<p>Given an integer array <code>nums</code>, find the <span data-keyword="subarray-nonempty">subarray</span> with the largest sum, and return <em>its sum</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [-2,1,-3,4,-1,2,1,-5,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The subarray [4,-1,2,1] has the largest sum 6.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The subarray [1] has the largest sum 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,4,-1,7,8]
<strong>Output:</strong> 23
<strong>Explanation:</strong> The subarray [5,4,-1,7,8] has the largest sum 23.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution using the <strong>divide and conquer</strong> approach, which is more subtle.</p>

class Solution {
public:
    int maxSubArray(vector<int>& v) {
         int dp[100005];

    dp[0] = v[0];

    int res = -999999999;

    for(int i = 0; i < v.size() - 1; i++)
    {
        if(v[i+1] + dp[i] > v[i+1])
        {
            dp[i+1] = dp[i] + v[i+1];
        }
        else
        {
            dp[i+1] = v[i+1];
        }
    }
    for(int i = 0; i < v.size(); i++)
    {
        res = max(res,dp[i]);
    }
    return res;
    }
};
71100