**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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LeetCode is for **software engineers who are looking to practice technical questions and advance their skills**. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either *Easy*, *Medium*, or *Hard*.

**LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:**

- Mathematics/Basic Logical Based Questions
- Arrays
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- Greedy Algorithms
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- Database
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- Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

** 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:**

Combinations– LeetCode Solutions

Combinations Solution in C++:

class Solution { public: vector> combine(int n, int k) { vector > ans; dfs(n, k, 1, {}, ans); return ans; } private: void dfs(int n, int k, int s, vector && path, vector >& 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> combine(int n, int k) { List

> ans = new ArrayList<>(); dfs(n, k, 1, new ArrayList<>(), ans); return ans; } private void dfs(int n, int k, int s, List

path, List > 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