LeetCode 410 - Split Array Largest Sum


http://bookshadow.com/weblog/2016/10/02/leetcode-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:
Given m satisfies the following constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.
Examples:
Input:
nums = [1,2,3,4,5]
m = 2

Output:
9

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5],
where the largest sum among the two subarrays is only 9.
X. Bisection method + greedy
Time complexity: O(log(sum(nums))*n)
https://leetcode.com/problems/split-array-largest-sum/discuss/89820/Explanation-%2B-Ruby-11-liner
For the binary search, we just need to be able to determine for a certain cap whether it's ok. Meaning whether we can split the array into m subarrays so that no subarray's sum is over this cap. To do that, we determine the minimum number of subarrays we can do with this cap. If it's smaller than or equal to m, this cap is ok. Wait, aren't we supposed to create exactly m subarrays? Yeah, but if we can do it with fewer, we can just split some up so we have exactly m.
So how to find the minimum number of subarrays we can do with a certain cap? We can do that greedily. Of course it's smart to stuff as many numbers into the first subarray as we can - then we don't need to try to fit them into the remaining subarrays. Same for the remaining subarrays - always stuff as many numbers into them as we can, only start a new subarray when the current number doesn't still fit into the previous subarray.
The initial range for possible caps is from the maximum in nums (because with a lower cap, we couldn't fit that number into any subarray) to the sum of nums (a larger cap would be useless - we'll never need more room than all numbers combined)
https://leetcode.com/problems/split-array-largest-sum/discuss/89819/C%2B%2B-Fast-Very-clear-explanation-Clean-Code-Solution-with-Greedy-Algorithm-and-Binary-Search
We can break this problem into two smaller problems:
  • Given an array (A), number of cuts (CUTS), and the Largest sum of sub-arrays (MAX). Can you use at most CUTS cuts to segment array A into CUTS + 1 sub-arrays, such that the sum of each sub-array is smaller or equal to MAX?
  • Given a lower bound (left), an upper bound (right), an unknown bool array (B), and an API uses i as input and tells you whether B[i] is true. If we know there exists an index k, that B[i] is false when i < k, and B[i] is true when i >= k. What is the fastest way to find this k (the lower bound)?

Solution to the first sub-problem (Skip this part if you already knew how to solve 1st sub-problem):

For the first question, we can follow these steps:
  • For each element in the array, if its value is larger than MAX, we know it's not possible to cut this array into groups that the sum of all groups are smaller than MAX. (Reason is straightforward, if A is [10, 2, 3, 5] and MAX is 6, even you have 3 cuts by which you can cut A as [[10], [2], [3], [5]], the group containing 10 will still be larger than 6).
  • Use greedy algorithm to cut A. Use an accumulator ACC to store the sum of the currently processed group, and process elements in A one by one. For each element num, if we add num with ACC and the new sum is still no larger than MAX, we update ACC to ACC + num, which means we can merge num into the current group. If not, we must use a cut before num to segment this array, then num will be the first element in the new group.
  • If we didn't go through A but already used up all cuts, then it's not possible only using CUTS cuts to segment this array into groups to make sure sum of each sub-array is smaller than MAX. Otherwise, if we can reach the end of A with cuts left (or use exactly CUTS cuts). It's possible to do so.
Then the first question is solved.

Solution to the second sub-problem(Skip this part if you already knew how to solve 2nd sub-problem):

  • The array B will be something like [false, false, ..., false, true, true, ..., true]. We want to find the index of the first true.
  • Use binary search to find this k. Keep a value mid, mid = (left + right) / 2. If B[mid] = false, then move the search range to the upper half of the original search range, a.k.a left = mid + 1, otherwise move search range to the lower half, a.k.a right = mid.

Why this algorithm is correct...

  • No matter how we cut the array A, the Largest sum of sub-arrays will fall into a range [left, right]. Left is the value of the largest element in this array. right is the sum of this array. (e.g., Given array [1, 2, 3, 4, 5], if we have 4 cuts and cut it as [[1], [2], [3], [4], [5]], the Largest sum of sub-arrays is 5, it cannot be smaller. And if we have 0 cut, and the only sub-array is [[1, 2, 3, 4, 5]], the Largest sum of sub-arrays is 15, it cannot be larger).
  • However, we cannot decide the number of cuts (CUTS), this is an given constraint. But we know there must be a magic number k, which is the smallest value of the Largest sum of sub-arrays when given CUTS cuts. When the Largest sum of sub-arrays is larger than k, we can always find a way to cut A within CUTS cuts. When the Largest sum of sub-arrays is smaller than k, there is no way to do this.

Example

For example, given array A [1, 2, 3, 4, 5]. We can use 2 cuts.
  • No matter how many cuts are allowed, the range of the possible value of the Largest sum of sub-arrays is [5, 15].
  • When given 2 cuts, we can tell the magic number k here is 6, the result of segmentation is [[1, 2, 3], [4], [5]].
  • When Largest sum of sub-arrays is in range [6, 15], we can always find a way to cut this array within two cuts. You can have a try.
  • However, when Largest sum of sub-arrays is in range [5, 5], there is no way to do this.
  • This mapped this problem into the second sub-problem. Bool array B here is [5:false, 6:true, 7:true, 8:true, ..., 15:true]. We want to find the index i of the first true in B, which is the answer of this entire question, and by solving the first sub-problem, we have an API that can tell us given an i (Largest sum of sub-arrays), whether B[i] is true (whether we can find a way to cut A to satisfy the constraint).


Below is the code with comment, just in case you don't have time to read the explanations above.
https://discuss.leetcode.com/topic/61315/java-easy-binary-search-solution-8ms
https://leetcode.com/problems/split-array-largest-sum/discuss/89817/Clear-Explanation%3A-8ms-Binary-Search-Java
  1. The answer is between maximum value of input array numbers and sum of those numbers.
  2. Use binary search to approach the correct answer. We have l = max number of array; r = sum of all numbers in the array;Every time we do mid = (l + r) / 2;
  3. Use greedy to narrow down left and right boundaries in binary search.
    3.1 Cut the array from left.
    3.2 Try our best to make sure that the sum of numbers between each two cuts (inclusive) is large enough but still less than mid.
    3.3 We'll end up with two results: either we can divide the array into more than m subarrays or we cannot.
    If we can, it means that the mid value we pick is too small because we've already tried our best to make sure each part holds as many non-negative numbers as we can but we still have numbers left. So, it is impossible to cut the array into m parts and make sure each parts is no larger than mid. We should increase m. This leads to l = mid + 1;
    If we can't, it is either we successfully divide the array into m parts and the sum of each part is less than mid, or we used up all numbers before we reach m. Both of them mean that we should lower mid because we need to find the minimum one. This leads to r = mid - 1;

  1. Given a result, it is easy to test whether it is valid or not.
  2. The max of the result is the sum of the input nums.
  3. The min of the result is the max num of the input nums.
    Given the 3 conditions above we can do a binary search. (need to deal with overflow)
    public int splitArray(int[] nums, int m) {
        long sum = 0;
        int max = 0;
        for(int num: nums){
            max = Math.max(max, num);
            sum += num;
        }
        return (int)binary(nums, m, sum, max);
    }
    
    private long binary(int[] nums, int m, long high, long low){
        long mid = 0;
        while(low < high){
            mid = (high + low)/2;
            if(valid(nums, m, mid)){
                //System.out.println(mid);
                high = mid;
            }else{
                low = mid + 1;
            }
        }
        return high;
    }
    
    private boolean valid(int[] nums, int m, long max){//canSplit
        int cur = 0;
        int count = 1;
        for(int num: nums){
            cur += num;
            if(cur > max){
                cur = num;
                count++;
                if(count > m){
                    return false;
                }
            }
        }
        return true;
    }

