What is the time complexity of Combination Sum II?

The optimal solution has a time complexity of O(2^n).

What data structures are used to solve Combination Sum II?

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 II?

Insight 1: Sorting the candidates array allows us to efficiently skip duplicate combinations using the condition `if i > start_index and candidates[i] == candidates[i-1]: continue`. Insight 2: Backtracking is a natural fit for exploring all possible combinations. It allows us to build up combinations incrementally and backtrack when a combination exceeds the target or reaches the end of the candidates list.

Which companies ask LeetCode 40. Combination Sum II in interviews?

This problem is frequently asked at Adobe, Amazon, Bloomberg, ByteDance, Infosys and 7 more companies.

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Combination Sum II - LeetCode 40 Solution

Combination Sum II - Complete Solution Guide

Combination Sum II is LeetCode problem 40, 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 a collection of candidate numbers ( candidates ) and a target number ( target ), find all unique combinations in candidates where the candidate numbers sum to target . Each number in candidates may only be used once in the combination. Note: The solution set must not contain duplicate combinations. Example 1: Input: candidates = [10,1,2,7,6,1,5], target = 8 Output: [ [1,1,6], [1,2,5], [1,7], [2,6] ] Example 2: Input: candidates = [2,5,2,1,2], target = 5 Output: [ [1,2,2], [5] ] Constraints

Detailed Explanation

The 'Combination Sum II' problem asks us to find all unique combinations of numbers from a given collection ('candidates') that add up to a specific target number. Each number in the 'candidates' list can only be used once in each combination, and the solution must not contain duplicate combinations. The candidates array can contain duplicate numbers, which is a key aspect of this problem that distinguishes it from 'Combination Sum I'. The goal is to return a list of lists, where each inner list represents a unique combination that sums to the target.

Solution Approach

The provided code utilizes a backtracking algorithm to find all unique combinations. The approach involves first sorting the input array 'candidates' to easily handle duplicate numbers and improve efficiency. Then, a recursive 'backtrack' function is used to explore potential combinations. This function builds up a combination by adding elements from the candidates array, checking at each step if the combination's sum equals the target. If it does, the combination is added to the result. If the sum exceeds the target, the current branch is abandoned. The crucial part is the handling of duplicates: the condition `if i > start_index and candidates[i] == candidates[i-1]: continue` prevents adding the same number multiple times in the same level of recursion, thus avoiding duplicate combinations.

Step-by-Step Algorithm

Step 1: Sort the 'candidates' array in ascending order. This step is crucial for efficiently skipping duplicate combinations.
Step 2: Define a recursive 'backtrack' function that takes the 'start_index', 'remaining_target', and 'current_path' as input. 'start_index' indicates from which index of 'candidates' to start considering numbers.
Step 3: In the 'backtrack' function, check if 'remaining_target' is equal to 0. If so, add a copy of the 'current_path' to the 'results' list and return. This signifies that a valid combination has been found.
Step 4: Iterate through the 'candidates' array from 'start_index' to the end. At each iteration, check if the current number is greater than 'remaining_target'. If so, break the loop because the remaining numbers will also be too large.
Step 5: Skip duplicate numbers. If the current number is the same as the previous number (and we are not at the starting index of the loop), skip it to avoid duplicate combinations. This is done with the condition `if i > start_index and candidates[i] == candidates[i-1]: continue`.
Step 6: Add the current number to the 'current_path'.
Step 7: Recursively call the 'backtrack' function with the updated 'start_index' (i + 1), 'remaining_target' (remaining_target - candidates[i]), and 'current_path'.
Step 8: Remove the last added number from the 'current_path' (backtrack) to explore other possible combinations.
Step 9: After the 'backtrack' function finishes exploring all possible combinations, return the 'results' list.

Key Insights

Insight 1: Sorting the candidates array allows us to efficiently skip duplicate combinations using the condition `if i > start_index and candidates[i] == candidates[i-1]: continue`.
Insight 2: Backtracking is a natural fit for exploring all possible combinations. It allows us to build up combinations incrementally and backtrack when a combination exceeds the target or reaches the end of the candidates list.
Insight 3: The `start_index` parameter in the recursive function is crucial for ensuring each number is used only once and for avoiding duplicate combinations. It indicates the index from which to start considering elements in the next level of recursion.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n)

Topics

This problem involves: Array, Backtracking.

Companies

Asked at: Adobe, Amazon, Bloomberg, ByteDance, Infosys, Meta, Microsoft, Oracle, Snap, Tesla, TikTok, Uber.

Combination Sum II | LeetCode 40

Medium

Array Backtracking

Time: O(2^n)Space: O(n)

Solution Code

Available in:

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

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> results = new ArrayList<>();
        Arrays.sort(candidates);

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

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

        for (int i = start_index; i < candidates.length; i++) {
            if (candidates[i] > remaining_target) {
                break;
            }

            if (i > start_index && candidates[i] == candidates[i - 1]) {
                continue;
            }

            current_path.add(candidates[i]);
            backtrack(i + 1, remaining_target - candidates[i], current_path, candidates, results);
            current_path.remove(current_path.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 40: Combination Sum II - Detailed explanation and solution approaches.

