LeetCode 410. Split Array Largest Sum

原题链接在这里：https://leetcode.com/problems/split-array-largest-sum/

题目：

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.

题解：

Let dp[i][j] denotes up to index j, with i cut, the minimum of largest sum of subarrays.

Thus with i = 0, that means there is no cut. Initialize dp[0][j] as sum of nums[0] to nums[j].

For dp[i][j], we could cut at (0, 1, .... j -1). And minimize Math.max(dp[i - 1][k], dp[0][j] - dp[0][k]).

Note: update part, i starts from 1, becasue we need dp[i - 1][k].

Time Complexity: O(m * n ^ 2). n = nums.length.

Space: O(m * n).

AC Java:

class Solution {
    public int splitArray(int[] nums, int m) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        
        int n = nums.length;
        int [][] dp = new int[m][n];
        dp[0][0] = nums[0];
        for(int j = 1; j < n; j++){
            dp[0][j] = dp[0][j - 1] + nums[j]; 
        }
        
        for(int i = 1; i < m; i++){
            for(int j = i; j < n; j++){
                int min = Integer.MAX_VALUE;
                for(int k = 0; k < j; k ++){
                    min = Math.min(min, Math.max(dp[i - 1][k], dp[0][j] - dp[0][k]));
                }
                
                dp[i][j] = min;
            }
        }
        
        return dp[m - 1][n - 1];
    }
}