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# Print all subsets of given size of a set

Generate all possible subsets of size r of the given array with distinct elements.

Examples:

Input  : arr[] = {1, 2, 3, 4}
r = 2
Output : 1 2
1 3
1 4
2 3
2 4
3 4
Input : arr[] = {10, 20, 30, 40, 50}
r = 3
Output : 10 20 30
10 20 40
10 20 50
10 30 40
10 30 50
10 40 50
20 30 40
20 30 50
20 40 50
30 40 50

This problem is the same Print all possible combinations of r elements in a given array of size n.
The idea here is similar to Subset Sum Problem. We, one by one, consider every element of the input array, and recur for two cases:

1. The element is included in the current combination (We put the element in data[] and increase the next available index in data[])
2. The element is excluded in the current combination (We do not put the element in and do not change the index)

When the number of elements in data[] becomes equal to r (size of a combination), we print it.

This method is mainly based on Pascal’s Identity, i.e. ncr = n-1cr + n-1cr-1

Implementation:

## C++

 // C++ Program to print all combination of size r in// an array of size n#include using namespace std; void combinationUtil(int arr[], int n, int r,                     int index, int data[], int i);  // The main function that prints all combinations of// size r in arr[] of size n. This function mainly// uses combinationUtil()void printCombination(int arr[], int n, int r){       // A temporary array to store all combination    // one by one    int data[r];      // Print all combination using temporary array 'data[]'    combinationUtil(arr, n, r, 0, data, 0);}  /* arr[]  ---> Input Array   n      ---> Size of input array   r      ---> Size of a combination to be printed   index  ---> Current index in data[]   data[] ---> Temporary array to store current combination   i      ---> index of current element in arr[]     */void combinationUtil(int arr[], int n, int r, int index,                     int data[], int i){    // Current combination is ready, print it    if (index == r) {        for (int j = 0; j < r; j++)            cout <<" "<< data[j];        cout <<"\n";        return;    }      // When no more elements are there to put in data[]    if (i >= n)        return;      // current is included, put next at next location    data[index] = arr[i];    combinationUtil(arr, n, r, index + 1, data, i + 1);      // current is excluded, replace it with next    // (Note that i+1 is passed, but index is not    // changed)    combinationUtil(arr, n, r, index, data, i + 1);}  // Driver program to test above functionsint main(){    int arr[] = { 10, 20, 30, 40, 50 };    int r = 3;    int n = sizeof(arr) / sizeof(arr[0]);    printCombination(arr, n, r);    return 0;} // This code is contributed by shivanisinghss2110

## C

 // C++ Program to print all combination of size r in// an array of size n#include void combinationUtil(int arr[], int n, int r,                     int index, int data[], int i); // The main function that prints all combinations of// size r in arr[] of size n. This function mainly// uses combinationUtil()void printCombination(int arr[], int n, int r){    // A temporary array to store all combination    // one by one    int data[r];     // Print all combination using temporary array 'data[]'    combinationUtil(arr, n, r, 0, data, 0);} /* arr[]  ---> Input Array   n      ---> Size of input array   r      ---> Size of a combination to be printed   index  ---> Current index in data[]   data[] ---> Temporary array to store current combination   i      ---> index of current element in arr[]     */void combinationUtil(int arr[], int n, int r, int index,                     int data[], int i){    // Current combination is ready, print it    if (index == r) {        for (int j = 0; j < r; j++)            printf("%d ", data[j]);        printf("\n");        return;    }     // When no more elements are there to put in data[]    if (i >= n)        return;     // current is included, put next at next location    data[index] = arr[i];    combinationUtil(arr, n, r, index + 1, data, i + 1);     // current is excluded, replace it with next    // (Note that i+1 is passed, but index is not    // changed)    combinationUtil(arr, n, r, index, data, i + 1);} // Driver program to test above functionsint main(){    int arr[] = { 10, 20, 30, 40, 50 };    int r = 3;    int n = sizeof(arr) / sizeof(arr[0]);    printCombination(arr, n, r);    return 0;}

