15. Partition Equal Subset Sum

The problem can be found at the following link: Question Link

My Approach

For solving the "Subset Sum Problem," which is a classic dynamic programming problem. We have to determine if it's possible to partition an array into two subsets such that the sum of elements in both subsets is equal.

• I start by calculating the total sum of all elements in the input array. If the sum is odd, it's impossible to partition the array into two subsets with equal sums, so I return false immediately.

• I create a 2D DP array dp, where dp[i][target] represents whether it's possible to achieve a sum of target using the first i elements of the array.

• I use a recursive function canPart to fill the DP array. For each element in the array, I have two options:

  • Take the element into the subset by subtracting its value from the target sum.

  • Skip the element and move to the next one.

  • I recursively explore both options, and if I find a path that leads to the target sum, I set dp[i][target] to true.

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity of this solution is O(N * target), where N is the number of elements in the input array, and target is the target sum we are trying to achieve.

• Auxiliary Space Complexity: The auxiliary space complexity is O(N * target) as well because of the DP array.

Code (C++)

class Solution {
public:
    bool canPart(int& N, int i, int arr[], int target, vector<vector<int>>& dp){
        if(target == 0)
            return true;
        
        if(i >= N)
            return false;
           
        if(target >= arr[i]){
            bool take = canPart(N,i+1,arr,target - arr[i], dp);
            if(take)
                return dp[i][target] = true;
        }
        
        bool notTake = canPart(N,i+1,arr,target, dp);
        return dp[i][target] = notTake;
    }
    
    bool equalPartition(int N, int arr[])
    {
        int sum = 0;
        for(int i = 0; i<N;++i){
            sum += arr[i];
        }
        if(sum % 2 != 0) // If the sum is odd, it's impossible to partition into equal subsets.
            return false;
            
        int target = sum/2;
        vector<vector<int>> dp(N,vector<int>(target+1,-1));
        return canPart(N,0,arr,target, dp);
    }
};

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