<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 = 10 ** 9 + 7
        n = len(nums)

        first = {}
        last = {}

        for i in range(n):
            num = nums[i]
            if num not in first:
                first[num] = i
            last[num] = i

        diff = [0] * n
        for num in first:
            diff[first[num]] += 1
        for num in last:
            diff[last[num]] -= 1

        m = 0
        cur = 0
        for i in range(n):
            cur += diff[i]
            if cur == 0:
                m += 1

        return 2 ** (m - 1) % MOD
755