<p>A split of an integer array is <strong>good</strong> if:</p>

<ul>
	<li>The array is split into three <strong>non-empty</strong> contiguous subarrays - named <code>left</code>, <code>mid</code>, <code>right</code> respectively from left to right.</li>
	<li>The sum of the elements in <code>left</code> is less than or equal to the sum of the elements in <code>mid</code>, and the sum of the elements in <code>mid</code> is less than or equal to the sum of the elements in <code>right</code>.</li>
</ul>

<p>Given <code>nums</code>, an array of <strong>non-negative</strong> integers, return <em>the number of <strong>good</strong> ways to split</em> <code>nums</code>. As the number may be too large, return it <strong>modulo</strong> <code>10<sup>9 </sup>+ 7</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only good way to split nums is [1] [1] [1].</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,2,2,5,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three good ways of splitting nums:
[1] [2] [2,2,5,0]
[1] [2,2] [2,5,0]
[1,2] [2,2] [5,0]
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no good way to split nums.</pre>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
    int binarysearch(int i, int j, vector<int> &prfsm, int w1)
    {
        int r = -1, n = prfsm.size();
        while(i <= j)
        {
            int mid = i + (j-i)/2;
            if(mid == n-1)
                break;
            if(prfsm[n-1]-prfsm[mid] >= prfsm[mid] - prfsm[w1])
            {
                r = mid;
                i = mid+1;
            }
            else
            {
                j = mid-1;
            }
        }
        return r;
    }
public:
    int waysToSplit(vector<int>& nums) {
        vector<int> prefixsm;
        int sum = 0, mod = 1e9 + 7, ans = 0, n = nums.size();
        for(int &num : nums)
        {
            sum += num;
            prefixsm.push_back(sum);
        }
        for(int i = 0; i < n-2; i++)
        {
            int fst = prefixsm[i];
            auto snditr = lower_bound(prefixsm.begin(), prefixsm.end(), 2*fst);
            if(snditr == prefixsm.end())
                return ans;
            int snd = snditr-prefixsm.begin();
            if(snd <= i)
                snd = i + 1;
            int trd = binarysearch(snd, n-1, prefixsm, i);
            if(trd == -1)
                continue;
            ans = (ans % mod + (trd-snd+1)%mod)%mod;
        }
        return ans;
    }
};

// brute force => prefix sum + i, j ::: O(n^2)
// optimized => prefix sum + i, log(n)[bs twice, once for 12 and then for 23] ::: O(nlogn)
92800