What is the time complexity of Combination Sum IV?

The optimal solution has a time complexity of O(target * len(nums)).

What is the space complexity of Combination Sum IV?

The space complexity is O(target).

What data structures are used to solve Combination Sum IV?

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

What are the key insights for solving Combination Sum IV?

Insight 1: Dynamic programming is effective here because the number of combinations for a given target can be built up from the number of combinations for smaller targets. Insight 2: The problem requires counting combinations where order matters. This dictates the DP state transition equation.

Which companies ask LeetCode 377. Combination Sum IV in interviews?

This problem is frequently asked at Snap.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Combination Sum IV - LeetCode 377 Solution

Combination Sum IV - Complete Solution Guide

Combination Sum IV is LeetCode problem 377, 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 nums and a target integer target , return the number of possible combinations that add up to target . The test cases are generated so that the answer can fit in a 32-bit integer. Example 1: Input: nums = [1,2,3], target = 4 Output: 7 Explanation: The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations. Example 2: Input: nums = [9], target = 3 Output:

Detailed Explanation

The problem "Combination Sum IV" asks us to find the number of different combinations of numbers from a given array `nums` that sum up to a target value `target`. The key point is that the order of the numbers in the combination matters. For example, (1, 2, 1) and (1, 1, 2) are considered distinct combinations. The input `nums` contains distinct positive integers, and the target is also a positive integer. The constraints ensure that the answer will fit in a 32-bit integer.

Solution Approach

The solution uses dynamic programming. We create a `dp` array of size `target + 1`, where `dp[i]` represents the number of combinations that sum up to `i`. We initialize `dp[0]` to 1 because there's one way to achieve a target of 0 (using no numbers). Then, we iterate through all possible target values from 1 to `target`. For each target value `i`, we iterate through each number `num` in the `nums` array. If `i - num` is non-negative, it means we can include `num` in a combination that sums up to `i`. The number of such combinations is the number of combinations that sum up to `i - num`, which is stored in `dp[i - num]`. So, we add `dp[i - num]` to `dp[i]`. Finally, `dp[target]` will contain the total number of combinations that sum up to `target`.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `target + 1` with all values set to 0.
Step 2: Set `dp[0] = 1` as the base case (one way to form a sum of 0: an empty combination).
Step 3: Iterate through target values from `i = 1` to `target`.
Step 4: For each target value `i`, iterate through each number `num` in the input array `nums`.
Step 5: If `i - num >= 0`, update `dp[i] += dp[i - num]` (add the number of combinations for `i - num` to the number of combinations for `i`).
Step 6: After iterating through all target values and numbers, return `dp[target]`.

Key Insights

Insight 1: Dynamic programming is effective here because the number of combinations for a given target can be built up from the number of combinations for smaller targets.
Insight 2: The problem requires counting combinations where order matters. This dictates the DP state transition equation.
Insight 3: The base case, when target is 0, has exactly one combination which is an empty combination

Complexity Analysis

Time Complexity: O(target * len(nums))

Space Complexity: O(target)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Snap.

Combination Sum IV | LeetCode 377

Medium

Array Dynamic Programming

Time: O(target * len(nums))Space: O(target)

Solution Code

Available in:

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];
        dp[0] = 1;

        for (int i = 1; i <= target; i++) {
            for (int num : nums) {
                if (i - num >= 0) {
                    dp[i] += dp[i - num];
                }
            }
        }

        return dp[target];
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 377: Combination Sum IV - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Snap

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:377

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

6 steps

1
Step 1: Initialize a DP array `dp` of size `target + 1` with all values set to 0.
2
Step 2: Set `dp[0] = 1` as the base case (one way to form a sum of 0: an empty combination).
3
Step 3: Iterate through target values from `i = 1` to `target`.
4
Step 4: For each target value `i`, iterate through each number `num` in the input array `nums`.
5
Step 5: If `i - num >= 0`, update `dp[i] += dp[i - num]` (add the number of combinations for `i - num` to the number of combinations for `i`).
6
Step 6: After iterating through all target values and numbers, return `dp[target]`.

💡

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(target * len(nums))

💾

Space Complexity

O(target)

📝

Why?

The time complexity stems from the nested loops, where the outer loop iterates `target` times and the inner loop iterates `len(nums)` times. The space complexity arises from storing the `dp` table of size `target + 1`.

📖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

Recursion with Memoization

We can solve this problem using recursion with memoization. The recursive function will take the current target as input. If the target is 0, return 1. If the target is negative, return 0. Otherwise, iterate through the `nums` array and recursively call the function with `target - num`. Store the results in a memoization table to avoid redundant calculations.

Time: O(target * len(nums))Space: O(target)

Trade-offs: Recursion with memoization can be more intuitive for some people, but it might have higher overhead due to function calls. In this case, iterative DP is often preferable.

✨

Key Insights

1Insight 1: Dynamic programming is effective here because the number of combinations for a given target can be built up from the number of combinations for smaller targets.
2Insight 2: The problem requires counting combinations where order matters. This dictates the DP state transition equation.
3Insight 3: The base case, when target is 0, has exactly one combination which is an empty combination

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `nums` is empty. The code correctly handles this case because the inner loop will not execute, and the `dp` array will remain with `dp[0] = 1` and the rest as 0, resulting in 0 combinations if the target is not 0, and 1 combination if the target is 0.
Edge case 2: `target` is 0. The code correctly handles this case by initializing `dp[0] = 1`, indicating that there is one way to achieve a target of 0.
Edge case 3: No combination sums up to the target. In this case, all `dp[i]` except `dp[0]` will remain 0, so `dp[target]` will be 0, which is the correct result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing `dp[0]` to 1. This is crucial because it represents the base case where the target is 0, and there's one way to achieve it (by not selecting any numbers).

✗Common mistake 2: Incorrectly calculating the DP state transition. The correct way is `dp[i] += dp[i - num]` because we are counting the number of combinations, not finding the minimum number of elements to reach the target.

✗Common mistake 3: Forgetting that order matters. If order didn't matter, we would need a completely different approach like standard combination sum.

✗Common mistake 4: Not considering the constraint that answer fits in 32-bit integer. The CPP version uses unsigned int as the intermediate results can exceed the range of signed integer.

✅

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: Coin change problem where the number of possible coin arrangements matters. For instance, in a vending machine, knowing the total number of possible ways to give change can be useful.

🏦Real-world scenario 2: In signal processing, finding the number of ways to decompose a signal into a sum of elementary signals with specific amplitudes.

💡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

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

⚠️ 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: June 18, 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

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 IV"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?