410. Split Array Largest Sum

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.

解法： 求一些最大值的最小值或一些最小值的最大值，是DP的关键词，所以想到DP。这道题的关键是想到让最后的元素分为一组，之前的元素分成 j - 1 组。 

0  1  2  3     0      0  0  0  0    1  7   0  7 ++ ++    2  2   0  9       3  5   0 14    4 10   0 24        1. dp[i][j]: first i elements, split into j groups, minimal of larget sum among the groups.    2. init: dp[i][j]: ++              dp[i][1]: first i elements 1 group, sum.              dp[i][0]: 0             dp[0][i]: 0    3. dp[i][j]:     e.g.    dp[4][3]: 7 2 5 10 into 3 groups的答案    即最后的元素分为一组，之前的元素分成2组的所有答案的最小 - min((i),(ii),(iii))    i) 7 | 2 5 10    max(dp[1][2], sum(2,5,10)) (dp[1][2]: invalid, continue)    ii) 7 2 | 5 10    max(dp[2][2], sum(5,10))    iii) 7 2  5 | 10   max(dp[3][2], sum(10))        formula: dp[i][j] = min{max(dp[k][j - 1], sum(k+1,i))} for k = [1,i-1]

 

1 public int splitArray(int[] nums, int m) { 2         int[][] dp = new int[nums.length + 1][m + 1]; 3         for (int i = 0; i < dp.length; i++) { 4             Arrays.fill(dp[i], Integer.MAX_VALUE); 5         } 6         // init 7         for (int i = 0; i < dp.length; i++) { 8             dp[i][0] = 0; 9         }10         for (int i = 0; i < dp[0].length; i++) {11             dp[0][i] = 0;12         }13         for (int i = 1; i < dp.length; i++) {14             dp[i][1] = dp[i - 1][1] + nums[i - 1];15         }16         17         for (int i = 2; i < dp.length; i++) {18             for (int j = 2; j < dp[0].length; j++) {19                 int ans = Integer.MAX_VALUE;20                 int totalSum = dp[i][1]; // for calculating sum(k+1,i). totalSum =  sum[0, i]21                 int cumSum = 0; // for calculating sum(k+1,i). cumSum of first k. 23                 for (int k = 1; k <= i - 1; k++) {24                     cumSum += nums[k - 1]; // outside, no matter what, add curSum25                     if (dp[k][j - 1] != Integer.MAX_VALUE) { 26                         ans = Math.min(ans, Math.max(dp[k][j - 1], totalSum - cumSum));27                     }28                 }29                 30                 dp[i][j] = ans;31                 32             }33         }34         return dp[dp.length - 1][dp[0].length - 1];35     }