https://discuss.leetcode.com/topic/61324/clear-explanation-8ms-binary-search-java
    1. The answer is between maximum value of input array numbers and sum of those numbers.
    2. Use binary search to approach the correct answer. We have l = max number of array; r = sum of all numbers in the array;Every time we do mid = (l + r) / 2;
    3. Use greedy to narrow down left and right boundaries in binary search.
      3.1 Cut the array from left.
      3.2 Try our best to make sure that the sum of numbers between each two cuts (inclusive) is large enough but still less than mid.
      3.3 We'll end up with two results: either we can divide the array into more than m subarrays or we cannot.
    http://blog.csdn.net/mebiuw/article/details/52724293
    /** * 这个数肯定介于最大的那一个单值和所有元素只和的中间 * */ public int splitArray(int[] nums, int m) { long sum = 0; int max = 0; for(int num: nums){ max = Math.max(max, num); sum += num; } return (int)binarySearch(nums, m, sum, max); } //二分查找 private long binarySearch(int[] nums, int m, long high, long low){ long mid = 0; while(low < high){ mid = (high + low)/2; //验证是否满足,也就是这么大的值有可能出现么 if(valid(nums, m, mid)){ high = mid; }else{ low = mid + 1; } } return high; } /** * 验证这个值是否合法 * */ private boolean valid(int[] nums, int m, long max){ int cur = 0; int count = 1; //是否有多余m个片段or区间，大于给定值的max的，如果有了，那么就不合法了，因为这样划分就不止m个，及max太小 for(int num: nums){ cur += num; if(cur > max){ cur = num; count++; if(count > m){ return false; } } } return true; }
    二分枚举答案（Binary Search）
    将数组nums拆分成m个子数组，每个子数组的和不小于sum(nums) / m，不大于sum(nums)
    又因为数组nums中只包含非负整数，因此可以通过二分法在上下界内枚举答案。
    时间复杂度O(n * log m)，其中n是数组nums的长度，m为数组nums的和
    http://www.cnblogs.com/grandyang/p/5933787.html
    https://all4win78.wordpress.com/2016/10/22/leetcode-410-split-array-largest-sum/
    另一种就是利用sum是整数的性质进项binary search。UB (upper bound) 是所有原array的所有数字的和，LB (lower bound) 是UB/m以数组中最大值两者之间的较大者。然后根据binary search的方式去验证给定的array能否依据这个largest sum进行分割。需要注意的是，这里sum需要使用long而不是int，不然会有overflow的问题。世间复杂度是O(n log sum) <= O(n log n*Integer.MAX_VALUE) <= O(n (logn + c)) = O(n logn)，额外的空间复杂度为O(1)。
    http://blog.csdn.net/mebiuw/article/details/52724293

    http://blog.csdn.net/u014688145/article/details/69525838
    如果直接按照上述的意思去写代码时，你会发现实操非常困难，所以必须对问题归简，这道题还可以这样理解，求最小的最大，关键在于最小，而不是最大！真正做约束的在于最小，针对每种划分情况，最大是每种划分的固有属性，所以完全可以不用考虑，否则增添理解题目的负担。那么最小的含义在于，对于m=2的情况，我们是尽可能的让左右两部分平衡，也就是说sum(左数组) = sum(右数组)，也就是平均分配。
    先说说我的想法吧，但我的代码没有AC，后来发现，这种做法没法实现m >= 3的情况，太可惜了，但m == 2是完全适用的。
    我的解法典型的给你nums,m想办法去求解minMaxsum，而大神的思路是假设我们已经在解空间里有了一系列minMaxsums，去搜索一个minMaxsum使得符合m，这让我非常震撼。
    很简单，可能的minMaxsum有哪些，中间的哪些minMaxsum我们是不知道的，这是问题的关键！所以这个问题就假设最极端的两头情况：
    • 当 m = 1 时，这种情况，minMaxsum = 整个数组之和。
    • 当 m = 数组长度时，这种情况，minMaxsum = 数组中最大的那个元素。
    如：[7,2,5,10,8] m = 1， 输出 minMaxsum = 32
    而：[7,2,5,10,8] m = 5， 输出 minMaxsum = 10
    这里让我对二分查找有了重新的认识：
    • 下标不一定是数组的下标，求解问题的解空间同样可以当作下标，满足状态递增就行。
    • 一定要符合状态空间中元素的有序性么？不一定，只要在搜索时，能够有约束条件排除左半或者右半即可。


