Skip to content
Related Articles

Related Articles

Improve Article

Combinations from n arrays picking one element from each array

  • Difficulty Level : Medium
  • Last Updated : 18 May, 2021

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 the total no of arrays. This array called indices helps us keep track of the index of the current element in each of the n arrays. Initially, it is initialized with all 0s indicating the current index in each array is that of the 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. moves 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++




// C++ 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);
}

Java




// Java program to find combinations from n
// arrays such that one element from each
// array is present
import java.util.*;
 
class GFG{
 
// Function to print combinations that contain
// one element from each of the given arrays
static void print(Vector<Integer> []arr)
{
     
    // Number of arrays
    int n = arr.length;
 
    // 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 (true)
    {
 
        // Print current combination
        for(int i = 0; i < n; i++)
            System.out.print(
                arr[i].get(indices[i]) + " ");
                 
        System.out.println();
 
        // 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 code
public static void main(String[] args)
{
     
    // Initializing a vector with 3 empty vectors
    @SuppressWarnings("unchecked")
    Vector<Integer> []arr = new Vector[3];
    for(int i = 0; i < arr.length; i++)
        arr[i] = new Vector<Integer>();
         
    // Now entering data
    // [[1, 2, 3], [4], [5, 6]]
    arr[0].add(1);
    arr[0].add(2);
    arr[0].add(3);
    arr[1].add(4);
    arr[2].add(5);
    arr[2].add(6);
 
    print(arr);
}
}
 
// This code is contributed by amal kumar choubey

Python3




# 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

C#




// C# program to find
// combinations from n
// arrays such that one
// element from each
// array is present
using System;
using System.Collections.Generic;
class GFG{
 
// Function to print combinations
// that contain one element from
// each of the given arrays
static void print(List<int> []arr)
{
  // Number of arrays
  int n = arr.Length;
 
  // 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 (true)
  {
    // Print current combination
    for(int i = 0; i < n; i++)
      Console.Write(arr[i][indices[i]] + " ");
     
    Console.WriteLine();
 
    // 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].Count))
      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 code
public static void Main(String[] args)
{
  // Initializing a vector
  // with 3 empty vectors
  List<int> []arr = new List<int>[3];
  for(int i = 0; i < arr.Length; i++)
    arr[i] = new List<int>();
 
  // Now entering data
  // [[1, 2, 3], [4], [5, 6]]
  arr[0].Add(1);
  arr[0].Add(2);
  arr[0].Add(3);
  arr[1].Add(4);
  arr[2].Add(5);
  arr[2].Add(6);
 
  print(arr);
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// Javascript program to find combinations from n
// arrays such that one element from each
// array is present
 
// Function to print combinations that contain
// one element from each of the given arrays
function print(arr)
{
     
    // Number of arrays
    let n = arr.length;
     
    // To keep track of next element in
    // each of the n arrays
    let indices = new Array(n);
     
    // Initialize with first element's index
    for(let i = 0; i < n; i++)
        indices[i] = 0;
     
    while (true)
    {
     
        // Print current combination
        for(let i = 0; i < n; i++)
            document.write(
            arr[i][indices[i]] + " ");
          
        document.write("<br>");
         
        // Find the rightmost array that has more
        // elements left after the current element
        // in that array
        let next = n - 1;
        while (next >= 0 && (indices[next] + 1 >=
                                 arr[next].length))
            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(let i = next + 1; i < n; i++)
            indices[i] = 0;
    }
}
 
// Driver code
 
// Initializing a vector with 3 empty vectors
let arr = new Array(3);
for(let i = 0; i < arr.length; i++)
    arr[i] = [];
 
// Now entering data
// [[1, 2, 3], [4], [5, 6]]
arr[0].push(1);
arr[0].push(2);
arr[0].push(3);
arr[1].push(4);
arr[2].push(5);
arr[2].push(6);
 
print(arr);
     
// This code is contributed by unknown2108
 
</script>

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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :