What is the time complexity of Combinations?

The optimal solution has a time complexity of O(C(n, k) * k).

What data structures are used to solve Combinations?

This problem uses Backtracking. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Combinations?

Insight 1: Backtracking is well-suited for generating combinations due to its ability to explore all possible choices and prune unnecessary branches. Insight 2: Maintaining a temporary 'path' to store the current combination being built, and then adding a copy of this path to the result once it reaches the desired size (k) is a key technique.

Which companies ask LeetCode 77. Combinations in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Google and 2 more companies.

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Combinations - LeetCode 77 Solution

Combinations - Complete Solution Guide

Combinations is LeetCode problem 77, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two integers n and k , return all possible combinations of k numbers chosen from the range [1, n] . You may return the answer in any order . 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. Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination. Example 2: Input: n = 1, k = 1 Output: [[1]] Explanation: There is 1 choose 1 = 1 total combination. Constraints: 1

Detailed Explanation

The problem requires generating all possible combinations of 'k' numbers chosen from a range of numbers from 1 to 'n'. A combination is a selection of items where the order does not matter. For example, if n=4 and k=2, we need to find all groups of 2 numbers selected from the set {1, 2, 3, 4}. The output should be a list of lists, where each inner list represents a unique combination. The constraints limit n to be between 1 and 20, and k to be between 1 and n.

Solution Approach

The solution utilizes a backtracking algorithm to explore the possible combinations. The algorithm recursively builds combinations by adding numbers from the range [1, n] to a 'path'. When the 'path' reaches a length of 'k', it's added to the 'result' list. Backtracking is achieved by adding a number to the 'path', making a recursive call to explore further combinations, and then removing the number from the 'path' (backtracking) to explore alternative combinations. A crucial optimization is the use of pruning to avoid unnecessary recursive calls. Before iterating through the numbers, the code calculates the maximum possible starting number, which ensures that enough numbers remain to form a complete combination of size 'k'.

Step-by-Step Algorithm

Step 1: Initialize an empty list called 'result' to store all the combinations.
Step 2: Initialize an empty list called 'path' to store the current combination being built.
Step 3: Define a recursive backtracking function that takes a 'start' index as input.
Step 4: Inside the backtracking function, check if the length of the 'path' is equal to 'k'. If it is, add a copy of the 'path' to the 'result' list and return.
Step 5: Calculate the 'end' index for the loop to incorporate pruning. `end = n - (k - len(path)) + 1`.
Step 6: Iterate from 'start' to 'end' (inclusive).
Step 7: In each iteration, add the current number 'i' to the 'path'.
Step 8: Recursively call the backtracking function with 'i + 1' as the new 'start' index.
Step 9: After the recursive call returns, remove the last element from the 'path' (backtracking).
Step 10: Call the backtracking function initially with 'start' set to 1.
Step 11: Return the 'result' list.

Key Insights

Insight 1: Backtracking is well-suited for generating combinations due to its ability to explore all possible choices and prune unnecessary branches.
Insight 2: Maintaining a temporary 'path' to store the current combination being built, and then adding a copy of this path to the result once it reaches the desired size (k) is a key technique.
Insight 3: Pruning the search space by calculating the maximum possible starting number based on the remaining numbers needed, significantly reduces the number of recursive calls and improves efficiency. The pruning calculation ensures that there are enough remaining numbers to complete a valid combination.

Complexity Analysis

Time Complexity: O(C(n, k) * k)

Space Complexity: O(k)

Topics

This problem involves: Backtracking.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Google, Meta, Microsoft.

Combinations | LeetCode 77

Medium

Backtracking

Time: O(C(n, k) * k)Space: O(k)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();

        backtrack(1, n, k, result, path);
        return result;
    }

    private void backtrack(int start, int n, int k, List<List<Integer>> result, List<Integer> path) {
        if (path.size() == k) {
            result.add(new ArrayList<>(path));
            return;
        }

        int end = n - (k - path.size()) + 1;
        for (int i = start; i <= end; i++) {
            path.add(i);
            backtrack(i + 1, n, k, result, path);
            path.remove(path.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 77: Combinations - Detailed explanation and solution approaches.

