# Can I Win

`Game Theory`, `Minimax`, `Memorized Search`, `Dynamic Programming`

Medium

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 & Analysis

boolean[] `{false, false, true, true, false}`, to integer with binary representation as`00110`.

--

By @leogogogo:

Most of the "Game Playing" problems on LeetCode 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.

-

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 that1)and2)are actually related because we can always get the2)by deducting the sum of chosen numbers from original desiredTotal.

Then the problem becomes how to describe the state using1).

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.

-

##### Time Complexity:

Thanks @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!)`)

##### Space Complexity:

`O(2^n)`

``````class Solution {
public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
int sum = maxChoosableInteger * (maxChoosableInteger + 1) / 2;
if (sum < desiredTotal) {
return false;
}
if (desiredTotal <= maxChoosableInteger) {
return true;
}
Map<Integer, Boolean> memo = new HashMap<Integer, Boolean>();
boolean[] used = new boolean[maxChoosableInteger + 1];
return helper(desiredTotal, memo, used);
}

private boolean helper(int desiredTotal, Map<Integer, Boolean> memo, boolean[] used) {
if (desiredTotal <= 0) {
return false;
}
int key = formatKey(used);
if (!memo.containsKey(key)) {

// try every unchosen number as next step
for (int i = 1; i < used.length; i++) {
if (!used[i]) {
used[i] = true;

// check whether this lead to a win (i.e. the other player lose)
if (!helper(desiredTotal - i, memo, used)) {
memo.put(key, true);
used[i] = false;
return true;
}
used[i] = false;
}
}
memo.put(key, false);
}
return memo.get(key);

}

private int formatKey(boolean[] used) {
int key = 0;
for (boolean b: used) {
key = key << 1;
if (b) {
key = key | 1;
}
}
return key;
}
}
``````