<p>Given a positive integer <code>n</code>, find the <strong>pivot integer</strong> <code>x</code> such that:</p>

<ul>
	<li>The sum of all elements between <code>1</code> and <code>x</code> inclusively equals the sum of all elements between <code>x</code> and <code>n</code> inclusively.</li>
</ul>

<p>Return <em>the pivot integer </em><code>x</code>. If no such integer exists, return <code>-1</code>. It is guaranteed that there will be at most one pivot index for the given input.</p>

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

<pre>
<strong>Input:</strong> n = 8
<strong>Output:</strong> 6
<strong>Explanation:</strong> 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> 1 is the pivot integer since: 1 = 1.
</pre>

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

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be proved that no such integer exist.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int pivotInteger(int n) {
        vector<int> dp(n);
        iota(dp.begin(),dp.end(),1);
        if(n==1) return 1;
        int sum = 0;
        long long s = accumulate(dp.begin(),dp.end(),0LL);
        for(auto x : dp) cout<<x<<' ';
        for(int i = 0;i<n;i++){
            sum+=dp[i];
            if(sum==accumulate(dp.begin()+i,dp.end(),0)){
                return i+1;
            }
            
         
        }
        return -1;
    }
};
9100