Split Array Largest Sum Leetcode Problem 410 [Python Solution]

Welcome back as we dismantle a challenging problem known as Split Array Largest Sum Leetcode problem.

This problem is categorized as “Hard” on LeetCode, and it falls under the domain of binary search.

So, if you’re ready to dive into this problem, let’s get started!

Problem Overview

Given an integer array nums and an integer k, our goal is to split nums into k non-empty subarrays in such a way that the largest sum of any subarray is minimized.

A subarray is essentially a contiguous part of the original array.

Let’s illustrate this with an example:

Example 1:

Input: nums = [7, 2, 5, 10, 8], k = 2
Output: 18

In this case, there are four ways to split nums into two subarrays.

The best approach is to split it into [7, 2, 5] and [10, 8], where the largest sum among the two subarrays is only 18. Now that we understand the problem, let’s explore it further.

Understanding Constraints

Before we delve into the solution, let’s break down the problem constraints:

• 1 <= nums.length <= 1000: The length of the input array is between 1 and 1000.
• 0 <= nums[i] <= 10^6: The values in the array are non-negative integers.
• 1 <= k <= min(50, nums.length): The number of subarrays k is a positive integer, with a maximum of 50.

Split Array Largest Sum LeetCode Problem Solution

Bruteforce Approach

One way to approach this problem is through brute force, where we explore all possible combinations of splitting the array.

We can use backtracking to evaluate every potential split.

However, this method is not the most efficient and might not be accepted on LeetCode due to performance constraints.

The time complexity for this approach is O(n^2 * m), where n is the size of the input array and m is the value of k.

Here’s a simple example of what the code might look like:

def splitArray(nums, k):
    def canSplit(largest):
        subarray = 1
        curSum = 0
        for n in nums:
            curSum += n
            if curSum > largest:
                subarray += 1
                curSum = n
        return subarray <= k

    result = float('inf')

    for i in range(1, len(nums) + 1):
        for j in range(sum(nums)):
            if canSplit(j):
                result = min(result, j)

    return result

Because this code is a brute force solution, it might not be efficient enough for larger inputs or get accepted on LeetCode.

Therefore, we’ll explore a more efficient approach next.

An Efficient Approach: Binary Search

The efficient approach to solving this problem involves using binary search.

This approach is more complex, but it provides a significant improvement in performance.

The main idea is to use binary search to find the minimum value that can be the largest sum among the subarrays.

1. We initialize the left boundary as the maximum value in nums (smallest possible result) and the right boundary as the sum of all elements in nums (largest possible result).
2. We then perform a binary search within this range to find the smallest maximum sum.
3. To check if a given mid value is valid, we use a greedy algorithm.

We iterate through the nums array and keep adding elements to a current subarray until the sum exceeds mid.

At this point, we increment the subarray count and start a new subarray.

If the subarray count exceeds k, we know that the current mid value is too small, so we update our binary search range.

4. If the subarray count does not exceed k, it means we can split the array into k subarrays where the maximum sum is less than or equal to mid.

In this case, we update our result and search for smaller values by narrowing the right boundary of the binary search.

Here’s the efficient Python code solution:

def splitArray(nums, k):
    def canSplit(largest):
        subarray = 1
        curSum = 0
        for n in nums:
            curSum += n
            if curSum > largest:
                subarray += 1
                curSum = n
        return subarray <= k

    left, right = max(nums), sum(nums)
    result = right

    while left <= right:
        mid = left + ((right - left) // 2)
        if canSplit(mid):
            result = mid
            right = mid - 1
        else:
            left = mid + 1

    return result

This binary search solution ensures that we find the minimum largest sum efficiently.

Now, let’s dive into the time and space complexity analysis.

Time and Space Complexity

• Time Complexity: The time complexity of the binary search solution is O(n * log(s)), where n is the size of the input array nums, and s is the sum of all elements in nums.

The binary search narrows down the search range, and for each midpoint calculation, we perform a linear-time check to validate if we can split the array into k subarrays.

This is an efficient algorithm that works well even for large input arrays.

• Space Complexity: The space complexity is O(1) since we are using a constant amount of extra space, regardless of the input size.

Reasoning Behind Our Approach

The binary search approach to solving the Split Array Largest Sum problem is an elegant solution that effectively narrows down the search space to find the minimum largest sum of subarrays.

It combines binary search techniques with a greedy algorithm to efficiently solve a challenging problem.

We start with a range of possible results, with the smallest and largest values that the result could be.

Through binary search, we iteratively refine this range, finding the minimum value that satisfies the problem’s constraints.

The greedy algorithm ensures that we can split the array into k subarrays while keeping the maximum sum as small as possible.

This approach minimizes the largest sum, allowing us to efficiently determine the optimal splitting of the input array.

The code is structured to provide a clear and effective solution, making it a valuable addition to your problem-solving toolkit.

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Conclusion

In this blog post, we explored the Split Array Largest Sum problem, a challenging LeetCode problem that falls under the category of binary search.

We discussed two approaches: a brute force method and an efficient binary search solution.

The binary search approach is the recommended one, as it offers better performance for larger inputs.

By understanding the problem constraints and leveraging binary search, we can find the minimum largest sum of subarrays, efficiently solving the problem.

This solution combines a clever algorithm with code that is both concise and readable.

I hope you found this explanation helpful, and

it aids you in your problem-solving journey.

If you have any questions or suggestions, please feel free to comment, ask questions, or share this content.

Happy coding!

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Thank you for reading and happy coding!