## Java

 // Java program to print all combination of size// r in an array of size nimport java.io.*; class Permutation {     /* arr[]  ---> Input Array    data[] ---> Temporary array to store current combination    start & end ---> Starting and Ending indexes in arr[]    index  ---> Current index in data[]    r ---> Size of a combination to be printed */    static void combinationUtil(int arr[], int n, int r,                          int index, int data[], int i)    {        // Current combination is ready to be printed,        // print it        if (index == r) {            for (int j = 0; j < r; j++)                System.out.print(data[j] + " ");            System.out.println("");            return;        }         // When no more elements are there to put in data[]        if (i >= n)            return;         // current is included, put next at next        // location        data[index] = arr[i];        combinationUtil(arr, n, r, index + 1,                               data, i + 1);         // current is excluded, replace it with        // next (Note that i+1 is passed, but        // index is not changed)        combinationUtil(arr, n, r, index, data, i + 1);    }     // The main function that prints all combinations    // of size r in arr[] of size n. This function    // mainly uses combinationUtil()    static void printCombination(int arr[], int n, int r)    {        // A temporary array to store all combination        // one by one        int data[] = new int[r];         // Print all combination using temporary        // array 'data[]'        combinationUtil(arr, n, r, 0, data, 0);    }     /* Driver function to check for above function */    public static void main(String[] args)    {        int arr[] = { 10, 20, 30, 40, 50 };        int r = 3;        int n = arr.length;        printCombination(arr, n, r);    }}/* This code is contributed by Devesh Agrawal */

## Python3

 # Python3 program to print all# subset combination of n# element in given set of r element . # arr[] ---> Input Array# data[] ---> Temporary array to#             store current combination# start & end ---> Starting and Ending#                  indexes in arr[]# index ---> Current index in data[]# r ---> Size of a combination#        to be printeddef combinationUtil(arr, n, r,                    index, data, i):    # Current combination is    # ready to be printed,    # print it    if(index == r):        for j in range(r):            print(data[j], end = " ")        print(" ")        return     # When no more elements    # are there to put in data[]    if(i >= n):        return     # current is included,    # put next at next    # location    data[index] = arr[i]    combinationUtil(arr, n, r,                    index + 1, data, i + 1)         # current is excluded,    # replace it with    # next (Note that i+1    # is passed, but index    # is not changed)    combinationUtil(arr, n, r, index,                    data, i + 1)  # The main function that# prints all combinations# of size r in arr[] of# size n. This function# mainly uses combinationUtil()def printcombination(arr, n, r):     # A temporary array to    # store all combination    # one by one    data = list(range(r))         # Print all combination    # using temporary    # array 'data[]'    combinationUtil(arr, n, r,                    0, data, 0)  # Driver Codearr = [10, 20, 30, 40, 50] r = 3n = len(arr)printcombination(arr, n, r) # This code is contributed# by Ambuj sahu

## C#

 // C# program to print all combination// of size r in an array of size nusing System; class GFG {     /* arr[] ---> Input Array    data[] ---> Temporary array to store    current combination start & end --->    Starting and Ending indexes in arr[]    index ---> Current index in data[]    r ---> Size of a combination to be    printed */    static void combinationUtil(int []arr,                  int n, int r, int index,                          int []data, int i)    {                 // Current combination is ready to        // be printed, print it        if (index == r)        {            for (int j = 0; j < r; j++)                Console.Write(data[j] + " ");                             Console.WriteLine("");                         return;        }         // When no more elements are there        // to put in data[]        if (i >= n)            return;         // current is included, put next        // at next location        data[index] = arr[i];        combinationUtil(arr, n, r, index + 1,                                data, i + 1);         // current is excluded, replace        // it with next (Note that i+1        // is passed, but index is not        // changed)        combinationUtil(arr, n, r, index,                                data, i + 1);    }     // The main function that prints all    // combinations of size r in arr[] of    // size n. This function mainly uses    // combinationUtil()    static void printCombination(int []arr,                                int n, int r)    {                 // A temporary array to store all        // combination one by one        int []data = new int[r];         // Print all combination using        // temporary array 'data[]'        combinationUtil(arr, n, r, 0, data, 0);    }     /* Driver function to check for    above function */    public static void Main()    {        int []arr = { 10, 20, 30, 40, 50 };        int r = 3;        int n = arr.Length;                 printCombination(arr, n, r);    }} // This code is contributed by vt_m.



