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DPL16 Partition A Set Into Two Subsets With Minimum Absolute Sum Difference 🔥

We are given an array ‘ARR’ with N positive integers. We need to partition the array into two subsets such that the absolute difference of the sum of elements of the subsets is minimum.
We need to return only the minimum absolute difference of the sum of elements of the two partitions.

Again This question is a slight modification of the problem discussed in LDP14. We need to partition the array into two subsets such that the absolute difference of the sum of elements of the subsets is minimum.

Algorithm / Intuition

Before discussing the approach for this question, it is important to understand what we did in the previous question of the DPL14. There we found whether or not a subset exists in an array with a given target sum. We used a dp array to get to our answer.

what a Subsequence/Subset? A subset/subsequence is a contiguous or non-contiguous part of an array, where elements appear in the same order as the original array.
For example, for the array: [2,3,1] , the subsequences will be [{2},{3},{1},{2,3},{2,1},{3,1},{2,3,1}] but {3,2} is not a subsequence because its elements are not in the same order as the original array.

We used to return dp[n-1][k] as our answer. One interesting thing that is happening is that for calculating our answer for dp[n-1][k], we are also solving multiple sub-problems and at the same time storing them as well. We will use this property to solve the question

In this question, we need to partition the array into two subsets( say with sum S1 and S2) and we need to return the minimized absolute difference of S1 and S2. But do we need two variables for it? The answer is No. We can use a variable totSum, which stores the sum of all elements of the input array, and then we can simply say S2 = totSum – S1. Therefore we only need one variable S1.

Now, what values can S1 take? Well, it can go anywhere from 0 (no elements in S1) to totSum( all elements in S1). If we observe the last row of the dp array which we had discussed above, it gives us the targets for which there exists a subset. We will set its column value to totSum, to find the answer from 0(smaller limit of S1) to totSum (the larger limit of S1).

Our work is very simple, using the last row of the dp array, we will first find which all S1 values are valid. Using the valid S1 values, we will find S2 (totSum – S1). From this S1 and S2, we will find their absolute difference. We will return the minimum value of this absolute difference as our answer.

Now remaining Entire 👇 Process will Be Same As Like in DPL14

Recursice Approch

Steps to form the Recursive Solution
Step 1: Express the problem in terms of indexes.

The array will have an index but there is one more parameter “target”. We are given the initial problem to find whether there exists in the whole array a subsequence whose sum is equal to the target.
So, we can say that initially, we need to fun(n-1, target) which means that we need to find whether there exists a subsequence in the array from index 0 to n-1, whose sum is equal to the target. Similarly, we can generalize it for any index ind as follows:

Our Base

If target == 0, it means that we have already found the subsequence from the previous steps, so we can return true.

If ind==0, it means we are at the first element, so we need to return arr[ind]==target. If the element is equal to the target we return true else false.

The Recursive Function is

Step 2: Try out all possible choices at a given index.

We need to generate all the subsequences. We will use the pick/non-pick technique as discussed in, That we All Ready Learn in the Recursion Series.

We have two choices:

Exclude the current element in the subsequence: We first try to find a subsequence without considering the current index element. For this, we will make a recursive call to f(ind-1,target).

Include the current element in the subsequence: We will try to find a subsequence by considering the current index as element as part of subsequence. As we have included arr[ind], the updated target which we need to find in the rest if the array will be target – arr[ind]. Therefore, we will call f(ind-1,target-arr[ind]).

Note: We will consider the current element in the subsequence only when the current element is less or equal to the target.

Step 3: Return (taken || notTaken)

As we are looking for only one subset, if any of the one among taken or not taken returns true, we can return true from our function. Therefore, we return ‘or(||)’ of both of them.

The final pseudocode after steps 1, 2, and 3:

Recursion Base Conditions

If target == 0, it means that we have already found the subsequence from the previous steps, so we can return true.

If ind==0, it means we are at the first element, so we need to return arr[ind]==target. If the element is equal to the target we return true else false.

