Count Number of Texts LeetCode solution in Python

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# Runtime: 2740 ms (Top 30.44%) | Memory: 19.5 MB (Top 65.22%)
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class Solution(object):
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    def countTexts(self, pressedKeys):
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        """
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        :type pressedKeys: str
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        :rtype: int
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        """
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        dp = [1] + [0] * len(pressedKeys)
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        mod = 10**9 + 7
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        for i, n in enumerate(pressedKeys):
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            dp[i + 1] = dp[i]
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            # check if is continous
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            if i >= 1 and pressedKeys[i - 1] == n:
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                dp[i + 1] += dp[i - 1]
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                dp[i + 1] %= mod
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                if i >= 2 and pressedKeys[i - 2] == n:
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                    dp[i + 1] += dp[i - 2]
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                    dp[i + 1] %= mod
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                    # Special case for '7' and '9' that can have 4 characters combination
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                    if i >= 3 and pressedKeys[i - 3] == n and (n == "7" or n == "9"):
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                        dp[i + 1] += dp[i - 3]
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                        dp[i + 1] %= mod
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        return dp[-1]