## PHP

 Input Arrayn ---> Size of input arrayr ---> Size of a combination to be printedindex ---> Current index in data[]data[] ---> Temporary array to storecurrent combinationi ---> index of current element in arr[] */function combinationUtil( \$arr, \$n, \$r, \$index,                    \$data, \$i){    // Current combination is ready, print it    if (\$index == \$r) {        for ( \$j = 0; \$j < \$r; \$j++)            echo \$data[\$j]," ";        echo "\n";        return;    }     // When no more elements are there to    // put in data[]    if (\$i >= \$n)        return;     // current is included, put next at    // next location    \$data[\$index] = \$arr[\$i];    combinationUtil(\$arr, \$n, \$r, \$index + 1,                              \$data, \$i + 1);     // current is excluded, replace it with    // next (Note that i+1 is passed, but    // index is not changed)    combinationUtil(\$arr, \$n, \$r, \$index,                            \$data, \$i + 1);} // Driver program to test above functions    \$arr = array( 10, 20, 30, 40, 50 );    \$r = 3;    \$n = count(\$arr);    printCombination(\$arr, \$n, \$r); // This code is contributed by anuj_67.?>

Output

10 20 30
10 20 40
10 20 50
10 30 40
10 30 50
10 40 50
20 30 40
20 30 50
20 40 50
30 40 50

Time complexity of this algorithm is O(n*r). The outer loop runs n times and the inner loop runs r times.
Auxiliary Space: O(r), the space complexity is O(r) because we are creating a temporary array of size r and storing the combinations
in it.

Approach 2: Using DP

The given program generates combinations of size r from an array of size n using a recursive approach. It does not use dynamic programming (DP) explicitly. However, dynamic programming can be applied to optimize the solution by avoiding redundant computations.

To implement a DP approach, we can use a 2D table to store the intermediate results and avoid recomputing the same combinations. Here’s an updated version of the program that incorporates dynamic programming:

## C++

 #include #include using namespace std; // Function to print all combinations of size r// using a dynamic programming approachvoid printCombination(int arr[], int n, int r){    vector> dp(n + 1, vector(r + 1, 0));     // Calculate the combinations using dynamic programming    for (int i = 0; i <= n; i++) {        for (int j = 0; j <= min(i, r); j++) {            if (j == 0 || j == i)                dp[i][j] = 1;            else                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];        }    }     // Print the combinations    for (int i = 0; i < dp[n].size(); i++) {        if (dp[n][i] == 0)            break;        for (int j = 0; j < dp[n][i]; j++) {            cout << arr[i] << " ";        }        cout << endl;    }} // Driver program to test above functionsint main(){    int arr[] = { 10, 20, 30, 40, 50 };    int r = 3;    int n = sizeof(arr) / sizeof(arr[0]);    printCombination(arr, n, r);    return 0;}

## Java

 import java.util.ArrayList;import java.util.List; public class GFG {    // Function to print all combinations of size r    // using a dynamic programming approach    public static void printCombination(int[] arr, int n, int r) {        List> dp = new ArrayList<>(n + 1);        for (int i = 0; i <= n; i++) {            dp.add(new ArrayList<>(r + 1));            for (int j = 0; j <= r; j++) {                dp.get(i).add(0);            }        }         // Calculate the combinations using dynamic programming        for (int i = 0; i <= n; i++) {            for (int j = 0; j <= Math.min(i, r); j++) {                if (j == 0 || j == i) {                    dp.get(i).set(j, 1);                } else {                    int val1 = dp.get(i - 1).get(j - 1);                    int val2 = dp.get(i - 1).get(j);                    dp.get(i).set(j, val1 + val2);                }            }        }         // Print the combinations        for (int i = 0; i < dp.get(n).size(); i++) {            int count = dp.get(n).get(i);            if (count == 0) {                break;            }            for (int j = 0; j < count; j++) {                System.out.print(arr[i] + " ");            }            System.out.println();        }    }     // Driver program to test above functions    public static void main(String[] args) {        int[] arr = { 10, 20, 30, 40, 50 };        int r = 3;        int n = arr.length;        printCombination(arr, n, r);    }}