Recursion Tree

Time & Space Complexity

Time Complexity: O(2^N)
Reason: Exponential Time we find out the all the Possible Path
Space Complexity: O(N)
Reason: We are using a recursion stack space(O(N))

Memoization Approch

If we observe in the recursion tree, we will observe a many number of overlapping subproblems. Therefore the recursive solution can be memoized for to reduce the time complexity.

Steps to convert Recursive code to memoization solution:

Create a dp array of size [n][totSum+1]. The size of the input array is ‘n’, so the index will always lie between ‘0’ and ‘n-1’. The target can take any value between ‘0’ and ‘totSum’. Therefore we take the dp array as dp[n][totSum+1]

We initialize the dp array to -1.

Whenever we want to find the answer of particular parameters (say f(ind,target)), we first check whether the answer is already calculated using the dp array(i.e dp[ind][target]!= -1 ). If yes, simply return the value from the dp array.

If not, then we are finding the answer for the given value for the first time, we will use the recursive relation as usual but before returning from the function, we will set dp[ind][target] to the solution we get.

When we get the dp array, we will use its last row to find the absolute minimum difference of two partitions.

Time & Space Complexity

Time Complexity: O(N*totSum) +O(N) +O(N)
Reason: There are two nested loops that account for O(N*totSum), at starting we are running a for loop to calculate totSum and at last a for loop to traverse the last row.
Space Complexity: O(N*totSum) + O(N)
Reason: We are using an external array of size ‘N * totSum’ and a stack space of O(N).

Tabulation Approch

Tabulation is a ‘bottom-up’ approach where we start from the base case and reach the final answer that we want and Memoization is the Top-down Approch.

In Tabulation Approch We Just Creat a DP Array Same as Memoization and Simply Convert the Recurance Relation into the form of the Looping

Steps to convert Recursive Solution to Tabulation one.

To convert the memoization approach to a tabulation one, create a dp array with the same size as done in memoization. We can set its type as bool and initialize it as false.

First, we need to initialize the base conditions of the recursive solution.

If target == 0, ind can take any value from 0 to n-1, therefore we need to set the value of the first column as true.

The first row dp[0][] indicates that only the first element of the array is considered, therefore for the target value equal to arr[0], only cell with that target will be true, so explicitly set dp[0][arr[0]] =true, (dp[0][arr[0]] means that we are considering the first element of the array with the target equal to the first element itself). Please note that it can happen that arr[0]>target, so we first check it: if(arr[0]<=target) then set dp[0][arr[0]]=true.

After that , we will set our nested for loops to traverse the dp array and following the logic discussed in the recursive approach, we will set the value of each cell. Instead of recursive calls, we will use the dp array itself.

When we get the dp array, we will use its last row to find the absolute minimum difference between two partitions.

Time & Space Complexity

Time Complexity: O(N*totSum) +O(N) +O(N)
Reason: There are two nested loops that account for O(N*totSum), at starting we are running a for loop to calculate totSum, and at last a for loop to traverse the last row.
Space Complexity: O(N*totSum)
Reason: We are using an external array of size ‘N * totSum’. Stack Space is eliminated.

Space Optimization

If we closelly Observed if any Tabulation Approch we used the Some Limited Stuff like: dp[ind][target] = dp[ind-1][target] || dp[ind-1][target-arr[ind]] for the finding the our ans then definetly here Spaced Optimization is Possible in that types of Problems. Always Remember

Golden Rule

If we required only Prev row and Prev Collom then definetly we can Space Optimized

We see that we only need the previous row and column, in order to calculate curr[i]. Therefore we can space optimize it.

Initially, we can take a dummy row ( say prev) and initialize it as False.

Now the current row(say temp) only needs the previous row value and the current row’s value in order to curr[i].

Whenever we create a new row ( say cur), we need to explicitly set its first element is true according to our base condition.

At last Min will give us the required answer.

Time & Space Complexity

Time Complexity: O(N*totSum) +O(N) +O(N)
Reason: There are two nested loops that account for O(N*totSum), at starting we are running a for loop to calculate totSum and at last a for loop to traverse the last row.
Space Complexity: O(totSum)
Reason: We are using an external array of size ‘totSum+1’ to store only one row.