<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 isposs(vector<int>v,int n,int m,int mid)
  {
    int prnct=1,wt=0;
  
    for(int i=0;i<n;i++)
    {
      if(wt+v[i]<=mid)
      {
        wt+=v[i];
      }
      else
      {
        prnct++;
        if(prnct>m||v[i]>mid)
        return false;
        wt=v[i];
      }
    }
    return true;
  }

    int splitArray(vector<int>& a, int k) {
        int n=a.size(),ans;
          long long sum=0;
        for(int i=0;i<n;i++)
{
  sum+=a[i];
}

int s=0,e=sum;
int mid=s+(e-s)/2;
while(s<=e)
{
  if(isposs(a,n,k,mid))
  {
    ans=mid;
    e=mid-1;
  }
  else
  s=mid+1;
  mid=(s+e)/2;
}
        return ans;
    }
};
9700