Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

Example:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

Combination Algorithm using Bitmasking

The combination can also be done in Recursive backtracking. However, it can also be implemented using the Bitmasking algorithm.

The idea is to bruteforce all possible configurations (of bitmasks) in O(2^N) where N is the length of the given input set. Then once the configuration has k bit sets, we output the corresponding configuration.

The following is the Python combination implementation using bitmasking.

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class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        for b in (range(1 << n)):
            if bin(b).count('1') == k:
                cur = []
                for i in range(n):
                    if (b & (1 << i)) > 0:
                        cur.append(i + 1)
                ans.append(cur)
        return ans

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        for b in (range(1 << n)):
            if bin(b).count('1') == k:
                cur = []
                for i in range(n):
                    if (b & (1 << i)) > 0:
                        cur.append(i + 1)
                ans.append(cur)
        return ans

and with slight changes – reversing the bit searching – still works

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class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        for b in reversed(range(1 << n)):
            if bin(b).count('1') == k:
                cur = []
                for i in range(n):
                    if (b & (1 << (n - i - 1))) > 0:
                        cur.append(i + 1)
                ans.append(cur)
        return ans

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        for b in reversed(range(1 << n)):
            if bin(b).count('1') == k:
                cur = []
                for i in range(n):
                    if (b & (1 << (n - i - 1))) > 0:
                        cur.append(i + 1)
                ans.append(cur)
        return ans

The recursive algorithm in C++: Recursive Combination Algorithm Implementation in C++.

Also, another interesting read: combination

–EOF (The Ultimate Computing & Technology Blog) —