<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:
    def waysToSplit(self, nums: List[int]) -> int:
        ans = 0
        n = len(nums)
        if(sum(nums)==0):
            return ((n-2)*(n-1)//2)%(10**9+7)

        pre = [0 for i in range(n)]
        pre[0] = nums[0]
        for i in range(1, n):
            pre[i] = pre[i-1] + nums[i]

        def binary1(l, r, x, sub):
            if(pre[l]-sub>=x):
                return l

            if(pre[r]-sub<x):
                return r+1
            
            if(l==r):
                return l

            m = (l+r)//2
            if(pre[m]-sub>=x):
                return binary1(l, m, x, sub)

            else:
                return binary1(m+1, r, x, sub)

        def binary2(l, r, x, sub):
            if(pre[l]-sub>x):
                return l-1

            if(pre[r]-sub<=x):
                return r

            if(l==r):
                return l

            m = (l+r)//2
            if(pre[m]-sub>x):
                return binary2(l, m-1, x, sub)

            else:
                return binary2(m+1, r, x, sub)

        ans = 0
        for i in range(n-2):
            a = binary1(i+1, n-1, pre[i], pre[i])
            b = binary2(i+1, n-1, (pre[-1]-pre[i])//2, pre[i])
            if(a==n or b==i):
                continue

            if(a>b):
                continue

            ans += b-a+1

        return ans%(10**9+7)
        



            

2024