Given two numbers n and k and you have to find all possible combination of k numbers from 1…n.
Examples:
Input : n = 4
k = 2
Output : 1 2
1 3
1 4
2 3
2 4
3 4
Input : n = 5
k = 3
Output : 1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
We have discussed one approach in the below post.
Print all possible combinations of r elements in a given array of size n
In this, we use DFS based approach. We want all numbers from 1 to n. We first push all numbers from 1 to k in tmp_vector and as soon as k is equal to 0, we push all numbers from tmp_vector to ans_vector. After this, we remove the last element from tmp_vector and make all remaining combination.
C++
#include <bits/stdc++.h>
using namespace std;
void makeCombiUtil(vector<vector< int > >& ans,
vector< int >& tmp, int n, int left, int k)
{
if (k == 0) {
ans.push_back(tmp);
return ;
}
for ( int i = left; i <= n; ++i)
{
tmp.push_back(i);
makeCombiUtil(ans, tmp, n, i + 1, k - 1);
tmp.pop_back();
}
}
vector<vector< int > > makeCombi( int n, int k)
{
vector<vector< int > > ans;
vector< int > tmp;
makeCombiUtil(ans, tmp, n, 1, k);
return ans;
}
int main()
{
int n = 5;
int k = 3;
vector<vector< int > > ans = makeCombi(n, k);
for ( int i = 0; i < ans.size(); i++) {
for ( int j = 0; j < ans[i].size(); j++) {
cout << ans.at(i).at(j) << " " ;
}
cout << endl;
}
return 0;
}
|
Java
import java.util.*;
public class Main
{
static Vector<Vector<Integer>> ans = new Vector<Vector<Integer>>();
static Vector<Integer> tmp = new Vector<Integer>();
static void makeCombiUtil( int n, int left, int k)
{
if (k == 0 ) {
ans.add(tmp);
for ( int i = 0 ; i < tmp.size(); i++)
{
System.out.print(tmp.get(i) + " " );
}
System.out.println();
return ;
}
for ( int i = left; i <= n; ++i)
{
tmp.add(i);
makeCombiUtil(n, i + 1 , k - 1 );
tmp.remove(tmp.size() - 1 );
}
}
static Vector<Vector<Integer>> makeCombi( int n, int k)
{
makeCombiUtil(n, 1 , k);
return ans;
}
public static void main(String[] args)
{
int n = 5 ;
int k = 3 ;
ans = makeCombi(n, k);
}
}
|
Python3
ans = []
tmp = []
def makeCombiUtil(n, left, k):
if (k = = 0 ):
ans.append(tmp)
for i in range ( len (tmp)):
print (tmp[i], end = " " )
print ()
return
for i in range (left, n + 1 ):
tmp.append(i)
makeCombiUtil(n, i + 1 , k - 1 )
tmp.pop()
def makeCombi(n, k):
makeCombiUtil(n, 1 , k)
return ans
n = 5
k = 3
ans = makeCombi(n, k)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static List<List< int >> ans = new List<List< int >>();
static List< int > tmp = new List< int >();
static void makeCombiUtil( int n, int left, int k)
{
if (k == 0) {
ans.Add(tmp);
for ( int i = 0; i < tmp.Count; i++)
{
Console.Write(tmp[i] + " " );
}
Console.WriteLine();
return ;
}
for ( int i = left; i <= n; ++i)
{
tmp.Add(i);
makeCombiUtil(n, i + 1, k - 1);
tmp.RemoveAt(tmp.Count - 1);
}
}
static List<List< int >> makeCombi( int n, int k)
{
makeCombiUtil(n, 1, k);
return ans;
}
static void Main()
{
int n = 5;
int k = 3;
ans = makeCombi(n, k);
}
}
|
Javascript
<script>
let ans = [];
let tmp = [];
function makeCombiUtil(n, left, k)
{
if (k == 0) {
ans.push(tmp);
for (let i = 0; i < tmp.length; i++)
{
document.write(tmp[i] + " " );
}
document.write( "</br>" );
return ;
}
for (let i = left; i <= n; ++i)
{
tmp.push(i);
makeCombiUtil(n, i + 1, k - 1);
tmp.pop();
}
}
function makeCombi(n, k)
{
makeCombiUtil(n, 1, k);
return ans;
}
let n = 5;
let k = 3;
ans = makeCombi(n, k);
</script>
|
Output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
Time Complexity: O((nCk)*k), where nCk is all possible subsets and k to copy subsets into ans vector.
Space Complexity: O((nCk)*k), to store all n C k subset in the ans vector of size k.
This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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