    搜索解空间，找对应的m，符合情况就输出，很好的一个思路。
    public int splitArray(int[] nums, int m) { long sum = 0; int max = 0; for(int num: nums){ max = Math.max(max, num); sum += num; } return (int)binary(nums, m, sum, max); } private long binary(int[] nums, int m, long high, long low){ long mid = 0; while(low < high){ mid = (high + low)/2; if(valid(nums, m, mid)){ high = mid; }else{ low = mid + 1; } } return high; } private boolean valid(int[] nums, int m, long max){ int cur = 0; int count = 1; for(int num: nums){ cur += num; if(cur > max){ cur = num; count++; if(count > m){ return false; } } } return true; }

    X. DP O(mn) TODO
    https://leetcode.com/problems/split-array-largest-sum/discuss/89822/DP-O(nm)-Solution
    The O(n^2m) DP solution to this problem is pretty obvious, where the transition function can be written as
    dp[i][j] = min{max{dp[k][j-1], subsum(k+1, i)}}, 0 <= k < i
    where dp[i][j] is the optimal result for splitting nums[:i+1] into j subarrays. And then for each i, j you would require O(n) time to find the k that minimizes dp[i][j], which makes this an overall O(n^2m) algorithm..
    The key to reducing time complexity down to O(nm) is the monotonic properties that dp and subsum hold. Mathematically I found myself hard to explain this well, but intuitively, if the last subarray for dp[i][j] is nums[k], nums[k+1], ..., nums[i], then for dp[i][j+1] the last subarray would always be some nums[k+x], ..., nums[i], x>=0, because if you were cutting an array evenly into j + 1 continuous subarray, the last subarray would always be smaller than it would had been using one less cut. So every time you find a k that minimizes dp[i][j] you only need to consider subarray starting from or after k when computing dp[i][j+1].
    class Solution(object):
    def splitArray(self, nums, m):
        n = len(nums)
        if not n:
            return 0
        pre_sum = [0] * (n + 1)
        for i in range(n):
            pre_sum[i + 1] = pre_sum[i] + nums[i]
        sub_sum = lambda i, j: pre_sum[j + 1] - pre_sum[i]

        dp = [[0 for _ in range(m + 1)] for _ in range(n)]
        for i in range(n):
            dp[i][1] = pre_sum[i + 1]
            k = 0
            for j in range(2, min(m + 1, i + 2)):
                while k < i - 1 and max(dp[k][j-1], sub_sum(k+1, i)) > max(dp[k+1][j-1], sub_sum(k+2,i)):
                    k += 1
                dp[i][j] = max(dp[k][j - 1], sub_sum(k+1, i))
        return dp[n - 1][m]

    X. DP O(n^2m)
    http://hehejun.blogspot.com/2018/02/leetcodesplit-array-largest-sum.html
    假设前i个数字被切分成j个subarray，j个subarray的区间和中最大值为x，我们用DP[i][j]表示所有切分的可能中，x的最小值。我们可以有递推公式：

    • dp[i][j] = min(max(dp[k][j - 1], sum(k, i - 1))), where k <= i - 1
    • base case：dp[i][1] = sum(0, i - 1)
    我们可以先求出presum array，这样当我们需要计算sum(k, i - 1)可以很快算得。假设输入数组array长度为n，要求切分为m个subarray，算法时间复杂度和空间复杂度均为O(n * m)，代码如下：
        int splitArray(vector<int>& nums, int m) {
            int len = nums.size(), sum = 0;
            //dp[i][j] represents minimum largest sum by dividing first i characters into j subarrays
            //dp[i][j] = min(max(dp[k][j - 1], sum(k + 1, i)))
            //base case: dp[i][1] = presum[i] - 0(since we need to get dp[k][j - 1] when calculate dp[i][j]), dp[i][0] = INT_MAX, dp[0][i] = INT_MAX;
            vector<vector<int>> dp(len + 1, vector<int>(m + 1, INT_MAX));
            vector<int> preSums(1, 0);
            for(auto& num : nums){sum += num;preSums.push_back(sum);};
            for(int i = 0; i < len; ++i)
            {
                dp[i + 1][1] = preSums[i + 1] - preSums[0];
                for(int j = 2; j <= min(i + 1, m); ++j)
                {
                    for(int k = i; k >= j - 1; --k)
                    {
                        dp[i + 1][j] = min(max(dp[k][j - 1], preSums[i + 1] - preSums[k]), dp[i + 1][j]);
                    }
                }
            }
            return dp[len][m];
        }

    https://zhuhan0.blogspot.com/2017/08/leetcode-410-split-array-largest-sum.html
    1. Sub-problem: minimize the largest sum among a (a < m) sub-arrays of a sub-array of nums.
    2. Function: m is the rows and nums is the columns. f[i][j] = min(max(f[i - 1][k] + nums[k] + nums[k + 1] + ... + nums[j])).
    3. Initialization: f[1][i] = prefix sum of nums[i].
    4. Answer: f[m][nums.length].

