Combinations from n arrays picking one element from each array

• Difficulty Level : Medium
• Last Updated : 20 Jul, 2022

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.

Implementation:

C++

 // C++ program to find combinations from n// arrays such that one element from each// array is present#include  using namespace std; // function to print combinations that contain// one element from each of the given arraysvoid print(vector >& 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 functionint main(){    // initializing a vector with 3 empty vectors    vector > arr(3, vector(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 presentimport java.util.*; class GFG{ // Function to print combinations that contain// one element from each of the given arraysstatic void print(Vector []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 codepublic static void main(String[] args){         // Initializing a vector with 3 empty vectors    @SuppressWarnings("unchecked")    Vector []arr = new Vector[3];    for(int i = 0; i < arr.length; i++)        arr[i] = new Vector();             // 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 print combinations that contain# one element from each of the given arraysdef 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):         # print current 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 vectorsarr = [[] 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 presentusing System;using System.Collections.Generic;class GFG{ // Function to print combinations// that contain one element from// each of the given arraysstatic void print(List []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 codepublic static void Main(String[] args){  // Initializing a vector  // with 3 empty vectors  List []arr = new List[3];  for(int i = 0; i < arr.Length; i++)    arr[i] = new List();   // 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



Output

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