<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 {
    bool isValid(vector<int>& nums, int k, int n, int maxSumAllowed) {
        int currSum = 0;
        int i = 0;
        while(i < n) {
            if(currSum + nums[i] <= maxSumAllowed) {
                currSum += nums[i++];
            } else {
                currSum = nums[i++];
                k--;
                if(k == 0)
                    return false;
            }
        }

        return k >= 1;
    }

public:
    int splitArray(vector<int>& nums, int k) {
        int n = nums.size();
        int totalSum = accumulate(nums.begin(), nums.end(), 0);

        int ans = 0;
        int low = *max_element(nums.begin(), nums.end());
        int high = totalSum;
        
        while(low <= high) {
            int mid = low + (high - low) / 2;
            if(isValid(nums, k, n, mid)) {
                ans = mid;
                high = mid-1;
            } else {
                low = mid+1;
            }
        }

        return ans;   
    }
};
9200