<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>

from typing import List

class Solution:
    def numberOfGoodPartitions(self, nums: List[int]) -> int:
        # Initialize a dictionary to store the last occurrence index of each number
        last_occurrence = {num: i for i, num in enumerate(nums)}
        
        # Initialize a set to keep track of numbers in the current subarray
        current_set = set()
        
        # Initialize a counter for the number of good partitions
        good_partition_count = 0
        
        # Iterate through the array
        for i, num in enumerate(nums):
            # Add the current number to the set
            current_set.add(num)
            
            # If the current number is at its last occurrence, remove it from the set
            if last_occurrence[num] == i:
                current_set.remove(num)
            
            # If the set is empty, it means we can make a partition here
            if len(current_set) == 0:
                good_partition_count += 1
        
        # Calculate the number of good partitions using the formula 2^(count - 1)
        # and return the result modulo 10^9 + 7
        return pow(2, good_partition_count - 1, 1_000_000_007)

# Example usage:
# solution = Solution()
# print(solution.numberOfGoodPartitions([1, 2, 3, 4]))  # Output: 8
# print(solution.numberOfGoodPartitions([1, 1, 1, 1]))  # Output: 1
# print(solution.numberOfGoodPartitions([1, 2, 1, 3]))  # Output: 2
737