Combination Sum IV 🎯

Welcome to a fascinating journey through the world of Data Structures and Algorithms! Today, let's dive into a classic problem called Combination Sum IV. This problem is a wonderful blend of Data Structures, Algorithms, and problem-solving skills. Let's get started!

Understanding the Problem 📝

Combination Sum IV is a variation of the Combination Sum problem. The main difference lies in the fact that we are now dealing with unique combinations. In simpler terms, once an element is included in our solution, it cannot be used again.

Given a number array nums and a target number target, find all unique combinations in nums that add up to target. The numbers in nums may be used multiple times, but each combination must be unique.

Breaking Down the Solution 💡

To solve the Combination Sum IV problem, we'll use a Depth-First Search (DFS) algorithm with a backtracking approach. Let's break down the solution step by step:

Initialize an empty list results to store our combinations.
Define a helper function dfs(remain, target, nums, index, combinations).
- remain: the remaining target number to achieve.
- target: the total target number.
- nums: the number array.
- index: the current index in the number array.
- combinations: the current combination we are building.
Base Case: If remain equals 0, append the current combination to results and return.
Recursive Step: For each number num from index to the end of nums, if num is less than or equal to remain, do:
- Increment remain by num.
- Include num in the current combination combinations.
- Recursively call dfs(remain, target, nums, index, combinations).
- Backtrack by decrementing remain by num and removing num from combinations.

Practical Example ✅

Let's try solving a problem using the approach discussed above.

Problem: Find all unique combinations that add up to 7 from the number array [1, 2, 3].

python

def combinationSum4(nums, target):
    def dfs(remain, target, nums, index, combinations):
        if not remain:
            results.append(combinations)
            return

        for i in range(index, len(nums)):
            if nums[i] <= remain:
                dfs(remain - nums[i], target, nums, i + 1, combinations + [nums[i]])

    results = []
    nums.sort()
    dfs(target, target, nums, 0, [])
    return results

nums = [1, 2, 3]
target = 7
print(combinationSum4(nums, target))

The output will be:

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

Challenges 💡

Now that you have a good grasp of the problem and its solution, try applying this approach to solve the following challenge:

Problem: Given the number array [4, 1, 2, 1, 2, 3] and the target number 9, find all unique combinations that add up to 9.

Combination Sum IV 🎯

Combination Sum IV 🎯

Understanding the Problem 📝

Breaking Down the Solution 💡

Practical Example ✅

Challenges 💡

Quiz 📝