    Time complexity: O(mn^2).
    If the prefix sums are stored in a separate array, the space complexity can be optimized by using one-dimensional DP.
        public int splitArray(int[] nums, int m) {
        if (m <= 0) {
            return -1;
        }
        if (nums.length == 0) {
            return 0;
        }
        
        int[][] f = new int[m][nums.length];
        f[0][0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            f[0][i] = f[0][i - 1] + nums[i];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = i; j < nums.length; j++) {
                int min = Integer.MAX_VALUE;
                for (int k = i - 1; k < j; k++) {
                    min = Math.min(min, Math.max(f[i - 1][k], f[0][j] - f[0][k]));
                }
                f[i][j] = min;
            }
        }
        return f[m - 1][nums.length - 1];
    }

    http://www.cnblogs.com/grandyang/p/5933787.html
    我们建立一个二维数组dp，其中dp[i][j]表示将数组中前j个数字分成i组所能得到的最小的各个子数组中最大值，初始化为整型最大值，如果无法分为i组，那么还是保持为整型最大值。为了能快速的算出子数组之和，我们还是要建立累计和数组，难点就是在于要求递推公式了。我们来分析，如果前j个数字要分成i组，那么i的范围是什么，由于只有j个数字，如果每个数字都是单独的一组，那么最多有j组；如果将整个数组看为一个整体，那么最少有1组，所以i的范围是[1, j]，所以我们要遍历这中间所有的情况，假如中间任意一个位置k，dp[i-1][k]表示数组中前k个数字分成i-1组所能得到的最小的各个子数组中最大值，而sums[j]-sums[k]就是后面的数字之和，我们取二者之间的较大值，然后和dp[i][j]原有值进行对比，更新dp[i][j]为二者之中的较小值，这样k在[1, j]的范围内扫过一遍，dp[i][j]就能更新到最小值，我们最终返回dp[m][n]即可，博主认为这道题所用的思想应该是之前那道题Reverse Pairs中解法二中总结的分割重现关系(Partition Recurrence Relation)，由此看来很多问题的本质都是一样，但是披上华丽的外衣，难免会让人有些眼花缭乱了
        int splitArray(vector<int>& nums, int m) {
        int n = nums.size();
        vector<int> sums(n + 1, 0);
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MAX));
        dp[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            sums[i] = sums[i - 1] + nums[i - 1];
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                for (int k = i - 1; k < j; ++k) {
                    int val = max(dp[i - 1][k], sums[j] - sums[k]);
                    dp[i][j] = min(dp[i][j], val);
                }
            }
        }
        return dp[m][n];
    }

    DP的话实现起来稍微麻烦一点，需要保存一个2d的array，大小为n*m，每个位置（i，j）保存从数列位置i开始到数列尾端，分割成j份的largest sum。同时空间和时间复杂度在m比较大的情况下比较高，并不是很建议勇这种方法implement，但是可以作为一种思路。类似的题目可以参考Burst Balloons。
    https://discuss.leetcode.com/topic/66289/java-dp
    public int splitArray(int[] nums, int m) {
  int[][] memo = new int[nums.length][m + 1];
  int[] sum = new int[nums.length];
  sum[nums.length - 1] = nums[nums.length - 1];
  for(int i = nums.length - 2; i >= 0; i--){
   sum[i] = sum[i + 1] + nums[i];
  }
     return findSA(nums, 0, m, sum, memo);
 }

 public int findSA(int[] nums, int start, int m, int[] sums, int[][] memo){
  if(m == 1) return sums[start];
  if(memo[start][m] > 0)
   return memo[start][m];
  int min = Integer.MAX_VALUE, sum = 0;
  for(int i = start; i <= nums.length - m; i++){
   sum += nums[i];
   min = Math.min(Math.max(sum, findSA(nums, i + 1, m - 1, sums, memo)), min);
  }
  memo[start][m] = min;
  return memo[start][m];
 }

