# Solving Leetcode Interviews in Seconds with AI: Split Array Largest Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "410" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any subarray is minimized. Return the minimized largest sum of the split. A subarray is a contiguous part of the array.   Example 1:  Input: nums = [7,2,5,10,8], k = 2 Output: 18 Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [7,2,5] and [10,8], where the largest sum among the two subarrays is only 18.  Example 2:  Input: nums = [1,2,3,4,5], k = 2 Output: 9 Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.    Constraints:  1 <= nums.length <= 1000 0 <= nums[i] <= 106 1 <= k <= min(50, nums.length)  

	# Explanation
	Here's the breakdown of the solution:

*   **Binary Search:** We use binary search to find the minimized largest sum. The search space is between the maximum element in `nums` (lower bound) and the sum of all elements in `nums` (upper bound).
*   **Feasibility Check:** For a given "mid" value (potential largest sum), we check if it's possible to split the array into `k` or fewer subarrays such that the sum of each subarray is no more than `mid`.
*   **Adjust Search Space:** Based on the feasibility check, we adjust the binary search range. If it's possible to split into `k` or fewer subarrays, we try a smaller `mid` (move `right` pointer). Otherwise, we try a larger `mid` (move `left` pointer).

*   **Runtime Complexity:** O(N log(Sum of nums)), where N is the length of `nums`. **Storage Complexity:** O(1).

	
	# Code
	```python
	def splitArray(nums, k):
    """
    Splits an array into k subarrays such that the largest sum of any subarray is minimized.

    Args:
        nums: A list of integers.
        k: The number of subarrays to split into.

    Returns:
        The minimized largest sum of the split.
    """

    def is_possible(max_sum):
        """
        Checks if it is possible to split the array into k or fewer subarrays such that the sum of each subarray is no more than max_sum.
        """
        subarray_count = 1
        current_sum = 0
        for num in nums:
            if current_sum + num <= max_sum:
                current_sum += num
            else:
                subarray_count += 1
                current_sum = num
                if current_sum > max_sum:
                    return False  # Single element exceeds max_sum
        return subarray_count <= k

    left = max(nums)
    right = sum(nums)
    ans = right

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

    return ans
	```