410. Split Array Largest Sum

Problem Description

Given an integer array nums and an integer k, we need to split nums into k non-empty subarrays such that the largest sum of any subarray is minimized. The goal is to find this minimized largest sum.

Solution: Binary Search

This problem can be efficiently solved using binary search. The core idea is that the maximum sum of any subarray monotonically increases as we increase the allowed maximum sum for any subarray. This allows us to use binary search to find the minimum possible maximum sum.

Algorithm:

1. Find search space: The minimum possible maximum sum is the maximum element in nums (because at least one subarray will have this element), and the maximum possible maximum sum is the total sum of all elements in nums (in the worst case, we have k=1).

2. Binary Search: Perform binary search within this range. For each mid (potential maximum sum):

   • Check feasibility: Determine if it's possible to split the array into k or fewer subarrays such that the sum of each subarray is at most mid. This can be done greedily: iterate through nums, adding elements to the current subarray until the sum exceeds mid. Each time this happens, we create a new subarray. If we end up with more than k subarrays, then mid is too small.

3. Update boundaries: Based on the feasibility check, update the search space (left or right boundary).

4. Return result: The final left boundary (after the binary search converges) will be the minimum possible maximum sum.

Code Implementation (Python)

import bisect
 
class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        def check(mx):
            s, cnt = 0, 1  # s tracks current subarray sum, cnt counts subarrays
            for x in nums:
                s += x
                if s > mx:
                    s = x
                    cnt += 1
            return cnt <= k
 
        left, right = max(nums), sum(nums)
        #Efficient way to use bisect_left - avoids explicit while loop
        return left + bisect_left(range(left, right + 1), True, key=check)
 

Explanation of Python Code:

• check(mx): This function efficiently checks if it's possible to split the array such that the maximum subarray sum is at most mx. It iterates through nums, creating new subarrays whenever the current sum exceeds mx. It returns True if the number of subarrays is less than or equal to k, otherwise False.

• bisect_left: This function from the bisect module is used to efficiently find the index where True is first encountered in a sorted sequence. This eliminates the need for a manual while loop in binary search, significantly improving performance.

• left and right: These variables define the search space for the binary search.

• The final line performs a concise binary search using bisect_left. It implicitly searches for the smallest mx for which check(mx) returns True.

Time and Space Complexity

• Time Complexity: O(N log S), where N is the length of nums and S is the sum of all elements in nums. The binary search takes O(log S) iterations, and each iteration involves a linear scan of nums in check().

• Space Complexity: O(1). The algorithm uses only a constant amount of extra space.

Code Implementation in other languages (Conceptual Outline)

The logic remains the same for other languages. The key differences lie in the syntax and available library functions. For instance:

• Java: You'd use a while loop for binary search and implement check() similarly. Java doesn't have a direct equivalent of bisect_left, so a manual while loop will be required.

• C++: Similar to Java, you'd manually implement binary search and the check() function.

• JavaScript/TypeScript: Similar to Python, you could use a while loop for binary search, or a more advanced approach using Array.reduce to make the check function concise.

The core algorithm (binary search and greedy check) is language-agnostic and can be adapted easily to other programming languages, maintaining the same time and space complexities.