# Find all combinations of k-bit numbers with n bits set where 1 <= n <= k in sorted order

Given a number k, find all the possible combinations of k-bit numbers with n-bits set where 1 <= n <= k. The solution should print all numbers with one set bit first, followed by numbers with two bits set,.. up to the numbers whose all k-bits are set. If two numbers have the same number of set bits, then smaller number should come first.

Examples:

```Input: K = 3
Output:
001 010 100
011 101 110
111

Input: K = 4
Output:
0001 0010 0100 1000
0011 0101 0110 1001 1010 1100
0111 1011 1101 1110
1111

Input: K = 5
Output:
00001 00010 00100 01000 10000
00011 00101 00110 01001 01010 01100 10001 10010 10100 11000
00111 01011 01101 01110 10011 10101 10110 11001 11010 11100
01111 10111 11011 11101 11110
11111
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We need to find all the possible combinations of k-bit numbers with n set bits where 1 <= n <= k. If we carefully analyze, we see that problem can further be divided into sub-problems. We can find all combinations of length k with n ones by prefixing 0 to all combinations of length k-1 with n ones and 1 to all combinations of length k-1 with n-1 ones. We can use Dynamic Programming to save solutions of sub-problems.

Below is C++ implementation of above idea –

 `// C++ program find all the possible combinations of  ` `// k-bit numbers with n-bits set where 1 <= n <= k ` `#include ` `#include ` `using` `namespace` `std; ` `// maximum allowed value of K ` `#define K 16 ` ` `  `// DP lookup table ` `vector DP[K][K]; ` ` `  `// Function to find all combinations k-bit numbers with  ` `// n-bits set where 1 <= n <= k ` `void` `findBitCombinations(``int` `k) ` `{ ` `    ``string str = ``""``; ` `     `  `    ``// DP[k] will store all k-bit numbers   ` `    ``// with 0 bits set (All bits are 0's) ` `    ``for` `(``int` `len = 0; len <= k; len++)  ` `    ``{ ` `        ``DP[len].push_back(str); ` `        ``str = str + ``"0"``; ` `    ``} ` `     `  `    ``// fill DP lookup table in bottom-up manner ` `    ``// DP[k][n] will store all k-bit numbers   ` `    ``// with n-bits set ` `    ``for` `(``int` `len = 1; len <= k; len++) ` `    ``{ ` `        ``for` `(``int` `n = 1; n <= len; n++) ` `        ``{ ` `            ``// prefix 0 to all combinations of length len-1  ` `            ``// with n ones ` `            ``for` `(string str : DP[len - 1][n]) ` `                ``DP[len][n].push_back(``"0"` `+ str); ` ` `  `            ``// prefix 1 to all combinations of length len-1  ` `            ``// with n-1 ones ` `            ``for` `(string str : DP[len - 1][n - 1]) ` `                ``DP[len][n].push_back(``"1"` `+ str); ` `        ``} ` `    ``} ` `     `  `    ``// print all k-bit binary strings with ` `    ``// n-bit set ` `    ``for` `(``int` `n = 1; n <= k; n++)  ` `    ``{ ` `        ``for` `(string str : DP[k][n]) ` `            ``cout << str << ``" "``; ` ` `  `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `k = 5; ` `    ``findBitCombinations(k); ` ` `  `    ``return` `0; ` `} `

Output:

```00000
00001 00010 00100 01000 10000
00011 00101 00110 01001 01010 01100 10001 10010 10100 11000
00111 01011 01101 01110 10011 10101 10110 11001 11010 11100
01111 10111 11011 11101 11110
11111
```

## tags:

Dynamic Programming Dynamic Programming