Next Permutation LeetCode solution in Python

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class Solution:
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    def nextPermutation(self, nums) -> None:
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        firstDecreasingElement = -1
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        toSwapWith = -1
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        lastIndex = len(nums) - 1
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        # Looking for an element that is less than its follower
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        for i in range(lastIndex, 0, -1):
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            if nums[i] > nums[i - 1]:
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                firstDecreasingElement = i - 1
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                break
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        # If there is not any then reverse the array to make initial permutation
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        if firstDecreasingElement == -1:
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            for i in range(0, lastIndex // 2 + 1):
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                nums[i], nums[lastIndex - i] = nums[lastIndex - i], nums[i]
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            return
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        # Looking for an element to swap it with firstDecreasingElement
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        for i in range(lastIndex, 0, -1):
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            if nums[i] > nums[firstDecreasingElement]:
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                toSwapWith = i
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                break
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        # Swap found elements
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        nums[firstDecreasingElement], nums[toSwapWith] = (
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            nums[toSwapWith],
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            nums[firstDecreasingElement],
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        )
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        # Reverse elements from firstDecreasingElement to the end of the array
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        left = firstDecreasingElement + 1
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        right = lastIndex
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        while left < right:
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            nums[left], nums[right] = nums[right], nums[left]
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            left += 1
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            right -= 1