<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:
        MOD = 10**9 + 7
        n = len(nums)
        
        if n < 3:
            return 0
        
        # Step 1: Calculate prefix sum
        prefix_sum = [0] * (n + 1)
        for i in range(n):
            prefix_sum[i + 1] = prefix_sum[i] + nums[i]
        
        count = 0
        
        # Step 2: Iterate over each possible left partition end
        for i in range(1, n - 1):
            left_sum = prefix_sum[i]
            
            # Step 3: Binary search for the left boundary of mid subarray
            low = i + 1
            high = n - 1
            
            # Find the smallest j such that prefix_sum[j] - prefix_sum[i] >= left_sum
            while low <= high:
                mid = (low + high) // 2
                mid_sum = prefix_sum[mid] - prefix_sum[i]
                if mid_sum >= left_sum:
                    high = mid - 1
                else:
                    low = mid + 1
            
            left_bound = low
            
            # Step 4: Binary search for the right boundary of mid subarray
            high = n - 1
            while low <= high:
                mid = (low + high) // 2
                mid_sum = prefix_sum[mid] - prefix_sum[i]
                right_sum = prefix_sum[n] - prefix_sum[mid]
                if mid_sum <= right_sum:
                    low = mid + 1
                else:
                    high = mid - 1
            
            right_bound = high
            
            # Step 5: Count valid splits
            if left_bound <= right_bound:
                count = (count + (right_bound - left_bound + 1)) % MOD
        
        return count
1636