    time complexity of naive dp is O(n2 * m)
        public int splitArray(int[] nums, int K) {
        int[] s = new int[nums.length];
        s[0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            s[i] = nums[i] + s[i - 1];
        }
        
        for (int k = 2; k <= K; k++) {
            for (int i = nums.length - 1; i >= k - 1; i--) {
                int min = Integer.MAX_VALUE;
                int left = nums[i];
                for (int p = i - 1; p >= k - 2; p--) {
                    min = Math.min(min, Math.max(s[p], left));
                    left += nums[p];
                    if (left >= min) {
                        break;
                    }
                }
                s[i] = min;
            }
        }
        
        return s[nums.length - 1];
    }
    https://discuss.leetcode.com/topic/61405/dp-java
    This is obviously not as good as the binary search solutions; but it did pass OJ.
    dp[s,j] is the solution for splitting subarray n[j]...n[L-1] into s parts.
    dp[s+1,i] = min{ max(dp[s,j], n[i]+...+n[j-1]) }, i+1 <= j <= L-s
    This solution does not take advantage of the fact that the numbers are non-negative (except to break the inner loop early). That is a loss. (On the other hand, it can be used for the problem containing arbitrary numbers)
    public int splitArray(int[] nums, int m)
{
    int L = nums.length;
    int[] S = new int[L+1];
    S[0]=0;
    for(int i=0; i<L; i++)
        S[i+1] = S[i]+nums[i];

    int[] dp = new int[L];
    for(int i=0; i<L; i++)
        dp[i] = S[L]-S[i];

    for(int s=1; s<m; s++)
    {
        for(int i=0; i<L-s; i++)
        {
            dp[i]=Integer.MAX_VALUE;
            for(int j=i+1; j<=L-s; j++)
            {
                int t = Math.max(dp[j], S[j]-S[i]);
                if(t<=dp[i])
                    dp[i]=t;
                else
                    break;
            }
        }
    }

    return dp[0];
}
    Elegant solution. Based on your solution I did some optimization, to short cut some un-necessary search:
    public static int splitArray(int[] nums, int m) {
    int[] dp = new int[nums.length];

    for(int i = nums.length-1; i>=0; i--) 
        dp[i] = i== nums.length -1 ? nums[i] : dp[i+1] + nums[i];
    for(int im = 2; im <= m; im ++) {
        int maxPart = nums.length + 1 - im;
        for(int i=0; i<maxPart; i++) {
            dp[i] = Integer.MAX_VALUE;
            int leftSum = 0;
            for(int p=i; p<maxPart; p++) {
                leftSum += nums[p];
                if(leftSum > dp[i])
                    break;  // There's no more better soluiton, stop the search.
                int val = Math.max(leftSum, dp[p+1]);
                if(val < dp[i])
                    dp[i] = val;
            }
            if(im == m)  // The last round, get first one is enough
                break;
        }
    }        
    return dp[0];
}
    You can add the sum on the run, you can compare the sum only since sum is accending and dp[j+1] is descending. Unlike the binary search solution, O(nlogk) (k: related to the range of num), this one is not related to the num range, O(n^2*m), I think it could do better for certain test cases, such as [700000000,200000000,500000000,1000000000,800000000], 2
    public int splitArray(int[] nums, int m) {
    int n = nums.length;
    int[] dp = new int[n+1];
    for (int i = n-1; i >= m-1; --i) {
        dp[i] = dp[i+1] + nums[i];
    }
    for (int k = 2; k <= m; ++k) {
        for (int i = m-k; i <= n-k; ++i) {
            dp[i] = Integer.MAX_VALUE;
            for (int j = i, sum = 0; j <= n-k; ++j) {
                sum += nums[j];
                if (sum >= dp[i]) break;
                dp[i] = Math.max(sum, dp[j+1]);
            }
            if (k == m) break;
        }
    }
    return dp[0];
}
    X. DP O(n) space
    https://leetcode.com/problems/split-array-largest-sum/discuss/89816/DP-Java
    dp[s,j] is the solution for splitting subarray n[j]...n[L-1] into s parts.
    dp[s+1,i] = min{ max(dp[s,j], n[i]+...+n[j-1]) }, i+1 <= j <= L-s
    This solution does not take advantage of the fact that the numbers are non-negative (except to break the inner loop early). That is a loss. (On the other hand, it can be used for the problem containing arbitrary numbers)
    public int splitArray(int[] nums, int m)
{
    int L = nums.length;
    int[] S = new int[L+1];
    S[0]=0;
    for(int i=0; i<L; i++)
        S[i+1] = S[i]+nums[i];

