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++ 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 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 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 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 ):
# 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 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# 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 |
<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> |
1 4 5 1 4 6 2 4 5 2 4 6 3 4 5 3 4 6