Topics

Backtracking

Companies

AdobeAmazonAppleBloombergGoogleMetaMicrosoft

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:77

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize an empty list called 'result' to store all the combinations.
2
Step 2: Initialize an empty list called 'path' to store the current combination being built.
3
Step 3: Define a recursive backtracking function that takes a 'start' index as input.
4
Step 4: Inside the backtracking function, check if the length of the 'path' is equal to 'k'. If it is, add a copy of the 'path' to the 'result' list and return.
5
Step 5: Calculate the 'end' index for the loop to incorporate pruning. `end = n - (k - len(path)) + 1`.
6
Step 6: Iterate from 'start' to 'end' (inclusive).
7
Step 7: In each iteration, add the current number 'i' to the 'path'.
8
Step 8: Recursively call the backtracking function with 'i + 1' as the new 'start' index.
9
Step 9: After the recursive call returns, remove the last element from the 'path' (backtracking).
10
Step 10: Call the backtracking function initially with 'start' set to 1.
11
Step 11: Return the 'result' list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(C(n, k) * k)

💾

Space Complexity

O(k)

📝

Why?

The time complexity is proportional to the number of combinations (C(n, k)) because the algorithm needs to generate all valid combinations. The factor of 'k' comes from copying each valid combination to the result. The space complexity is dominated by the depth of the recursion and the size of 'path'. Because of pruning, the search space is greatly reduced.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Approach using Bit Masking

An alternative approach is to use bit masking to generate combinations. Each bit in a binary number represents whether an element is included in the combination or not. We iterate through all possible binary numbers with 'k' bits set, and convert them to the corresponding combinations.

Time: O(C(n, k) * n)Space: O(k)

Trade-offs: The iterative approach might be more difficult to understand and implement compared to the recursive backtracking approach. While it avoids recursion overhead, converting bitmasks to combinations can involve more operations. Backtracking is generally preferred for its clearer structure and ease of modification, especially with the pruning optimization.

✨

Key Insights

1Insight 1: Backtracking is well-suited for generating combinations due to its ability to explore all possible choices and prune unnecessary branches.
2Insight 2: Maintaining a temporary 'path' to store the current combination being built, and then adding a copy of this path to the result once it reaches the desired size (k) is a key technique.
3Insight 3: Pruning the search space by calculating the maximum possible starting number based on the remaining numbers needed, significantly reduces the number of recursive calls and improves efficiency. The pruning calculation ensures that there are enough remaining numbers to complete a valid combination.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 1, k = 1. The algorithm correctly returns [[1]].
Edge case 2: n = 4, k = 4. The algorithm correctly returns [[1, 2, 3, 4]].
Edge case 3: n = 4, k = 1. The algorithm correctly returns [[1], [2], [3], [4]].

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not making a copy of the 'path' before adding it to the 'result' list. If the 'path' is added directly, subsequent modifications to the 'path' will affect the lists already added to the 'result'.

✗Common mistake 2: Forgetting to backtrack by removing the last element from the 'path' after the recursive call returns. Without backtracking, the 'path' will keep growing, leading to incorrect combinations.

✗Common mistake 3: Not including pruning to reduce the search space. This leads to increased execution time, particularly when n is significantly larger than k.

✗Common mistake 4: Incorrectly calculating the end index for the pruning loop, which can either exclude valid combinations or fail to prune properly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Feature selection in machine learning. Finding the best combination of features to use in a model.

🏦Real-world scenario 2: Selecting items for a menu. Generating different meal combinations from a set of available ingredients.

🎮Real-world scenario 3: Lottery number generation. Creating possible lottery ticket combinations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Backtracking = DFS + pruning. The key is identifying when to backtrack. Always restore state after recursion.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Combinations"?

For this medium-level Backtracking problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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