**LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++**

Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

**all Test Cases Passed,**you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the **Combinations** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Combinations– LeetCode Problem

Combinations– LeetCode Problem

**Problem:**

Given two integers `n`

and `k`

, return *all possible combinations of* `k`

*numbers out of the range* `[1, n]`

.

You may return the answer in **any order**.

**Example 1:**

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

**Example 2:**

Input:n = 1, k = 1Output:[[1]]

**Constraints:**

`1 <= n <= 20`

`1 <= k <= n`

Combinations– LeetCode Solutions

Combinations Solution in C++:

class Solution { public: vector<vector<int>> combine(int n, int k) { vector<vector<int>> ans; dfs(n, k, 1, {}, ans); return ans; } private: void dfs(int n, int k, int s, vector<int>&& path, vector<vector<int>>& ans) { if (k == 0) { ans.push_back(path); return; } for (int i = s; i <= n; ++i) { path.push_back(i); dfs(n, k - 1, i + 1, move(path), ans); path.pop_back(); } } };

Combinations in Java:

class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> ans = new ArrayList<>(); dfs(n, k, 1, new ArrayList<>(), ans); return ans; } private void dfs(int n, int k, int s, List<Integer> path, List<List<Integer>> ans) { if (k == 0) { ans.add(new ArrayList<>(path)); return; } for (int i = s; i <= n; ++i) { path.add(i); dfs(n, k - 1, i + 1, path, ans); path.remove(path.size() - 1); } } }

Combinations in Python:

class Solution: def combine(self, n: int, k: int) -> List[List[int]]: ans = [] def dfs(n: int, k: int, s: int, path: List[int]) -> None: if k == 0: ans.append(path) return for i in range(s, n + 1): dfs(n, k - 1, i + 1, path + [i]) dfs(n, k, 1, []) return ans

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