<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 ispossible(vector<int> arr,int n,int m,int mid)
{
    int studentcount = 1;
    int pagesum = 0;
    for(int i=0; i<n;i++)
    {
        if(pagesum + arr[i] <= mid)
        {
        pagesum += arr[i];
        }
        else
        {
            studentcount++;
            if(studentcount > m || arr[i]>mid)
            {
                return false;
            }
            pagesum = arr[i];
        }
    }
    return true;
}
    int splitArray(vector<int>& nums, int k) {
      int s = 0;
      int sum = 0;
      for(int i = 0;i<nums.size();i++)
      {
        sum+= nums[i];
      }
      int e = sum;
      int ans = -1;
      int mid = s + (e-s)/2;
      while(s<=e)
      {
        if(ispossible(nums,nums.size(),k,mid))
        {
         ans = mid;
         e = mid-1;
        }
        else
        {
            s = mid +1;
        }
        mid = s + (e-s)/2;
      }
      return ans;
    }
};
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