largest sum java_LeetCode 410. 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:

1 classSolution {2 public int splitArray(int[] nums, intm) {3 if(nums == null || nums.length == 0){4 return 0;5 }6

7 int n =nums.length;8 int [][] dp = new int[m][n];9 dp[0][0] = nums[0];10 for(int j = 1; j < n; j++){11 dp[0][j] = dp[0][j - 1] +nums[j];12 }13

14 for(int i = 1; i < m; i++){15 for(int j = i; j < n; j++){16 int min =Integer.MAX_VALUE;17 for(int k = 0; k < j; k ++){18 min = Math.min(min, Math.max(dp[i - 1][k], dp[0][j] - dp[0][k]));19 }20

21 dp[i][j] =min;22 }23 }24

25 return dp[m - 1][n - 1];26 }27 }