Solving nSum Problems Efficiently: A Universal Approach in Python

When it comes to LeetCode problems like 2Sum, 3Sum, and 4Sum, many programmers struggle to find a solution that is both efficient and flexible enough to handle different variants of the same problem. While solving each one individually is possible, a universal solution that can handle any number of sums (nSum) is ideal. In this post, we will dive deep into a Python implementation of a general solution that can tackle these problems, scaling from 2Sum to 3Sum, 4Sum, and even higher-order sums.

By the end of this post, you’ll understand how to solve any nSum problem efficiently, using a combination of the two-pointer method and recursion, while avoiding duplicates. Let’s get started!

Understanding the Problem

The nSum problem asks you to find combinations of n numbers from an array that sum up to a specific target value. For instance, in the 2Sum problem, you are asked to find pairs of numbers that add up to a given target. In the 3Sum and 4Sum problems, you need to extend this logic to find triplets and quadruplets, respectively.

The challenge becomes more complex as the number of elements increases. Additionally, the solutions must avoid returning duplicate combinations. This makes nSum problems a perfect case for recursion and two-pointer techniques, which we will leverage to create a universal solution.

Problem Breakdown

• Input: An unsorted array of integers nums and an integer target.
• Output: All unique combinations of n elements from nums that add up to the target.

The Universal nSum Solution in Python

Let’s take a look at the Python implementation of a universal solution that can handle any nSum problem. The key is to build upon the simpler 2Sum solution and then recursively expand to handle 3Sum, 4Sum, and beyond.

def nSum(nums, n, start, target):
    res = []
    sz = len(nums)

    # If n < 2 or the array size is less than n, return an empty result
    if n < 2 or sz < n:
        return res

    # Special handling for 2Sum (using the two-pointer method)
    if n == 2:
        lo, hi = start, sz - 1
        while lo < hi:
            sum = nums[lo] + nums[hi]
            left, right = nums[lo], nums[hi]
            if sum < target:
                lo += 1
            elif sum > target:
                hi -= 1
            else:
                res.append([left, right])
                # Skip duplicates
                while lo < hi and nums[lo] == left:
                    lo += 1
                while lo < hi and nums[hi] == right:
                    hi -= 1
    else:
        # General handling for n > 2, recursively solving the (n-1)Sum problem
        for i in range(start, sz):
            # Skip duplicates
            if i > start and nums[i] == nums[i - 1]:
                continue
            sub_res = nSum(nums, n - 1, i + 1, target - nums[i])
            for sub in sub_res:
                res.append([nums[i]] + sub)

    return res

# Example usage: solving the 4Sum problem
def fourSum(nums, target):
    nums.sort()  # Sort the array
    return nSum(nums, 4, 0, target)

# Example usage: solving the 3Sum problem
def threeSum(nums):
    nums.sort()  # Sort the array
    return nSum(nums, 3, 0, 0)

# Test case
nums = [1,1,1,2,2,3,3]
target = 4
print(fourSum(nums, target))  # Output results for 4Sum

Key Concepts and Approach

1. Two-Pointer Method for 2Sum

When n == 2, we solve the problem using the two-pointer technique. Here’s how it works:

• Sort the array to make sure we can efficiently navigate using pointers.
• Two pointers start at both ends of the sorted array. If the sum of the elements at these pointers is less than the target, move the left pointer to the right (to increase the sum). If the sum is greater, move the right pointer to the left (to decrease the sum).
• Avoid duplicates by skipping over any repeated elements after finding a valid pair.

This method runs in O(N) time, making it very efficient.

2. Recursive Approach for nSum

For n > 2, we use recursion. The idea is to:

• Fix one number in the array and reduce the problem to an (n-1)Sum problem.
• Recursively solve the smaller problem, and then combine the fixed number with the result of the (n-1)Sum function.

Each recursive call reduces the size of the problem until it eventually becomes a 2Sum problem, which is solved using the two-pointer method.

3. Avoiding Duplicates

To ensure no duplicate combinations are returned:

• Skip over any repeated numbers when fixing numbers in the recursive step.
• Skip over duplicates in the two-pointer step when generating pairs.

Time Complexity Analysis

Let’s break down the time complexity:

• Sorting the array takes O(NlogN).
• Each recursive step loops over the array, resulting in a complexity of O(N) per level of recursion.
• For 2Sum, the complexity is O(N). For 3Sum, it’s O(N^2) because we loop through the array and for each element, we solve a 2Sum problem. For 4Sum, it’s O(N^3), and so on.

Thus, the total time complexity of solving an nSum problem is O(N^(n-1)), where n is the number of elements in the sum.

Example Output

Let’s see the output for a test case:

nums = [1,1,1,2,2,3,3]
target = 4
print(fourSum(nums, target))

Output:

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

A Universal nSum Solution

This implementation is flexible enough to handle any nSum problem, from 2Sum to 100Sum (if required). Here’s why:

• Base case: When n == 2, we efficiently solve it using the two-pointer method.
• Recursive step: For any n > 2, we break the problem down into a smaller (n-1)Sum problem, recursively solving it.

The beauty of this approach is that it doesn’t matter whether you’re solving 3Sum or 4Sum—this one function can handle it all.

Conclusion

In this post, we walked through the creation of a universal Python solution for the nSum problem. By combining the two-pointer method with recursion, we can efficiently solve 2Sum, 3Sum, 4Sum, and beyond, all while avoiding duplicates. This approach not only helps solve these specific problems but also illustrates how breaking down complex problems into simpler subproblems can be a powerful problem-solving strategy.

If you have any questions or thoughts on optimizing this approach further, feel free to leave a comment. Happy coding!

Key Takeaways:

1. The two-pointer method is an efficient way to solve the 2Sum problem in linear time.
2. The recursive approach allows us to extend this logic to handle any nSum problem.
3. Handling duplicate values is critical to ensure correctness and efficiency in your solutions.

Further Reading:

• LeetCode Problem 15: 3Sum
• LeetCode Problem 18: 4Sum