    int[] dp = new int[L];
    for(int i=0; i<L; i++)
        dp[i] = S[L]-S[i];

    for(int s=1; s<m; s++)
    {
        for(int i=0; i<L-s; i++)
        {
            dp[i]=Integer.MAX_VALUE;
            for(int j=i+1; j<=L-s; j++)
            {
                int t = Math.max(dp[j], S[j]-S[i]);
                if(t<=dp[i])
                    dp[i]=t;
                else
                    break;
            }
        }
    }

    return dp[0];
}
    Based on your solution I did some optimization, to short cut some un-necessary search:
    public static int splitArray(int[] nums, int m) {
    int[] dp = new int[nums.length];

    for(int i = nums.length-1; i>=0; i--) 
        dp[i] = i== nums.length -1 ? nums[i] : dp[i+1] + nums[i];
    for(int im = 2; im <= m; im ++) {
        int maxPart = nums.length + 1 - im;
        for(int i=0; i<maxPart; i++) {
            dp[i] = Integer.MAX_VALUE;
            int leftSum = 0;
            for(int p=i; p<maxPart; p++) {
                leftSum += nums[p];
                if(leftSum > dp[i])
                    break;  // There's no more better soluiton, stop the search.
                int val = Math.max(leftSum, dp[p+1]);
                if(val < dp[i])
                    dp[i] = val;
            }
            if(im == m)  // The last round, get first one is enough
                break;
        }
    }        
    return dp[0];
}

    X.  DFS + Cache
        public int splitArray(int[] nums, int m) {
        int n = nums.length;
        int[] presum = new int[n+1];
        presum[0] = 0;
        
        for (int i = 1; i <= n; i++) {
            presum[i] += nums[i-1] + presum[i-1];
        }
        
        int[][] visited = new int[n][m+1];
        return dfs(0, m, nums, presum, visited);
    }
    
    private int dfs(int start, int m, int[] nums, int[] presum, int[][] visited) {
        if (m == 1) {
            return presum[nums.length] - presum[start];
        }
        
        if (visited[start][m] != 0) {
            return visited[start][m];
        }
        
        int maxSum = Integer.MAX_VALUE;
        
        for (int i = start; i < nums.length-1; i++) {
            int l = presum[i+1] - presum[start];
            int rightIntervalMax = dfs(i+1, m-1, nums, presum, visited);
            maxSum = Math.min(maxSum, Math.max(l, rightIntervalMax));
            
        }
        
        visited[start][m] = maxSum;
        return maxSum;
    }

    X. brute force
    https://discuss.leetcode.com/topic/64189/java-recursive-dp-having-trouble-in-iterative-dp
    public int splitArray(int[] nums, int m) {
 if (nums.length == 0 || nums == null || m == 0)
  return Integer.MAX_VALUE;
 return splitArray(nums, m, 0);
}

public int splitArray(int[] nums, int m, int start) {
 if (nums.length == 0 || nums == null || m == 0)
  return Integer.MAX_VALUE;
 if (start > nums.length)
  return Integer.MAX_VALUE;
 if (m == 1) {
  int sum = 0;
  for (int i = start; i < nums.length; i++)
   sum += nums[i];
  return sum;
 }
 int sum = 0;
 int split = 0;
 int min = Integer.MAX_VALUE;
 for (int i = start; i < nums.length; i++) {
  sum += nums[i];
  split = Math.max(sum, splitArray(nums, m - 1, i + 1));
  min = Math.min(min, split);
 }
 return min;
}

    http://www.geeksforgeeks.org/split-array-two-equal-sum-subarrays/
    Given an array of integers greater than zero, find if it is possible to split it in two subarrays (without reordering the elements), such that the sum of the two subarrays is the same.

