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.

Thinking: binary answer: if segments smaller space, close to n, the lowest sum is max (A [i]), if there is only one interval, then that sum [0 ~ n-1];

Then becomes between the lower bound and the upper bound binary search the solution space, if the average value is a number K, greater than the number K of the number of sections can not split into sections, it is greater than K m represents may further increase, smaller than m representatives, K too, should be a small point;

Note that the last judgment when the first judgment start, and then determine the end; there it is possible to sum burst, into long on it.

Time: Log (sum (Ai)) * n Space: The (N)

class Solution {
    public int splitArray(int[] nums, int m) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        long start = 0, end = 0;
        for(int i = 0; i < nums.length; i++) {
            start = Math.max(start, nums[i]);
            end += nums[i];
        }
        while(start + 1 < end) {
            long mid = start + (end - start) / 2;
            if(cansplit(nums, mid) > m) {
                start = mid;
            } else {
                // cansplit(nums, mid) < m;
                end = mid;
            }
        }
        // first check start; then check end;
        if(cansplit(nums, start) > m) {
            return (int)end;
        }
        return (int)start;
    }
    
    private int cansplit(int[] nums, long sumlimit) {
        int count = 0;
        int i = 0;
        long sum = 0;
        while(i < nums.length) {
            while(i < nums.length && sum + nums[i] <= sumlimit) {
                sum += nums[i];
                i++;
            }
            count++;
            sum = 0;
        }
        return count;
    }
}

Ideas 2: This problem can dp, DFS + memo cache

Leftsum prefixsum preceding paragraph may be determined, followed by a cut subproblems m - 1 segment;

Time: O (m * n ^ 2) Space: O (m * n); Dp can be extended to have a negative case, if all is positive, binary search or above predominates;

class Solution {
    public int splitArray(int[] nums, int m) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        int n = nums.length;
        int[] prefixsum = new int[n+1];
        prefixsum[0] = 0;
        for(int i = 1; i <= n; i++) {
            prefixsum[i] = prefixsum[i - 1] + nums[i - 1];
        }
        
        int[][] cache = new int[n][m+1];
        return splitArrayHelper(0, m, cache, nums, prefixsum);
    }
    
    private int splitArrayHelper(int j, int m, int[][] cache, int[] nums,
                                int[] prefixsum) {
        int n = nums.length;
        if(m == 1) {
            return prefixsum[n] - prefixsum[j];
        }
        if(cache[j][m] != 0) {
            return cache[j][m];
        }
        int res = Integer.MAX_VALUE;
        for(int k = j; k < n - 1; k++) {
            // j ~k sum;
            int leftsum = prefixsum[k+1] - prefixsum[j];
            // k+1 ~ n sum;
            int rightsum = splitArrayHelper(k+1, m - 1, cache, nums, prefixsum);
            res = Math.min(res, Math.max(leftsum, rightsum));
        }
        cache[j][m] = res;
        return res;
    }
}