What is the time complexity of Combination Sum?

The optimal solution has a time complexity of O(N^(T/M)).

What is the space complexity of Combination Sum?

The space complexity is O(T/M).

What data structures are used to solve Combination Sum?

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

What are the key insights for solving Combination Sum?

Insight 1: Backtracking is a suitable approach for exploring the solution space due to the combinatorial nature of the problem. Insight 2: The 'start' index in the backtracking function is crucial to avoid generating duplicate combinations (e.g., [2, 2, 3] and [2, 3, 2] are considered duplicates).

Which companies ask LeetCode 39. Combination Sum in interviews?

This problem is frequently asked at Adobe, Airbnb, Amazon, Apple, Autodesk and 22 more companies.

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Combination Sum - LeetCode 39 Solution

Combination Sum - Complete Solution Guide

Combination Sum is LeetCode problem 39, 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 an array of distinct integers candidates and a target integer target , return a list of all unique combinations of candidates where the chosen numbers sum to target . You may return the combinations in any order . The same number may be chosen from candidates an unlimited number of times . Two combinations are unique if the frequency of at least one of the chosen numbers is different. The test cases are generated such that the number of unique combinations that sum up to target is less tha

Detailed Explanation

The 'Combination Sum' problem asks us to find all unique combinations of numbers from a given array (candidates) that sum up to a specific target value. The same number from the candidate array can be used multiple times within a single combination. The input consists of an array of distinct integers 'candidates' and an integer 'target'. The output should be a list of lists, where each inner list represents a unique combination that sums to the target. A key constraint is that the number of unique combinations is limited to less than 150 for any given input, ensuring backtracking does not explode in terms of memory.

Solution Approach

The solution utilizes a backtracking algorithm to systematically explore all possible combinations of candidates. The core idea is to recursively try including each candidate number into a combination, and if the sum becomes equal to the target, we add the combination to the results. If the sum exceeds the target, we backtrack and try a different path. The 'start' index is used to ensure that we only consider candidates from the current index onwards, preventing duplicate combinations.

Step-by-Step Algorithm

Step 1: Initialize an empty list `results` to store the valid combinations.
Step 2: Define a recursive `backtrack` function that takes the remaining sum, the current combination `combo`, and the starting index `start` in the `candidates` array as parameters.
Step 3: In the `backtrack` function, check if the remaining sum is 0. If it is, add a copy of the current combination `combo` to the `results` list and return.
Step 4: If the remaining sum is negative, it means the current combination exceeds the target, so return.
Step 5: Iterate through the `candidates` array from the `start` index to the end.
Step 6: For each candidate number, add it to the current combination `combo`.
Step 7: Recursively call the `backtrack` function with the updated remaining sum (remaining - candidate), the updated combination `combo`, and the same `start` index (to allow using the same candidate multiple times).
Step 8: After the recursive call returns, remove the last added candidate from the `combo` (backtracking step) to explore other possibilities.
Step 9: Call the `backtrack` function initially with the target sum, an empty combination list, and a starting index of 0.
Step 10: Return the `results` list containing all valid combinations.

Key Insights

Insight 1: Backtracking is a suitable approach for exploring the solution space due to the combinatorial nature of the problem.
Insight 2: The 'start' index in the backtracking function is crucial to avoid generating duplicate combinations (e.g., [2, 2, 3] and [2, 3, 2] are considered duplicates).
Insight 3: We can repeatedly use the same candidate number, so we pass the current index 'i' instead of 'i+1' in the recursive call.

Complexity Analysis

Time Complexity: O(N^(T/M))

Space Complexity: O(T/M)

Topics

This problem involves: Array, Backtracking.

Companies

Asked at: Adobe, Airbnb, Amazon, Apple, Autodesk, Bloomberg, ByteDance, Citadel, Confluent, DE Shaw, Huawei, Infosys, J.P. Morgan, LinkedIn, Meta, Microsoft, Oracle, Pinterest, Salesforce, ServiceNow, Snap, TikTok, Uber, Walmart Labs, Yahoo, Zoho, tcs.

Combination Sum | LeetCode 39

Medium

Array Backtracking

Time: O(N^(T/M))Space: O(T/M)

Solution Code

Available in:

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

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> results = new ArrayList<>();

        backtrack(candidates, target, new ArrayList<>(), 0, results);
        return results;
    }

    private void backtrack(int[] candidates, int remaining, List<Integer> combo, int start, List<List<Integer>> results) {
        if (remaining == 0) {
            results.add(new ArrayList<>(combo));
            return;
        }

        if (remaining < 0) {
            return;
        }

        for (int i = start; i < candidates.length; i++) {
            int candidate = candidates[i];
            combo.add(candidate);
            backtrack(candidates, remaining - candidate, combo, i, results);
            combo.remove(combo.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 39: Combination Sum - Detailed explanation and solution approaches.

