<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:
    #有两个切割位置l,r  -->  l <= r - l <= n - r 可以变形成 -- l >= 2r-n 和 r >= 2l
    #根据上述这就可以看出来需要前缀和 -- 隐藏条件 1 <= l < r
    #解题思路 ---- 枚举 r，通过二分可以得到使不等式成立的 l 的范围，累加这个范围的长度即为答案。
        mod = 10 ** 9 + 7
        # prefix = list(accumulate(nums, initial = 0))
        n = len(nums)
        prefix = [0] * (n+1)
        cur = 0
        for index, num in enumerate(nums):
            cur += num
            prefix[index+1] = cur
        res = 0

        for right in range(n):
        # r >= 2l and 2l >= 4r - 2n ----> r >= 4r - 2n ---> 2n >= 3r --- 2n < 3r break
            if 3*prefix[right] > 2 * prefix[n]: break
            l = bisect_left(prefix, 2*prefix[right]-prefix[n], 1, right)
            r = bisect_right(prefix, prefix[right]//2, l, right)
            res += r - l
        return res % mod
809