leetcode410. Dynamic programming of the maximum value of the split array

• hard:https://leetcode.cn/problems/split-array-largest-sum/

• Given an array nums of nonnegative integers and an integer m , 你需要将这个数组分成 m 个非空的连续子数组.

• Design an algorithm to make m 个子数组各自和this 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)

answer

• Let dp[i][j] represent 前 i 个数分割为 j 组the minimum value of the largest continuous subarray sum that can be obtained from the array

• Determine the filling transfer equation (considering that dp[i][j] needs to traverse all the situations that are divided into j-1 groups):
  dp [ i ] [ j ] = mink = 0 i − 1 { max ( dp [ k ] [ j − 1 ] , sub ( k + 1 , i ) ) } = mink = 0 i − 1 { max ( dp [ k ] [ j − 1 ] , sum ( nums [ k + 1 … j ] ) ) } dp[i] [j]= min_{k=0}^{i−1} \{max(dp[k][j−1],sub(k+1,i))\}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = min_{k=0}^{ i−1} \{max(dp[k][j−1] ,sum(nums[k+1...j]))\}dp[i][j]=mink=0i−1​{ max(dp[k][j−1],sub(k+1,i))}                               =mink=0i−1​{ max(dp[k][j−1],sum(nums[k+1…j]))}

• Determining Boundaries: The Form Filling Method

        nums = [7,2,5,10,8]，m=2。

i\j	0	1	2
0	Cannot be divided into 0 groups	INT_MAX	INT_MAX
1	Cannot be divided into 0 groups	7	1 number cannot be divided into 2 groups (i<j)
2	Cannot be divided into 0 groups	7+2	$min([7],[2])=2$
3	Cannot be divided into 0 groups	7+2+5	$min\left[\begin{array}{c}max(dp[1][1],[2,5])\\ max(dp[2][1],[5])\end{array}\right]=7$
4	Cannot be divided into 0 groups	7+2+5+10	$min\left[\begin{array}{c}max\left(dp\right[1\left]\right[1],[2,5,10])\\ max(dp[2][1],[5,10])\\ max(dp[3][1],[10])\end{array}\right]=The\; first\; 14$ * numbers are divided into one group and the rest
5	Cannot be divided into 0 groups	7+2+5+10+8	$min\left[\begin{array}{c}max(dp[1][1],[2,5,10,8])\\ max(dp[2][1],[5,10,8])\\ max(dp[3][1],[10,8])\\ max(dp[4][1],[8])\end{array}\right]=18$

code

class Solution {
    
    
public:
    int splitArray(vector<int>& nums, int m) {
    
    
        int n = nums.size();
        vector<vector<long long>> dp(n + 1, vector<long long>(m + 1, LLONG_MAX));
        vector<long long> sub(n + 1, 0);
        for (int i = 0; i < n; i++) {
    
    
            sub[i + 1] = sub[i] + nums[i];
        }
        dp[0][0] = 0;
        
        for (int i = 1; i <= n; i++) {
    
    
            for (int j = 1; j <= min(i, m); j++) {
    
    
                for (int k = 0; k < i; k++) {
    
    
                    dp[i][j] = min(dp[i][j], max(dp[k][j - 1], sub[i] - sub[k]));
                }
            }
        }
        return (int)dp[n][m];
    }
};