Topics

Array Backtracking

Companies

AdobeAirbnbAmazonAppleAutodeskBloombergByteDanceCitadel

Stats

Difficulty:Medium

Languages:4

Companies:27

Problem #:39

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

10 steps

1
Step 1: Initialize an empty list `results` to store the valid combinations.
2
Step 2: Define a recursive `backtrack` function that takes the remaining sum, the current combination `combo`, and the starting index `start` in the `candidates` array as parameters.
3
Step 3: In the `backtrack` function, check if the remaining sum is 0. If it is, add a copy of the current combination `combo` to the `results` list and return.
4
Step 4: If the remaining sum is negative, it means the current combination exceeds the target, so return.
5
Step 5: Iterate through the `candidates` array from the `start` index to the end.
6
Step 6: For each candidate number, add it to the current combination `combo`.
7
Step 7: Recursively call the `backtrack` function with the updated remaining sum (remaining - candidate), the updated combination `combo`, and the same `start` index (to allow using the same candidate multiple times).
8
Step 8: After the recursive call returns, remove the last added candidate from the `combo` (backtracking step) to explore other possibilities.
9
Step 9: Call the `backtrack` function initially with the target sum, an empty combination list, and a starting index of 0.
10
Step 10: Return the `results` list containing all valid combinations.

💡

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(N^(T/M))

💾

Space Complexity

O(T/M)

📝

Why?

The time complexity arises from the nature of the backtracking algorithm, which explores all possible paths in the decision tree. The branching factor at each node is equal to the number of candidates (N). The depth of the recursion tree is at most T/M, where T is the target and M is the minimum candidate. The space complexity is primarily determined by the maximum depth of the recursion stack during backtracking, and to a lesser extent by the space required to store intermediate combinations.

📖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

Dynamic Programming

Dynamic programming can be used to solve this problem, but it might be less intuitive and efficient than backtracking. DP solutions are generally better when you want to find the *minimum* number of combinations. In this case, we want *all* combinations, which makes backtracking more direct.

Time: O(T * N), where T is the target and N is the number of candidates.Space: O(T), where T is the target.

Trade-offs: DP might be preferable if we needed to find the *minimum* number of combinations to reach the target instead of finding all combinations, or if we expected a very large number of combinations, potentially exceeding memory limits with backtracking.

✨

Key Insights

1Insight 1: Backtracking is a suitable approach for exploring the solution space due to the combinatorial nature of the problem.
2Insight 2: The 'start' index in the backtracking function is crucial to avoid generating duplicate combinations (e.g., [2, 2, 3] and [2, 3, 2] are considered duplicates).
3Insight 3: We can repeatedly use the same candidate number, so we pass the current index 'i' instead of 'i+1' in the recursive call.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `candidates` is empty. The solution handles this correctly because the loop in the backtrack function won't execute, and an empty list will be returned.
Edge case 2: `target` is less than the smallest candidate. The solution handles this because if the `remaining` becomes negative, the backtrack function returns, effectively pruning that branch of the search tree.
Edge case 3: `target` is 0. The initial check `if remaining == 0` covers this case and adds an empty list to the results if the initial target is 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to use the 'start' index in the recursive call, leading to duplicate combinations. For example, [2, 2, 3] and [2, 3, 2] would be considered distinct if the starting index wasn't correctly managed.

✗Common mistake 2: Not creating a *copy* of the `combo` list before adding it to the `results` list. If you add `combo` directly, all the lists in `results` will point to the same `combo` object, and all lists will eventually be equal to whatever `combo` looks like at the end of the execution.

✗Common mistake 3: Incorrectly handling the backtracking step, i.e., forgetting to remove the last added element from the `combo` before returning from the recursive call. This can lead to incorrect combinations being generated.

✅

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: Inventory management - Determining the possible combinations of products to fulfill an order with a specific total value or weight, given a set of available products and their corresponding prices or weights.

🏦Real-world scenario 2: Financial modeling - Identifying different combinations of investments to reach a specific investment goal, considering varying investment returns and risk levels.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: July 10, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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 "Combination Sum"?

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

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