LeetCode 410. Split Array Largest Sum DP 二分

Given an array 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.

Note:
If n is the length of array, assume the following constraints are satisfied:

• 1 ≤ n ≤ 1000
• 1 ≤ m ≤ min(50, n)

 

Examples:

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.

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這好像是一道很著名的問題，明顯的解法就是DP，但是剛開始腦子進水，用f(x,y,p)表示從下標x到下標y分p份時候最大字段和的最小結果，然後生生搞成了5維DP。降維也很簡單，利用f(x,p)表示從下標0到下標x分成p份時候最大字段和的最小結果。那麼

f(x,p) = min{max[f(x0,p-1), sum(nums[x0+1...x])]}，其中0<=x0<x，所以有三個變量：x0依賴於x,x依賴於p，所以三重循環，就先p，再x，最內層x0，最好f(x,p)寫成f(p,x)，最終codes偷懶沒有改：

class Solution:
    def splitArray(self, nums, m: int) -> int:
        l = len(nums)
        f = [[0 for j in range(m+1)] for i in range(l+1)]
        accu = [nums[0] for i in range(l)]
        for i in range(1,l):
            accu[i] = accu[i-1]+nums[i]
        for i in range(l):
            f[i][1] = accu[i]

        for p in range(2,m+1):
            for x in range(p-1,l):
                f[x][p] = accu[l-1]
                for x0 in range(x):
                    f[x][p] = min(f[x][p], max(f[x0][p-1],accu[x]-accu[x0]))
        return f[l-1][m]

s = Solution()
print(s.splitArray(nums = [7,2,5,10,8], m = 2))

DP的問題也很明顯，並沒有利用單調性的特點，既然單調性，那就二分走起啊：

class Solution:
    def get_partion_cnt(self, nums, threshold):
        res, cur = 0, 0
        for num in nums:
            if (cur + num <= threshold):
                cur += num
            else:
                res += 1
                cur = num
        if (cur > 0):
            res += 1
        return res

    def splitArray(self, nums, m: int) -> int:
        total, ma = sum(nums), max(nums)
        # ma<=res<=total
        l, r = ma, total
        while (l <= r):
            mid = l + ((r - l) >> 1)
            pa = self.get_partion_cnt(nums, mid)
            if (pa <= m):
                r = mid - 1
            else:
                l = mid + 1
        return l