首页 > 代码库 > [LintCode] Coins in a Line II

[LintCode] Coins in a Line II

2024-09-29 22:05:02   218人阅读

There are n coins with different value in a line. Two players take turns to take one or two coins from left side until there are no more coins left. The player who take the coins with the most value wins.

Could you please decide the first player will win or lose?

Example

Given values array A = [1,2,2], return true.

Given A = [1,2,4], return false.

 

Solution 1. Recursion 

In order for the first player to win, the coin value he gets must be bigger than a half of the total value of n coins. 

For a given start index that the first player can pick one or two coins at,  he can either pick one coin of values[startIdx] or two coins of values[startIdx] + values[startIdx + 1].  

Given that the second player plays optimally too, we have the following optimal substructure.

As we can see from the optimal substructure, there exists many overlapping subproblems that are redundantly recomputed, making this recursive solution inefficient.

26         int pickOneVal = Math.min(fPMaxValue(values, startIdx + 2, currVal), 
27                          fPMaxValue(values, startIdx + 3, currVal)) + values[startIdx];
28         int pickTwoVal = Math.min(fPMaxValue(values, startIdx + 3, currVal),
29                          fPMaxValue(values, startIdx + 4, currVal)) + values[startIdx] + values[startIdx + 1];
30         return Math.max(pickOneVal, pickTwoVal);

 1 public class Solution {
 2     public boolean firstWillWin(int[] values) {
 3         if(values == null || values.length == 0){
 4             return false;
 5         }
 6         if(values.length <= 2){
 7             return true;
 8         }
 9         int sum = 0; 
10         for(int i = 0; i < values.length; i++){
11             sum += values[i];
12         }
13         return fPMaxValue(values, 0, 0) > sum / 2;
14     }
15     private int fPMaxValue(int[] values, int startIdx, int currVal){
16         int diff = values.length - startIdx;
17         if(diff == 2 || diff == 3){
18             return currVal + values[startIdx] + values[startIdx + 1];
19         }
20         if(diff == 1){
21             return currVal + values[startIdx];
22         }
23         if(diff <= 0){
24             return currVal;
25         }
26         int pickOneVal = Math.min(fPMaxValue(values, startIdx + 2, currVal), 
27                          fPMaxValue(values, startIdx + 3, currVal)) + values[startIdx];
28         int pickTwoVal = Math.min(fPMaxValue(values, startIdx + 3, currVal),
29                          fPMaxValue(values, startIdx + 4, currVal)) + values[startIdx] + values[startIdx + 1];
30         return Math.max(pickOneVal, pickTwoVal);
31     }
32 }

 

Solution 2. Dynamic Programming 

Since solution 1 has both optimal substructure and overlapping subproblems, we can apply dynamic programming to avoid redundant recomputation of subproblems.

Dp state: dp[i] is the most value player one can get when there is i coins left.

Dp function: 

dp[i] = Math.max(Math.min(dp[i - 2], dp[i - 3]) + values[n - i],  Math.min(dp[i - 3], dp[i - 4]) + values[n - i] + values[n - i + 1]);

 

 1 public class Solution {
 2     public boolean firstWillWin(int[] values) {
 3         if(values == null || values.length == 0){
 4             return false;
 5         }
 6         if(values.length <= 2){
 7             return true;
 8         }
 9         int sum = 0; 
10         for(int i = 0; i < values.length; i++){
11             sum += values[i];
12         }
13         int n = values.length;
14         int[] dp = new int[n + 1];
15         dp[0] = 0;
16         dp[1] = values[n - 1];
17         dp[2] = values[n - 2] + values[n - 1];
18         dp[3] = values[n - 3] + values[n - 2];
19         for(int i = 4; i <= n; i++){
20             dp[i] = Math.max(Math.min(dp[i - 2], dp[i - 3]) + values[n - i], 
21                              Math.min(dp[i - 3], dp[i - 4]) + values[n - i] + values[n - i + 1]);
22         }
23         return dp[n] > sum / 2;
24     }
25 }

 

[LintCode] Coins in a Line II


联系
我们