<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:
        tmp = {}
        res = []
        for i, num in enumerate(nums):
            if num not in tmp:
                tmp[num] = i
                res.append([i,i])
            else:
                res.append([tmp[num],i])
        new = self.merge(res)
        #print (new)
        n = len(new)
        return pow(2,n-1,10**9 + 7)

    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals = sorted(intervals, key =lambda x: x[0]) #sorting intervals by index 0
        
        mergedIntervals = []
        
        for interval in intervals:
		    #check if the list of mergedIntervals is empty or if the current interval does not overlap with the previous interval, simply append it.
            if not mergedIntervals or mergedIntervals[-1][1] < interval[0]:
                mergedIntervals.append(interval)
            else:
			# else, there is an overlap, so we merge the current and previous intervals.
                mergedIntervals[-1][1] = max(mergedIntervals[-1][1], interval[1])
        return mergedIntervals
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