Topics

Array Backtracking

Companies

AdobeAmazonBloombergByteDanceInfosysMetaMicrosoftOracle

Stats

Difficulty:Medium

Languages:4

Companies:12

Problem #:40

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

9 steps

1
Step 1: Sort the 'candidates' array in ascending order. This step is crucial for efficiently skipping duplicate combinations.
2
Step 2: Define a recursive 'backtrack' function that takes the 'start_index', 'remaining_target', and 'current_path' as input. 'start_index' indicates from which index of 'candidates' to start considering numbers.
3
Step 3: In the 'backtrack' function, check if 'remaining_target' is equal to 0. If so, add a copy of the 'current_path' to the 'results' list and return. This signifies that a valid combination has been found.
4
Step 4: Iterate through the 'candidates' array from 'start_index' to the end. At each iteration, check if the current number is greater than 'remaining_target'. If so, break the loop because the remaining numbers will also be too large.
5
Step 5: Skip duplicate numbers. If the current number is the same as the previous number (and we are not at the starting index of the loop), skip it to avoid duplicate combinations. This is done with the condition `if i > start_index and candidates[i] == candidates[i-1]: continue`.
6
Step 6: Add the current number to the 'current_path'.
7
Step 7: Recursively call the 'backtrack' function with the updated 'start_index' (i + 1), 'remaining_target' (remaining_target - candidates[i]), and 'current_path'.
8
Step 8: Remove the last added number from the 'current_path' (backtrack) to explore other possible combinations.
9
Step 9: After the 'backtrack' function finishes exploring all possible combinations, return the 'results' 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(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because, in the worst-case, we explore all possible subsets of the input array. The space complexity is determined by the maximum depth of the recursion tree, which can be at most 'n' (the length of the 'candidates' array). Additionally, we are storing a combination which is O(n) in space, however this can be seen as auxiliary space and should be accounted for in the space complexity. Also in the C code provided, the result array is initialized with a large size initially and then reallocated at the end which would also increase the space complexity to potentially `O(n * m)` in the worst case where n is the length of `candidates` and m is the number of combinations. So, the O(n) stated in the problem information is not completely accurate. It relies on an assumption that the number of combinations is a constant multiple of n, which isn't generally the case.

📖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

It is theoretically possible to solve this problem iteratively, but it would be significantly more complex to implement than the backtracking approach. It would involve maintaining a stack of states to represent the current combination and remaining target, and carefully handling the duplicate skipping logic.

Time: O(2^n)Space: O(n)

Trade-offs: The iterative approach would avoid the overhead of recursive function calls, but the increased complexity and difficulty in managing the state would likely outweigh any performance benefits. Backtracking is generally preferred for its clarity and simplicity in this case.

✨

Key Insights

1Insight 1: Sorting the candidates array allows us to efficiently skip duplicate combinations using the condition `if i > start_index and candidates[i] == candidates[i-1]: continue`.
2Insight 2: Backtracking is a natural fit for exploring all possible combinations. It allows us to build up combinations incrementally and backtrack when a combination exceeds the target or reaches the end of the candidates list.
3Insight 3: The `start_index` parameter in the recursive function is crucial for ensuring each number is used only once and for avoiding duplicate combinations. It indicates the index from which to start considering elements in the next level of recursion.

📚

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: Empty 'candidates' array. The code correctly handles this case because the loop in the 'backtrack' function won't execute, and an empty list will be returned.
Edge case 2: 'target' is 0. The code correctly handles this case because the 'backtrack' function immediately adds an empty list to the 'results' if the 'remaining_target' is initially 0.
Edge case 3: 'candidates' array contains duplicate numbers. The `if i > start_index and candidates[i] == candidates[i-1]: continue` condition efficiently skips duplicate numbers, preventing duplicate combinations in the result.
Edge case 4: 'candidates' array contains numbers greater than target, this is handled by `if candidates[i] > remaining_target: break` which avoids unnecessary recursive calls.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the 'candidates' array. Without sorting, the duplicate skipping logic will not work correctly, leading to duplicate combinations in the result.

✗Common mistake 2: Not including the 'start_index' in the recursive function call. Without the 'start_index', each number in the 'candidates' array can be used multiple times in a combination, violating the problem constraint.

✗Common mistake 3: Not handling the duplicate skipping logic correctly. The `if i > start_index and candidates[i] == candidates[i-1]: continue` condition must be carefully implemented to ensure only one instance of each duplicate number is considered at each level of recursion.

✗Common mistake 4: Forgetting to create a *copy* of currentPath when adding it to results. If you directly add `currentPath`, any later modifications to `currentPath` will change the contents of the lists in `results` as well, resulting in incorrect output.

✅

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: Recommendation systems. Finding combinations of products or services that meet a specific budget or user preference.

🏦Real-world scenario 2: Financial portfolio optimization. Selecting a combination of assets that meet a target return with a given risk tolerance.

🎮Real-world scenario 3: Inventory management. Determining combinations of items to fulfill an order with a limited number of available items.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 11, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Combination Sum II"?

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.

🤖 Need help with this problem?