Combinations from n arrays picking one element from each array

Given a list of arrays, find all combinations where each combination contains one element from each given array.

Examples:

Input : [ [1, 2], [3, 4] ]
Output : 1 3  
         1 4 
         2 3 
         2 4       

Input : [ [1], [2, 3, 4], [5] ]
Output : 1 2 5  
         1 3 5
         1 4 5           



We keep an array of size equal to total no of arrays. This array called indices helps us keep track of the index of current element in each of the n arrays. Initially it is initialized with all 0s indicating current index in each array is that of first element. We keep printing the combinations until no new combinations can be found. Starting from the rightmost array we check if more elements are there in that array. If yes, we increment the entry for that array in indices i.e. move to the next element in that array. We also make the current indices 0 in all the arrays to the right of this array. We keep moving left to check all arrays until one such array is found. If no more arrays are found we stop there.

C++

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// CPP program to find combinations from n 
// arrays such that one element from each 
// array is present
#include <bits/stdc++.h>
  
using namespace std;
  
// function to print combinations that contain
// one element from each of the given arrays
void print(vector<vector<int> >& arr)
{
    // number of arrays
    int n = arr.size();
  
    // to keep track of next element in each of
    // the n arrays
    int* indices = new int[n];
  
    // initialize with first element's index
    for (int i = 0; i < n; i++)
        indices[i] = 0;
  
    while (1) {
  
        // print current combination
        for (int i = 0; i < n; i++)
            cout << arr[i][indices[i]] << " ";
        cout << endl;
  
        // find the rightmost array that has more
        // elements left after the current element 
        // in that array
        int next = n - 1;
        while (next >= 0 && 
              (indices[next] + 1 >= arr[next].size()))
            next--;
  
        // no such array is found so no more 
        // combinations left
        if (next < 0)
            return;
  
        // if found move to next element in that 
        // array
        indices[next]++;
  
        // for all arrays to the right of this 
        // array current index again points to 
        // first element
        for (int i = next + 1; i < n; i++)
            indices[i] = 0;
    }
}
  
// driver function to test above function
int main()
{
    // initializing a vector with 3 empty vectors
    vector<vector<int> > arr(3, vector<int>(0, 0));
  
    // now entering data
    // [[1, 2, 3], [4], [5, 6]]
    arr[0].push_back(1);
    arr[0].push_back(2);
    arr[0].push_back(3);
    arr[1].push_back(4);
    arr[2].push_back(5);
    arr[2].push_back(6);
  
    print(arr);
}

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Python3

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# Python3 program to find combinations from n 
# arrays such that one element from each 
# array is present
  
# function to prcombinations that contain
# one element from each of the given arrays
def print1(arr):
      
    # number of arrays
    n = len(arr)
  
    # to keep track of next element 
    # in each of the n arrays
    indices = [0 for i in range(n)]
  
    while (1):
  
        # prcurrent combination
        for i in range(n):
            print(arr[i][indices[i]], end = " ")
        print()
  
        # find the rightmost array that has more
        # elements left after the current element
        # in that array
        next = n - 1
        while (next >= 0 and 
              (indices[next] + 1 >= len(arr[next]))):
            next-=1
  
        # no such array is found so no more
        # combinations left
        if (next < 0):
            return
  
        # if found move to next element in that
        # array
        indices[next] += 1
  
        # for all arrays to the right of this
        # array current index again points to
        # first element
        for i in range(next + 1, n):
            indices[i] = 0
  
# Driver Code
  
# initializing a vector with 3 empty vectors
arr = [[] for i in range(3)]
  
# now entering data
# [[1, 2, 3], [4], [5, 6]]
arr[0].append(1)
arr[0].append(2)
arr[0].append(3)
arr[1].append(4)
arr[2].append(5)
arr[2].append(6)
  
print1(arr)
  
# This code is contributed by mohit kumar

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Output:

1 4 5
1 4 6
2 4 5
2 4 6
3 4 5
3 4 6

This article is contributed by aditi sharma 2. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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