Python Program to get K length groups with given summation

Last Updated : 21 Apr, 2023

Given a list, our task is to write a Python program to extract all K length sublists with lead to given summation.

Input : test_list = [6, 3, 12, 7, 4, 11], N = 21, K = 4

Output : [(6, 6, 6, 3), (6, 6, 3, 6), (6, 3, 6, 6), (6, 7, 4, 4), (6, 4, 7, 4), (6, 4, 4, 7), (3, 6, 6, 6), (3, 3, 3, 12), (3, 3, 12, 3), (3, 3, 4, 11), (3, 3, 11, 4), (3, 12, 3, 3), (3, 7, 7, 4), (3, 7, 4, 7), (3, 4, 3, 11), (3, 4, 7, 7), (3, 4, 11, 3), (3, 11, 3, 4), (3, 11, 4, 3), (12, 3, 3, 3), (7, 6, 4, 4), (7, 3, 7, 4), (7, 3, 4, 7), (7, 7, 3, 4), (7, 7, 4, 3), (7, 4, 6, 4), (7, 4, 3, 7), (7, 4, 7, 3), (7, 4, 4, 6), (4, 6, 7, 4), (4, 6, 4, 7), (4, 3, 3, 11), (4, 3, 7, 7), (4, 3, 11, 3), (4, 7, 6, 4), (4, 7, 3, 7), (4, 7, 7, 3), (4, 7, 4, 6), (4, 4, 6, 7), (4, 4, 7, 6), (4, 11, 3, 3), (11, 3, 3, 4), (11, 3, 4, 3), (11, 4, 3, 3)]

Explanation : All groups of length 4 and sum 21 are printed.

Input : test_list = [6, 3, 12, 7, 4, 11], N = 210, K = 4

Output : []

Explanation : Since no 4 size group sum equals 210, no group is printed as result.

Method : Using sum() + product() + loop

In this all possible sublists of length K are computed using product(), sum() is used to compare the sublist's sum with the required summation.

Python3

# Python3 code to demonstrate working of # K length groups with given summation # Using sum + product() from itertools import product  # initializing list test_list = [6, 3, 12, 7, 4, 11]               # printing original list print("The original list is : " + str(test_list))  # initializing Summation  N = 21  # initializing size  K = 4  # Looping for each product and comparing with required summation res = [] for sub in product(test_list, repeat = K):   if sum(sub) == N:     res.append(sub)          # printing result print("The sublists with of given size and sum : " + str(res))

Output:

The original list is : [6, 3, 12, 7, 4, 11]

The sublists with of given size and sum : [(6, 6, 6, 3), (6, 6, 3, 6), (6, 3, 6, 6), (6, 7, 4, 4), (6, 4, 7, 4), (6, 4, 4, 7), (3, 6, 6, 6), (3, 3, 3, 12), (3, 3, 12, 3), (3, 3, 4, 11), (3, 3, 11, 4), (3, 12, 3, 3), (3, 7, 7, 4), (3, 7, 4, 7), (3, 4, 3, 11), (3, 4, 7, 7), (3, 4, 11, 3), (3, 11, 3, 4), (3, 11, 4, 3), (12, 3, 3, 3), (7, 6, 4, 4), (7, 3, 7, 4), (7, 3, 4, 7), (7, 7, 3, 4), (7, 7, 4, 3), (7, 4, 6, 4), (7, 4, 3, 7), (7, 4, 7, 3), (7, 4, 4, 6), (4, 6, 7, 4), (4, 6, 4, 7), (4, 3, 3, 11), (4, 3, 7, 7), (4, 3, 11, 3), (4, 7, 6, 4), (4, 7, 3, 7), (4, 7, 7, 3), (4, 7, 4, 6), (4, 4, 6, 7), (4, 4, 7, 6), (4, 11, 3, 3), (11, 3, 3, 4), (11, 3, 4, 3), (11, 4, 3, 3)]

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

Approach#2: Using recursion: The approach uses backtracking to generate all possible groups of length K with sum N. It starts with an empty group and recursively tries to add numbers from the given list to form a group. If the group becomes of length K and its sum is N, it is added to the list of groups.

Define a backtrack function that takes curr_group, remaining_sum, and remaining_len as parameters.
Check if remaining_sum is equal to 0 and remaining_len is equal to 0. If yes, return the current group as a list of list.
Check if remaining_sum is less than 0 or remaining_len is less than 0. If yes, return an empty list.
Initialize an empty list of groups to store all possible groups.
Loop through the given list test_list using the enumerate() function.
Add the current number to the current group to form a new group and reduce the remaining_sum and remaining_len accordingly.
Recursively call the backtrack function with the new group, new_sum, new_len, and remaining_list.
Append the returned groups to the groups list.
Return the final groups list.

Python3

# Python program for the above approach  # Function to get the K length groups def get_k_length_groups(test_list, N, K):     # Function to recursively find the states     def backtrack(curr_group, remaining_sum, remaining_len):         if remaining_sum == 0 and remaining_len == 0:             return [curr_group]         if remaining_sum < 0 or remaining_len < 0:             return []         groups = []          # Update the group and call recursively         for i, num in enumerate(test_list):             new_group = curr_group + [num]             new_sum = remaining_sum - num             new_len = remaining_len - 1             remaining_list = test_list[i:]             groups += backtrack(new_group, new_sum, new_len)          # Return the groups         return groups      # Backtrack the current state     groups = backtrack([], N, K)      # Return the resultant groups     return groups   # Driver Code test_list = [6, 3, 12, 7, 4, 11] N = 21 K = 4 print(get_k_length_groups(test_list, N, K))

Output

[[6, 6, 6, 3], [6, 6, 3, 6], [6, 3, 6, 6], [6, 7, 4, 4], [6, 4, 7, 4], [6, 4, 4, 7], [3, 6, 6, 6], [3, 3, 3, 12], [3, 3, 12, 3], [3, 3, 4, 11], [3, 3, 11, 4], [3, 12, 3, 3], [3, 7, 7, 4], [3, 7, 4, 7], [3, 4, 3, 11], [3, 4, 7, 7], [3, 4, 11, 3], [3, 11, 3, 4], [3, 11, 4, 3], [12, 3, 3, 3], [7, 6, 4, 4], [7, 3, 7, 4], [7, 3, 4, 7], [7, 7, 3, 4], [7, 7, 4, 3], [7, 4, 6, 4], [7, 4, 3, 7], [7, 4, 7, 3], [7, 4, 4, 6], [4, 6, 7, 4], [4, 6, 4, 7], [4, 3, 3, 11], [4, 3, 7, 7], [4, 3, 11, 3], [4, 7, 6, 4], [4, 7, 3, 7], [4, 7, 7, 3], [4, 7, 4, 6], [4, 4, 6, 7], [4, 4, 7, 6], [4, 11, 3, 3], [11, 3, 3, 4], [11, 3, 4, 3], [11, 4, 3, 3]]

Time Complexity: O(N^K), because we are generating all possible groups of length K with sum N, and each number in the list, can either be included or excluded in a group.

Space Complexity: O(K) because we are using the curr_group list to store the current group, and its length is at most K. Additionally, the recursion depth can go up to K, so the space complexity is O(K).

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Python Program to get K length groups with given summation

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