviously we cannot, because the if we use boolean[] as a key, the reference to boolean[] won't reveal the actual content in boolean[].
Since in the problem statement, it says maxChoosableInteger will not be larger than 20, which means the length of our boolean[] array will be less than 20. Then we can use an Integer to represent this boolean[] array. How?
Say the boolean[] is {false, false, true, true, false}, then we can transfer it to an Integer with binary representation as 00110. Since Integer is a perfect choice to be the key of HashMap, then we now can "memorize" the sub-problems using Map<Integer, Boolean>.
The rest part of the solution is just simulating the game process using the top-down dp.

In Java, to denote boolean[], an easier way is to use Arrays.toString(boolean[]), which will transfer a boolean[] to sth like "[true, false, false, ....]", which is also not limited to how maxChoosableInteger is set, so it can be generalized to arbitrary large maxChoosableInteger.

    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal<=0) return true;
        if (maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return false;
        return canIWin(desiredTotal, new int[maxChoosableInteger], new HashMap<>());
    }
    private boolean canIWin(int total, int[] state, HashMap<String, Boolean> hashMap) {
        String curr=Arrays.toString(state);
        if (hashMap.containsKey(curr)) return hashMap.get(curr);
        for (int i=0;i<state.length;i++) {
            if (state[i]==0) {
                state[i]=1;
                if (total<=i+1 || !canIWin(total-(i+1), state, hashMap)) {
                    hashMap.put(curr, true);
                    state[i]=0;
                    return true;
                }
                state[i]=0;
            }
        }
        hashMap.put(curr, false);
        return false;
    }


    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal<=0) return true;
        if (maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return false;
        return canIWin(desiredTotal, maxChoosableInteger, 0, new HashMap<>());
    }
    private boolean canIWin(int total, int n, int state, HashMap<Integer, Boolean> hashMap) {
        if (hashMap.containsKey(state)) return hashMap.get(state);
        for (int i=0;i<n;i++) {
            if ((state&(1<<i))!=0) continue;
            if (total<=i+1 || !canIWin(total-(i+1), n, state|(1<<i), hashMap)) {
                hashMap.put(state, true);
                return true;
            }
        }
        hashMap.put(state, false);
        return false;
    }
}
http://blog.csdn.net/mebiuw/article/details/53266731
public boolean canIWin(int maxChoosableInteger, int desiredTotal) { if (desiredTotal<=0) return true; //如果1到最大能选的值所有和都不能满足目标值,那么肯定失败 if (maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return false; char state[] = new char[maxChoosableInteger]; //maybe just int or boolean array for(int i=0;i<maxChoosableInteger;i++) state[i] = '0'; return dfs(desiredTotal, state, new HashMap<>()); } private boolean dfs(int total, char[] state, HashMap<String, Boolean> hashMap) { String key= new String(state); if (hashMap.containsKey(key)) return hashMap.get(key); for (int i=0;i<state.length;i++) { if (state[i]=='0') { state[i]='1'; if (total<=i+1 || !dfs(total-(i+1), state, hashMap)) { hashMap.put(key, true); state[i]='0'; return true; } state[i]='0'; } } hashMap.put(key, false); return false; }
http://bookshadow.com/weblog/2016/11/20/leetcode-can-i-win/
记忆化搜索 + 位运算
由于maxChoosableInteger不大于20,因此可以通过整数state表示当前已经选择了哪些数字
state的第i位为1时,表示选择了数字i + 1
利用字典dp记录已经搜索过的状态
https://discuss.leetcode.com/topic/78928/brute-force-and-memoization
  1. O(n!) brute force, n is maxChoosableInteger. T(n)=nT(n-1)
    bool canIWin(int maxChoosableInteger, int desiredTotal) {
        if(!desiredTotal) return 1;
        return canWin(~0<<maxChoosableInteger, maxChoosableInteger, desiredTotal);
    }
    bool canWin(int pool, int maxint, int tot) {
        if(tot<=0) return 0;
        for(int i=0;i<maxint;i++) {
            int mask = 1<<i;
            if(pool & mask) continue;
            pool|=mask;
            if(!canWin(pool,maxint, tot-i-1)) return 1;
            pool^=mask;
        }
        return 0;
    }
  1. O(2^n) Memoization. There is redundant computation in #1. A state with a pool and total may be computed many times. So we can cache the state and reuse it. At first glance, it seems that a state is determined by two values, the pool and the total. However, since the initial total is known, the remaining total is known given the pool. So a state can be identified by the pool only.
    bool canIWin(int maxChoosableInteger, int desiredTotal) {
        if(!desiredTotal) return 1;
        if(maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return 0;
        unordered_map<int,char> mem;
        return canWin(~0<<maxChoosableInteger, maxChoosableInteger, desiredTotal, mem);
    }
    bool canWin(int pool, int maxint, int tot, unordered_map<int,char>& mem) {
        if(tot<=0) return 0;
        auto it = mem.find(pool);
        if(it != mem.end()) return it->second;
        for(int i=0;i<maxint;i++) {
            int mask = 1<<i;
            if(pool & mask) continue;
            pool|=mask;
            if(!canWin(pool,maxint,tot-i-1,mem)) return mem[pool^=mask]=1;
            pool^=mask;
        }
        return mem[pool] = 0;
    }
  1. Iterative dp. For most dp problems, the next step is to transform recursion with memoization to iterative dp. However, that does not help and is actually pretty bad for this problem. In iterative dp, we have to visit all the 2^n states to get the result. In #2 DFS with memoization, DFS terminates as soon as it finds a way to win. The worst case O(2^n) rarely happens. So if DFS has early termination condition, then it should be better than dp that visits all the states. Similar problems are word break and Concatenated Words.


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