Created
October 15, 2020 18:42
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class Solution { | |
vector<vector<int>> mem; | |
bool subsetSum(vector<int>& nums,int n,int pos,int sum) | |
{ | |
if(sum==0) //Sum found | |
return true; | |
else if(pos>=n || sum<0) //Out of bounds | |
return false; | |
if(mem[pos][sum]!=-1) | |
return mem[pos][sum]; | |
return mem[pos][sum] = subsetSum(nums,n,pos+1,sum-nums[pos]) || subsetSum(nums,n,pos+1,sum); | |
} | |
public: | |
bool canPartition(vector<int>& nums) { | |
int sum = 0; | |
int n = nums.size(); | |
for(int i=0;i<n;++i) | |
sum += nums[i]; | |
if(sum & 1) //ODD sum | |
return false; | |
mem.clear(); | |
mem.resize(n+1,vector<int>(sum/2+1,-1)); | |
return subsetSum(nums,n,0,sum/2); | |
} | |
}; | |
class Solution { | |
public: | |
bool canPartition(vector<int>& nums) { | |
int sum = 0; | |
int n = nums.size(); | |
for(int i=0;i<n;++i) | |
sum += nums[i]; | |
if(sum & 1) //ODD sum | |
return false; | |
bool dp[n+1][sum/2+1]; | |
//Fill DP table (Subset Sum Problem code) | |
for(int i=0;i<=n;++i) | |
for(int j=0;j<=sum/2;++j) | |
{ | |
if(i==0 || j==0) //Base Case | |
dp[i][j] = false; | |
else if(nums[i-1] > j) //SkIP case | |
dp[i][j] = dp[i-1][j]; | |
else if(nums[i-1]==j) | |
dp[i][j] = true; | |
else | |
dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i-1]]; | |
} | |
return dp[n][sum/2]; | |
} | |
}; | |
class Solution { | |
public boolean canPartition(int[] nums) { | |
int n = nums.length; | |
int sum = 0; | |
for(int i=0;i<n;++i) | |
sum += nums[i]; | |
if(sum%2 == 1) | |
return false; | |
boolean[][] dp = new boolean[n+1][sum/2+1]; | |
for(int i=0;i<=n;++i) | |
for(int j=0;j<=sum/2;++j) | |
{ | |
if(i==0 || j==0) //Base Case | |
dp[i][j] = false; | |
else if(nums[i-1] > j) //SkIP case | |
dp[i][j] = dp[i-1][j]; | |
else if(nums[i-1]==j) | |
dp[i][j] = true; | |
else | |
dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i-1]]; | |
} | |
return dp[n][sum/2]; | |
} | |
} |
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