Combinations Leetcode Problem 77 [Python Solution]

In this blog post, we're going to tackle the Combinations problem from LeetCode.

This problem falls under the category of backtracking and is rated as a medium difficulty problem.

We will provide a detailed Python solution for this problem, including a step-by-step explanation of our approach.

But first, let's understand the problem statement.

Problem Overview

Given two integers, n and k, our task is to return all possible combinations of k numbers chosen from the range [1, n].

It's important to note that we can return the answer in any order.

In other words, we need to generate all possible combinations of size k from the numbers 1 through n.

Here's an example to illustrate the problem:

Example 1:

Input: n = 4, k = 2

Output:

[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]


Explanation: There are 4 choose 2 = 6 total combinations.

It's important to note that combinations are unordered, meaning that [1, 2] and [2, 1] are considered to be the same combination.

Example 2:

Input: n = 1, k = 1

Output:

[[1]]


Explanation: There is only one combination here, as k is equal to 1.

Understanding Constraints

Before we dive into the solution, let's understand the constraints provided in the problem statement:

• 1 <= n <= 20
• 1 <= k <= n

Now that we have a clear understanding of the problem, let's move on to our Python solution.

Combinations LeetCode Problem Solution

We will approach this problem using backtracking.

Backtracking is a common technique used to solve combinatorial problems, where we explore all possible combinations by making decisions and undoing those decisions as needed.

1. Bruteforce Approach

Here's a Python function to solve this problem using backtracking:

from typing import List

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]:
        res = []
        
        def backtrack(start, comb):
            if len(comb) == k:
                res.append(comb.copy())
                return
            for i in range(start, n + 1):
                comb.append(i)
                backtrack(i + 1, comb)
                comb.pop()
        
        backtrack(1, [])
        return res

Let's break down the code step by step:

1. We create a class Solution with a method combine that takes two parameters: n and k.

This method will return a list of lists, representing all combinations.

1. We initialize an empty list res to store our result, which will be a list of all combinations.

2. We define an inner function called backtrack, which is where the magic happens.

This function takes two arguments: start (the current number we are considering) and comb (the current combination being built).

1. In the backtrack function, we have a base case: if the length of comb equals k, it means we have formed a valid combination of size k.

We make a copy of this combination and append it to the res list.

1. If we haven't reached the desired combination size yet, we enter a loop that iterates from start to n.

This loop represents our decision tree.

1. Inside the loop, we make a decision by appending the current number i to the comb list, effectively adding it to our combination.

2. We then recursively call the backtrack function with an updated start value (i + 1) and the current comb.

3. After the recursive call, we undo our decision by popping the last element from the comb list.

This backtracking step is crucial to explore all possible combinations.

1. Finally, we initiate the backtracking process by calling the backtrack function with an initial start of 1 and an empty comb list.

This sets the process in motion.

1. Once the backtracking is complete, we return the res list containing all the valid combinations.

2. Efficient Approach

This solution may appear brute force, but it is indeed the most efficient way to solve the problem since we must generate all possible combinations.

The time complexity is exponential, roughly O(n choose k), which is the number of possible combinations.

Now, let's provide the Python code for this solution:

from typing import List

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        res = []

        def backtrack(start, comb):
            if len(comb) == k:
                res.append(comb.copy())
                return
            for i in range(start, n + 1):
                comb.append(i)
                backtrack(i + 1, comb)
                comb.pop()

        backtrack(1, [])
        return res

Time and Space Complexity

Let's analyze the time and space complexity of our solution.

• Time Complexity: The time complexity of our solution is roughly O(n choose k), where n choose k represents the number of possible combinations.

In the worst case, this is an exponential time complexity, but it's necessary to explore all possible combinations.

• Space Complexity: The space complexity is O(k) because, in the worst case, we will have k elements in our combination.

The recursive stack space used by the backtracking algorithm is also O(k).

Reasoning Behind Our Approach

Our approach relies on the fundamental concept of backtracking, where we systematically explore all possible combinations by making decisions and undoing them when needed.

The key insight here is that we don't need to generate permutations, just combinations, which simplifies the process.

We start with an empty combination and systematically add numbers to it, making sure to explore all valid paths while avoiding duplicates.

The use of the backtrack function helps us maintain the current state of our combination and explore the decision tree effectively.

Related Interview Questions By Company:

• Simplify Path LeetCode
• Maximum Twin Sum Of A Linked List LeetCode
• Maximum Length Of A Concatenated String With Unique Characters

Related Interview Questions By Difficulty:

• Combination Sum LeetCode
• Combinations LeetCode
• Palindrome Partitioning LeetCode

Related Interview Questions By Category:

• Concatenation Of Array LeetCode
• Merge Two Binary Trees LeetCode
• Island Perimeter LeetCode

Conclusion

In this blog post, we've addressed the Combinations problem from LeetCode.

We've provided a Python solution that utilizes backtracking to generate all possible combinations of size k from the range [1, n].

While the time complexity is exponential, this approach is efficient and necessary for exploring all valid combinations.

We hope this explanation and solution were helpful in understanding how to tackle combinatorial problems like this one.

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For the original problem statement and to try out the code on LeetCode, visit the following link: Combinations LeetCode Problem.

Thank you for reading, and happy coding!