    An Efficient solution is to first compute the sum of the whole array from left to right. Now we traverse array from right and keep track of right sum, left sum can be computed by subtracting current element from whole sum.
    int findSplitPoint(int arr[], int n)
    {
        // traverse array element and compute sum
        // of whole array
        int leftSum = 0;
        for (int i = 0 ; i < n ; i++)
            leftSum += arr[i];
        // again traverse array and compute right sum
        // and also check left_sum equal to right
        // sum or not
        int rightSum = 0;
        for (int i=n-1; i >= 0; i--)
        {
            // add current element to right_sum
            rightSum += arr[i];
            // exclude current element to the left_sum
            leftSum -=  arr[i] ;
            if (rightSum == leftSum)
                return i ;
        }
        // if it is not possible to split array
        // into two parts.
        return -1;
    }

    A Simple solution is to run two loop to split array and check it is possible to split array into two parts such that sum of first_part equal to sum of second_part.
    int findSplitPoint(int arr[], int n)
    {
        int leftSum = 0 ;
        // traverse array element
        for (int i = 0; i < n; i++)
        {
            // add current element to left Sum
            leftSum += arr[i] ;
            // find sum of rest array elements (rightSum)
            int rightSum = 0 ;
            for (int j = i+1 ; j < n ; j++ )
                rightSum += arr[j] ;
            // split point index
            if (leftSum == rightSum)
                return i+1 ;
        }
        // if it is not possible to split array into
        // two parts
        return -1;
    }

    http://www.geeksforgeeks.org/allocate-minimum-number-pages/
    Given number of pages in n different books and m students. The books are arranged in ascending order of number of pages. Every student is assigned to read some consecutive books. The task is to assign books in such a way that the maximum number of pages assigned to a student is minimum.
    The idea is to use Binary Search. We fix a value for number of pages as mid of current minimum and maximum. We initialize minimum and maximum as 0 and sum-of-all-pages respectively. If a current mid can be a solution, then we search on the lower half, else we search in higher half.
    Now the question arises, how to check if a mid value is feasible or not? Basically we need to check if we can assign pages to all students in a way that the maximum number doesn’t exceed current value. To do this, we sequentially assign pages to every student while current number of assigned pages doesn’t exceed the value. In this process, if number of students become more than m, then solution is not feasible. Else feasible.
    // Utility function to check if current minimum value
    // is feasible or not.
    bool isPossible(int arr[], int n, int m, int curr_min)
    {
        int studentsRequired = 1;
        int curr_sum = 0;
        // iterate over all books
        for (int i = 0; i < n; i++)
        {
            // check if current number of pages are greater
            // than curr_min that means we will get the result
            // after mid no. of pages
            if (arr[i] > curr_min)
                return false;
            // count how many students are required
            // to distribute curr_min pages
            if (curr_sum + arr[i] > curr_min)
            {
                // increment student count
                studentsRequired++;
                // update curr_sum
                curr_sum = arr[i];
                // if students required becomes greater
                // than given no. of students,return false
                if (studentsRequired > m)
                    return false;
            }
            // else update curr_sum
            else
                curr_sum += arr[i];
        }
        return true;
    }
    // function to find minimum pages
    int findPages(int arr[], int n, int m)
    {
        long long sum = 0;
        // return -1 if no. of books is less than
        // no. of students
        if (n < m)
            return -1;
        // Count total number of pages
        for (int i = 0; i < n; i++)
            sum += arr[i];
        // initialize start as 0 pages and end as
        // total pages
        int start = 0, end = sum;
        int result = INT_MAX;
        // traverse until start <= end
        while (start <= end)
        {
            // check if it is possible to distribute
            // books by using mid is current minimum
            int mid = (start + end) / 2;
            if (isPossible(arr, n, m, mid))
            {
                // if yes then find the minimum distribution
                result = min(result, mid);
                // as we are finding minimum and books
                // are sorted so reduce end = mid -1
                // that means
                end = mid - 1;
            }
            else
                // if not possible means pages should be
                // increased so update start = mid + 1
                start = mid + 1;
        }
        // at-last return minimum no. of  pages
        return result;
    }


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