<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:
    number_their_interval = {}
    number_their_first_occurence = []

    for index, val in enumerate(nums):
      if val in number_their_interval:
        number_their_interval[val][1] = index
      else:
        number_their_interval[val] = [index, index]
        number_their_first_occurence.append(val)
    
    
    count_for_actual_interval = 1
    last_value = number_their_interval[number_their_first_occurence[0]][1]
    #print(number_their_interval)

    for num in number_their_first_occurence[1:]:
      next_interval = number_their_interval[num]
      if last_value < next_interval[0]:
        last_value = next_interval[1]
        count_for_actual_interval += 1
      else:
        last_value = max(next_interval[1], last_value)
    
    #print(count_for_actual_interval)
    dp = [-1] * (count_for_actual_interval + 1)
    dp[0] = 1

    def countPossibleCombinations(n):
      if dp[n] != -1:
        return dp[n]

      count_for_current_number = 0

      for i in range(1, n + 1):
        count_for_current_number += countPossibleCombinations(n - i)

      dp[n] = count_for_current_number
      return dp[n]

    

    #ans = countPossibleCombinations(count_for_actual_interval)

    #print(dp)

    return (2 ** (count_for_actual_interval - 1) ) % 1000000007


767