<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:
    bool func(vector<int> &nums,int n,int k,int mid){
        int sum=0,c=1;
        for(int i=0;i<n;i++){
            if(nums[i]>mid){
                return false;
            }
            else if(sum+nums[i]>mid){
                c++;
                sum=nums[i];
                if(c>k){
                    return false;
                }
            }
            else sum+=nums[i];
        }
        return true;
    }


    int splitArray(vector<int>& nums, int k) {
        int n=nums.size();
        int s=0,ans=-1;
        int e=accumulate(nums.begin(),nums.end(),0);
        while(s<=e){
            int mid=s+(e-s)/2;
            if(func(nums,n,k,mid)){
                ans=mid;
                e=mid-1;
            }
            else s=mid+1;
        }
        return ans;
    }
};
9200