## 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 = 2Output:18Explanation: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 = 2Output:9

**Example 3:**

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

**Constraints:**

`1 <= nums.length <= 1000`

`0 <= nums[i] <= 10`

^{6}`1 <= m <= min(50, nums.length)`

## Explanation

One approach to solving the problem is to use binary search. We can use binary search to search for a largest subarray sum which can make the array split into m parts. The lower bound of the largest subarray sum is the maximum number of the array when m equals the length of the array. The higher bound is the sum of the array when m equals 1. We can base on how many pieces split to adjust the largest subarray sum value.

## Python Solution

```
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
"""
:type nums: List[int]
:type m: int
:rtype: int
"""
n = len(nums)
start = max(nums)
end = sum(nums)
while start + 1< end:
mid = start + (end - start) // 2
if self.split_into_pieces(nums, mid) > m:
start = mid + 1
else:
end = mid
if self.split_into_pieces(nums, start) <= m:
return start
return end
def split_into_pieces(self, nums, largest_sum):
previous_sum = 0
pieces = 1
for num in nums:
if previous_sum + num > largest_sum:
previous_sum = num
pieces += 1
else:
previous_sum += num
return pieces
```

- Time Complexity: O(Nlog(N))
- Space Complexity: O(1)

class Solution {

public int firstMissingPositive(int[] A) {

int m = 1;

HashSet set = new HashSet();

for(int num: nums){

if(num>m){

set.add(num);

}else if(num==m){

m++;

while(set.contains(m)){

set.remove(m);

m++;

}

}

}

return m;

}

}