Cartesian Product of any number of sets

Given N number of sets. The task is to write a program to perform cartesian product of all the sets in a given order.

Example:

Input:
1st set: 1 2
2nd set: A 
3rd set: x 
4th set: 5 6
Output:
[['1', 'A', 'x', '5'],
 ['1', 'A', 'x', '6'],
 ['2', 'A', 'x', '5'],
 ['2', 'A', 'x', '6']]

Input:
1st set: 1 2
2nd set: A 
3rd set: x y z 
Output:
[['1', 'A', 'x'],
 ['1', 'A', 'y'],
 ['1', 'A', 'z'],
 ['2', 'A', 'x'],
 ['2', 'A', 'y'],
 ['2', 'A', 'z']]

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The approach is to compute the product of set-1 and set-2 at the beginning and then the resultant of set-1 and set-2 will have a product with set-3 and then the resultant of set-1, set-2, set-3 will have a Cartesian product with set-4 and so on till set-n.

Below is the implementation of the above approach.

# Python program for cartesian 
# product of N-sets  
  
# function to find cartesian product of two sets  
def cartesianProduct(set_a, set_b): 
    result =[] 
    for i in range(0, len(set_a)): 
        for j in range(0, len(set_b)): 
  
            # for handling case having cartesian 
            # prodct first time of two sets 
            if type(set_a[i]) != list:          
                set_a[i] = [set_a[i]] 
                  
            # coping all the members 
            # of set_a to temp 
            temp = [num for num in set_a[i]] 
              
            # add member of set_b to  
            # temp to have cartesian product      
            temp.append(set_b[j])              
            result.append(temp)   
              
    return result 
  
# Function to do a cartesian  
# product of N sets  
def Cartesian(list_a, n): 
      
    # result of cartesian product 
    # of all the sets taken two at a time 
    temp = list_a[0] 
      
    # do product of N sets  
    for i in range(1, n): 
        temp = cartesianProduct(temp, list_a[i]) 
          
    print(temp) 
  
# Driver Code 
list_a = [[1, 2],          # set-1 
          ['A'],          # set-2 
          ['x', 'y', 'z']]   # set-3 
            
# number of sets 
n = len(list_a)  
  
# Function is called to perform  
# the cartesian product on list_a of size n  
Cartesian(list_a, n) 

Output:

[[1, 'A', 'x'],
 [1, 'A', 'y'],
 [1, 'A', 'z'], 
 [2, 'A', 'x'], 
 [2, 'A', 'y'], 
 [2, 'A', 'z']]

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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