<p>Given an integer array <code>nums</code> and an integer <code>k</code>, split <code>nums</code> into <code>k</code> non-empty subarrays such that the largest sum of any subarray is <strong>minimized</strong>.</p>

<p>Return <em>the minimized largest sum of the split</em>.</p>

<p>A <strong>subarray</strong> is a contiguous part of the array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,2,5,10,8], k = 2
<strong>Output:</strong> 18
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5], k = 2
<strong>Output:</strong> 9
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= min(50, nums.length)</code></li>
</ul>

class Solution {
public:
    int mem[1001][51];
    int hp(vector<int>& pf, int ci, int rem) {
        int n = pf.size() - 1;
        if (mem[ci][rem] != -1) {
            return mem[ci][rem];
        }
        if (rem == 1) {
            return mem[ci][rem] = pf[n] - pf[ci];
        }
        int mnss = INT_MAX;
        for (int i = ci; i <= n - rem; ++i) {
            int ss = pf[i + 1] - pf[ci];
            int lg = max(ss, hp(pf, i + 1, rem - 1));
            mnss = min(mnss, lg);
            if (ss >= mnss) break;
        }
        return mem[ci][rem] = mnss;
    }
    int splitArray(vector<int>& nums, int k) {
        int n = nums.size();
        memset(mem, -1, sizeof(mem));
        vector<int> pf(n + 1, 0);
        for (int i = 0; i < n; ++i) {
            pf[i + 1] = pf[i] + nums[i];
        }
        return hp(pf, 0, k);
    }
};
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