<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisting of <strong>positive</strong> integers.</p>

<p>A partition of an array into one or more <strong>contiguous</strong> subarrays is called <strong>good</strong> if no two subarrays contain the same number.</p>

<p>Return <em>the <strong>total number</strong> of good partitions of </em><code>nums</code>.</p>

<p>Since the answer may be 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 = [1,2,3,4]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible good partition is: ([1,1,1,1]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,1,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def numberOfGoodPartitions(self, nums: List[int]) -> int:
        MOD = 1_000_000_007

        cnt, n = 0, len(nums)

        # put all the identical numbers in the same partition
        firstIdx = defaultdict(int)
        lastIdx = defaultdict(int)

        for i in range(n):
            lastIdx[nums[i]] = i

            if nums[i] not in firstIdx:
                firstIdx[nums[i]] = i
        
        # merge these intervals
        intervals = [(firstIdx[x], lastIdx[x]) for x in firstIdx]

        def merge(ls: list) -> list:
            ls.sort()
            res = []

            i = 0
            while i < len(ls):
                leftPoint = ls[i][0]
                rightPoint = ls[i][1]

                while i < len(ls) and ls[i][0] <= rightPoint:
                    rightPoint = max(rightPoint, ls[i][1])
                    i += 1
                res.append([leftPoint, rightPoint])
            return res

        # aggregate together the intervals and treat them as a single number
        intervals = merge(intervals)
        for i, interval in enumerate(intervals):
            cnt += 1

            # how many numbers are free between these intervals?
            if i + 1 < len(intervals):
                cnt += intervals[i + 1][0] - interval[1] - 1

        return pow(2, cnt - 1, MOD)
890