# LeetCode 410. Split Array Largest Sum

## Description

https://leetcode.com/problems/split-array-largest-sum/

Given an array `nums` which consists of non-negative integers and an integer `m`, you can split the array into `m` non-empty continuous subarrays.

Write an algorithm to minimize the largest sum among these `m` subarrays.

Example 1:

```Input: nums = [7,2,5,10,8], m = 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], m = 2
Output: 9
```

Example 3:

```Input: nums = [1,4,4], m = 3
Output: 4
```

Constraints:

• `1 <= nums.length <= 1000`
• `0 <= nums[i] <= 106`
• `1 <= m <= min(50, nums.length)`

## Python Solution

``````class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
if not nums:
return 0

n = len(nums)
start = max(nums)
end = sum(nums)

while start + 1 < end:
largest_sum = (start + end) // 2
if self.largest_sum_satisfy_m(nums, m, largest_sum):
end = largest_sum
else:
start = largest_sum

if self.largest_sum_satisfy_m(nums, m, start):
return start

return end

def largest_sum_satisfy_m(self, nums, m, largest_sum):
num_of_sub = 0
current_sum = 0

for num in nums:
if current_sum + num <= largest_sum:
current_sum += num
else:
num_of_sub += 1
current_sum = num
num_of_sub += 1

return num_of_sub <= m
``````
• Time Complexity: ~NlogX where X = sum(nums) – max(nums)
• Space Complexity: ~1.