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.
classSolution {publicintsplitArray(int[] nums,int m) {// When every element forms it's own grouplong min =Integer.MIN_VALUE;// When all the elements are in one grouplong max =0;for (int x : nums) { min =Math.max(min, x); max += x; }while (min < max) {// "mid" represents the max sum in each grouplong mid = min + (max - min) /2;int groups =numberOfGroups(nums, mid);if (groups > m) min = mid +1;else max = mid; }return (int) min; }publicintnumberOfGroups(int[] nums,long maxSum) {long currSum =0, groups =1;for (int x : nums) { currSum += x;if (currSum > maxSum) { currSum = x; groups++; } }return (int) groups; }}