## Python3

 def print_combination(arr, n, r):    # Create a 2D list to store combinations    dp = [[0] * (r + 1) for _ in range(n + 1)]     # Calculate the combinations using dynamic programming    for i in range(n + 1):        for j in range(min(i, r) + 1):            if j == 0 or j == i:                dp[i][j] = 1            else:                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]     # Print the combinations    for i in range(len(dp[n])):        if dp[n][i] == 0:            break        for j in range(dp[n][i]):            print(arr[i], end=" ")  # Print the element 'arr[i]' 'dp[n][i]' times        print()  # Move to the next line for the next element # Driver program to test the functionif __name__ == "__main__":    arr = [10, 20, 30, 40, 50]    r = 3  # Size of combinations    n = len(arr)    print_combination(arr, n, r)     # This code is contributed by shivamgupta0987654321

Output

10
20 20 20 20 20
30 30 30 30 30 30 30 30 30 30
40 40 40 40 40 40 40 40 40 40

Time complexity of this algorithm is O(n*r).The outer loop runs n times and the inner loop runs r times.

Auxiliary Space: O(n*r).

Follow the below steps to solve the above problem:

1) Generate all possible binary numbers with a length equal to the number of elements in the set. Each binary number will represent a potential subset.
2) Iterate through each binary number from 0 to 2^N – 1, where N is the number of elements in the set. This represents all possible subsets of the set.
3) For each binary number, check the number of set bits(1s). If the count of set bits is equal to the desired subset size, consider it as a valid subset.
4) To extract the elements of the subset, iterate through the bits of the binary number. If a bit is set (1), include the corresponding element from the set in the subset.
5) Print the desired output.

Below is the implementation of above approach:

## C++

 #include using namespace std; void printSubset(int arr[], int subset[], int r) {    for (int i = 0; i < r; i++)        cout << subset[i] << " ";    cout << endl;} void printCombination(int arr[], int n, int r) {    int totalSubsets = 1 << n;  // Total number of subsets is 2^n    for (int bitmask = 0; bitmask < totalSubsets; bitmask++) {        // Count the number of set bits in the bitmask        int count = 0;        int temp = bitmask;        while (temp > 0) {            count += temp & 1;            temp >>= 1;        }         if (count == r) {            int subset[r];            int index = 0;            for (int i = 0; i < n; i++) {                if (bitmask & (1 << i))                    subset[index++] = arr[i];            }            printSubset(arr, subset, r);        }    }} int main() {    int arr[] = {10, 20, 30, 40, 50};    int r = 3;    int n = sizeof(arr) / sizeof(arr[0]);     printCombination(arr, n, r);     return 0;}// This code is contributed by Yash Agarwal(yashagarwal2852002)

## C#

 using System; class GFG{    static void PrintSubset(int[] arr, int[] subset, int r)    {        for (int i = 0; i < r; i++)            Console.Write(subset[i] + " ");        Console.WriteLine();    }     static void PrintCombination(int[] arr, int n, int r)    {        int totalSubsets = 1 << n; // Total number of subsets is 2^n        for (int bitmask = 0; bitmask < totalSubsets; bitmask++)        {            // Count the number of set bits in the bitmask            int count = 0;            int temp = bitmask;            while (temp > 0)            {                count += temp & 1;                temp >>= 1;            }             if (count == r)            {                int[] subset = new int[r];                int index = 0;                for (int i = 0; i < n; i++)                {                    if ((bitmask & (1 << i)) > 0)                        subset[index++] = arr[i];                }                PrintSubset(arr, subset, r);            }        }    }     static void Main()    {        int[] arr = { 10, 20, 30, 40, 50 };        int r = 3;        int n = arr.Length;         PrintCombination(arr, n, r);     }}

Output

10 20 30
10 20 40
10 30 40
20 30 40
10 20 50
10 30 50
20 30 50
10 40 50
20 40 50
30 40 50

Time Complexity: O(2^n), where n is the number of elements in the given array.

Auxiliary Space: O(n+r)

Refer to the post below for more solutions and ideas to handle duplicates in the input array.
Print all possible combinations of r elements in a given array of size n.

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