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Python Program for Subset Sum Problem | DP-25

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Write a Python program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum.

Examples:

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True
Explanation: There is a subset (4, 5) with sum 9.

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
Explanation: There is no subset that adds up to 30.

Python Program for Subset Sum Problem using Recursion:

For the recursive approach, there will be two cases.

  • Consider the ‘last’ element to be a part of the subset. Now the new required sum = required sum – value of ‘last’ element.
  • Don’t include the ‘last’ element in the subset. Then the new required sum = old required sum.

In both cases, the number of available elements decreases by 1.

Step-by-step approach:

  • Build a recursive function and pass the index to be considered (here gradually moving from the last end) and the remaining sum amount.
  • For each index check the base cases and utilize the above recursive call.
  • If the answer is true for any recursion call, then there exists such a subset. Otherwise, no such subset exists.

Below is the implementation of the above approach.

Python3




# A recursive solution for subset sum
# problem
 
 
# Returns true if there is a subset
# of set[] with sun equal to given sum
def isSubsetSum(set, n, sum):
 
    # Base Cases
    if (sum == 0):
        return True
    if (n == 0):
        return False
 
    # If last element is greater than
    # sum, then ignore it
    if (set[n - 1] > sum):
        return isSubsetSum(set, n - 1, sum)
 
    # Else, check if sum can be obtained
    # by any of the following
    # (a) including the last element
    # (b) excluding the last element
    return isSubsetSum(
        set, n-1, sum) or isSubsetSum(
        set, n-1, sum-set[n-1])
 
 
# Driver code
if __name__ == '__main__':
    set = [3, 34, 4, 12, 5, 2]
    sum = 9
    n = len(set)
    if (isSubsetSum(set, n, sum) == True):
        print("Found a subset with given sum")
    else:
        print("No subset with given sum")
 
# This code is contributed by Nikita Tiwari.


Output

Found a subset with given sum

Time Complexity: O(2n)
Auxiliary space: O(n)

Python Program for Subset Sum Problem using Memoization:

As seen in the previous recursion method, each state of the solution can be uniquely identified using two variables – the index and the remaining sum. So create a 2D array to store the value of each state to avoid recalculation of the same state.

Below is the implementation of the above approach:

Python3




# Python program for the above approach
 
# Taking the matrix as globally
tab = [[-1 for i in range(2000)] for j in range(2000)]
 
 
# Check if possible subset with
# given sum is possible or not
def subsetSum(a, n, sum):
 
    # If the sum is zero it means
    # we got our expected sum
    if (sum == 0):
        return 1
 
    if (n <= 0):
        return 0
 
    # If the value is not -1 it means it
    # already call the function
    # with the same value.
    # it will save our from the repetition.
    if (tab[n - 1][sum] != -1):
        return tab[n - 1][sum]
 
    # If the value of a[n-1] is
    # greater than the sum.
    # we call for the next value
    if (a[n - 1] > sum):
        tab[n - 1][sum] = subsetSum(a, n - 1, sum)
        return tab[n - 1][sum]
    else:
 
        # Here we do two calls because we
        # don't know which value is
        # full-fill our criteria
        # that's why we doing two calls
        tab[n - 1][sum] = subsetSum(a, n - 1, sum)
        return tab[n - 1][sum] or subsetSum(a, n - 1, sum - a[n - 1])
 
 
# Driver Code
if __name__ == '__main__':
    n = 5
    a = [1, 5, 3, 7, 4]
    sum = 12
    if (subsetSum(a, n, sum)):
        print("YES")
    else:
        print("NO")
 
# This code is contributed by shivani.


Output

YES


Time Complexity: O(sum*n)
Auxiliary space: O(n)

Python Program for Subset Sum Problem using Dynamic Programming:

We can solve the problem in Pseudo-polynomial time we can use the Dynamic programming approach.

So we will create a 2D array of size (n + 1) * (sum + 1) of type boolean. The state dp[i][j] will be true if there exists a subset of elements from set[0 . . . i] with sum value = ‘j’.

The dynamic programming relation is as follows:

if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]

Below is the implementation of the above approach:

Python3




# A Dynamic Programming solution for subset
# sum problem Returns true if there is a subset of
# set[] with sun equal to given sum
 
 
# Returns true if there is a subset of set[]
# with sum equal to given sum
def isSubsetSum(set, n, sum):
 
    # The value of subset[i][j] will be
    # true if there is a
    # subset of set[0..j-1] with sum equal to i
    subset = ([[False for i in range(sum + 1)]
            for i in range(n + 1)])
 
    # If sum is 0, then answer is true
    for i in range(n + 1):
        subset[i][0] = True
 
    # If sum is not 0 and set is empty,
    # then answer is false
    for i in range(1, sum + 1):
        subset[0][i] = False
 
    # Fill the subset table in bottom up manner
    for i in range(1, n + 1):
        for j in range(1, sum + 1):
            if j < set[i-1]:
                subset[i][j] = subset[i-1][j]
            if j >= set[i-1]:
                subset[i][j] = (subset[i-1][j] or
                                subset[i - 1][j-set[i-1]])
 
    return subset[n][sum]
 
 
# Driver code
if __name__ == '__main__':
    set = [3, 34, 4, 12, 5, 2]
    sum = 9
    n = len(set)
    if (isSubsetSum(set, n, sum) == True):
        print("Found a subset with given sum")
    else:
        print("No subset with given sum")
 
# This code is contributed by
# sahil shelangia.


Output

Found a subset with given sum

Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum*n), as the size of the 2-D array is sum*n.

Python Program for Subset Sum Problem using Dynamic Programming with space optimization to linear:

In previous approach of dynamic programming we have derive the relation between states as given below:
if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]
If we observe that for calculating current dp[i][j] state we only need previous row dp[i-1][j] or dp[i-1][j-set[i-1]].
There is no need to store all the previous states just one previous state is used to compute result.

Step-by-step approach:

  • Define two arrays prev and curr of size Sum+1 to store the just previous row result and current row result respectively.
  • Once curr array is calculated then curr becomes our prev for the next row.
  • When all rows are processed the answer is stored in prev array.

Below is the implementation of the above approach:

Python3




# Returns True if there is a subset of set[]
# with a sum equal to the given sum
def isSubsetSum(nums, n, sum):
    # Create a list to store the previous row result
    prev = [False] * (sum + 1)
 
    # If sum is 0, then the answer is True
    prev[0] = True
 
    # If sum is not 0 and the set is empty,
    # then the answer is False
    for i in range(1, n + 1):
        curr = [False] * (sum + 1)
        for j in range(1, sum + 1):
            if j < nums[i - 1]:
                curr[j] = prev[j]
            if j >= nums[i - 1]:
                curr[j] = prev[j] or prev[j - nums[i - 1]]
        # Now curr becomes prev for (i+1)-th element
        prev = curr
 
    return prev[sum]
 
# Driver code
if __name__ == "__main__":
    nums = [3, 34, 4, 12, 5, 2]
    sum_value = 9
    n = len(nums)
    if isSubsetSum(nums, n, sum_value):
        print("Found a subset with the given sum")
    else:
        print("No subset with the given sum")


Output

Found a subset with given sum

Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum), as the size of the 1-D array is sum+1.

Please refer complete article on Subset Sum Problem | DP-25 for more details!



Last Updated : 10 Nov, 2023
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