# 410.split-array-largest-sum

## Statement

• Difficulty: Hard
• Tag: `贪心` `数组` `二分查找` `动态规划`

``````输入：nums = [7,2,5,10,8], m = 2

``````输入：nums = [1,2,3,4,5], m = 2

``````

``````输入：nums = [1,4,4], m = 3

``````

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

• Link: Split Array Largest Sum
• Difficulty: Hard
• Tag: `Greedy` `Array` `Binary Search` `Dynamic Programming`

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)`

## Solution

``````from bisect import bisect_left, bisect_right
from typing import List

class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
S = sum(nums) + 1

def get(x: int):
cur = 0
res = 1
for a in nums:
if cur + a > x:
cur = 0
res += 1
cur += a
return -res

return bisect_left(range(S + 1), -m, max(nums), S + 1, key=get)
``````