[416] Partition Equal Subset Sum

Claim

• All numbers in a non-empty array are positive integers. Can the elements of this array be divided into two parts so that the numbers in each part are equal
• Up to 200 numbers, each number up to 100

Examples

• [1,5,11,5], return true
• [1,2,3,5], return false

Ideas

• Pick a certain item out of n items and fill the sum / 2 backpack
• Status: F (n, C)
• 转移：F(i,c)=F(i-1,c) || F(i-1,c-w(i))
• Complexity: O (n * sum / 2) = O (n * sum)

achieve

Recursive + memory search

1  class Solution {
 2  private :
 3      // memo [i] [c] means whether the element with index [0 ... i] can completely fill a backpack with capacity c
 4      // -1 means not calculated, 0 means not fillable, 1 means fillable 
5      vector <vector < int >> memo;
 6      
7      // If you use nums [0 ... index], can you completely fill a backpack with sum capacity 
8      bool tryPartition ( const vector < int > & nums, int index, int sum) {
 9          
10          if (sum == 0 )
 11              return  true ;
 12              
13          if( sum < 0 || index < 0 )
14             return false;
15         
16         if( memo[index][sum] != -1 )
17             return memo[index][sum] == 1;
18                 
19         memo[index][sum] = ( tryPartition(nums, index-1, sum ) || 
20                  tryPartition(nums, index-1, sum-nums[index] ) ) ? 1 : 0;
21         return memo[index][sum] == 1;    
22     }    
23 public:
24     bool canPartition(vector<int>& nums) {
25         int sum = 0 ; 
26         for( int i = 0 ; i < nums.size() ; i ++ ){
27             assert( nums[i] > 0 );
28             sum += nums[i];
29         }
30         
31         if( sum%2 != 0 )
32             return false;
33         
34         memo = vector<vector<int>>( nums.size(), vector<int>(sum/2+1,-1));
35         return tryPartition( nums, nums.size()-1, sum/2 );
36     }
37 };

View Code

Dynamic programming

 1 class Solution {
 2     
 3 public:
 4     bool canPartition(vector<int>& nums) {
 5         int sum = 0 ; 
 6         for( int i = 0 ; i < nums.size() ; i ++ ){
 7             assert( nums[i] > 0 );
 8             sum += nums[i];
 9         }
10         
11         if( sum%2 != 0 )
12             return false;
13         
14         int n = nums.size();
15         int C = sum/2;
16         vector<bool> memo(C+1, false);
17         
18         for( int i = 0 ; i <= C ; i ++ )
19             memo[i] = ( nums[0] == i );
20         
21         for( int i = 1 ; i < n ; i ++ )
22             for( int j = C ; j >= nums[i] ; j -- )
23                 memo[j] = memo[j] || memo[j-nums[i]];
24         
25         return memo[C];        
26     }
27 };

View Code

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