416. Partition Equal Subset Sum
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Problem statement:
Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
- Each of the array element will not exceed 100.
- The array size will not exceed 200.
Example 1:
Input: [1, 5, 11, 5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
Solution: DFS with return value(TLE).
This is problem can be solved by DFS with or without a return value. I choose a the DFS template with a bool return value. It returns true once we find a subset.
Before the DFS, I sort the array in ascending order first.
In each DFS level, ignore the duplicated value.
One cur_sum < 0, return false.
cur_sum == 0, return true;
cur_sum ---> continue on DFS search.
However, it is a TLE solution, can not pass OJ.
Since there are two situations for each element: selected or not. Time complexity is O(2^n).
class Solution { public: bool canPartition(vector<int>& nums) { sort(nums.begin(), nums.end()); int sum = accumulate(nums.begin(), nums.end(), 0); return !(sum & 1) && can_partition_dfs(nums, sum / 2, 0); } bool can_partition_dfs(vector<int>& nums, int cur_sum, int idx){ if(cur_sum < 0){ return cur_sum == 0; } int can_partition = false; for(int i = idx; i < nums.size(); i++){ can_partition |= can_partition_dfs(nums, cur_sum - nums[i], i + 1); while(i + 1 < nums.size() && nums[i] == nums[i + 1]){ i++; } } return can_partition; } };
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