15.Partition Equal Subset Sum

Given a non-empty array nums 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.

Example 1:
Input: nums = [1,5,11,5]
Output: true
Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:
Input: nums = [1,2,3,5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

Solution : (Recursion)

class Solution
{
public:
    bool subsetSum(vector<int> v, int sum, int pos)
    {
        if(pos == v.size()){
            return false;
        }
        if (sum == 0)
        {
            return true;
        }

        if (subsetSum(v, sum - v[pos], pos + 1) || subsetSum(v, sum, pos + 1))
        {
            return true;
        }

        return false;
    }
    
    bool canPartition(vector<int> &nums)
    {
        int s = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            s = s + nums[i];
        }
        
        // Checking if solution exits
        if (s % 2 == 0)
        {
            //Subset sum(Finding  a subset with given sum)
            if (subsetSum(nums, s / 2, 0))
            {
                return true;
            }
            return false;
        }

        return false;
    }
};

Time Complexity: O(2^n)

Solution : (DP)

class Solution
{
public:
    bool subsetSum(vector<int> nums, int target)
    {
        vector<vector<bool>> v(nums.size() + 1, vector<bool>(target + 1, 0));
        v[0][0] = true;

        for (int i = 1; i <= nums.size(); i++)
        {
            int cur = nums[i - 1];

            for (int j = 0; j <= target; j++)
            {
                if (j < cur)
                {
                    v[i][j] = v[i - 1][j];
                }
                else
                {
                    v[i][j] = v[i - 1][j] || v[i - 1][j - cur];
                }
            }
        }
        return v[nums.size()][target];
    }

    bool canPartition(vector<int> &nums)
    {
        int s = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            s = s + nums[i];
        }

        if (s % 2 == 0)
        {
            //Subset sum(Finding  a subset with given sum)
            if (subsetSum(nums, s / 2))
            {
                return true;
            }
            return false;
        }

        return false;
    }
};

Time Complexity: O(n * m) Space Complexity: O(n * m)