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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + |
| 4 | + /* |
| 5 | + best solution |
| 6 | + bool cal(vector<int> nums, int sum, int i) { |
| 7 | + if (i == nums.size() || sum < nums[i]) return false; |
| 8 | + if (sum == nums[i]) return true; |
| 9 | + |
| 10 | + return cal(nums, sum - nums[i], i+1) || cal(nums, sum, i+1); |
| 11 | + } |
| 12 | + |
| 13 | + bool canPartition(vector<int>& nums) { |
| 14 | + sort(nums.rbegin(), nums.rend()); |
| 15 | + int sum = 0; |
| 16 | + for (auto num : nums) sum += num; |
| 17 | + if (sum%2) return false; |
| 18 | + sum /= 2; |
| 19 | + return cal(nums, sum, 0); |
| 20 | + } |
| 21 | + */ |
| 22 | + |
| 23 | + bool subsetSumDP(vector<int>& nums, int sum, int n) { |
| 24 | + |
| 25 | + //dp matrix |
| 26 | + //first col => true, first row => false |
| 27 | + |
| 28 | + vector<vector<bool>> dp(n + 1, vector<bool>(sum+1)); |
| 29 | + |
| 30 | + //sum > 0 but n is 0 is not possible.. first row of dp matrix |
| 31 | + for(int i=0; i<=sum; i++) dp[0][i] = false; |
| 32 | + |
| 33 | + //empty sum is subset => {} |
| 34 | + for(int i=0; i<=n; i++) dp[i][0] = true; |
| 35 | + |
| 36 | + for(int i=1; i<=n; i++) { |
| 37 | + for(int j=1; j<=sum; j++) { |
| 38 | + |
| 39 | + if(nums[i-1] <= j) //current value is less than sum => include or exclude |
| 40 | + dp[i][j] = dp[i-1][j - nums[i-1]] || dp[i-1][j]; |
| 41 | + |
| 42 | + else |
| 43 | + dp[i][j] = dp[i-1][j]; //current value > sum; definitely exclude |
| 44 | + } |
| 45 | + } |
| 46 | + return dp[n][sum]; |
| 47 | + } |
| 48 | + |
| 49 | + bool canPartition(vector<int>& nums) { |
| 50 | + auto sum = 0; |
| 51 | + sum = accumulate(nums.begin(), nums.end(), sum); |
| 52 | + cout<<sum<<"\n"; |
| 53 | + if(sum % 2 != 0) return false; |
| 54 | + return subsetSumDP(nums, sum/2, nums.size()); |
| 55 | + |
| 56 | + } |
| 57 | +}; |
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