410.split-array-largest-sum

Statement

简体中文English

Metadata

• Link: 分割数组的最大值
• Difficulty: Hard
• Tag: 贪心 数组 二分查找 动态规划

给定一个非负整数数组 nums 和一个整数 m ，你需要将这个数组分成 m 个非空的连续子数组。

设计一个算法使得这 m 个子数组各自和的最大值最小。

 

示例 1：

输入：nums = [7,2,5,10,8], m = 2
输出：18
解释：
一共有四种方法将 nums 分割为 2 个子数组。 
其中最好的方式是将其分为 [7,2,5] 和 [10,8] 。
因为此时这两个子数组各自的和的最大值为18，在所有情况中最小。

示例 2：

输入：nums = [1,2,3,4,5], m = 2
输出：9

示例 3：

输入：nums = [1,4,4], m = 3
输出：4

 

提示：

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

Metadata

• 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

Python 3

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)

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最后更新: October 11, 2023