Find and Replace Pattern LeetCode solution in Python

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# Runtime: 47 ms (Top 36.0%) | Memory: 16.40 MB (Top 44.2%)
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class Solution:
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    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
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        d = {}
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        for i, v in enumerate(pattern):
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            if v in d:
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                d[v].append(i)
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            else:
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                d |= {v: [i]}
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        # DICTIONARY CONTAINING LETTERS AND THEIR INDICES
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        ans = []
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        for i in words:
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            e = {}
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            for j, v in enumerate(i):
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                if v in e:
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                    e[v].append(j)
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                else:
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                    e |= {v: [j]}
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            # DICTIONARY CONTAINING LETTERS OF INDICES OF CURRENT WORD
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            for u, v in zip(d.values(), e.values()):
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                # COMPARING EACH VALUE
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                if u != v:
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                    break
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            # IF SUCCESSFUL APPEND TO ANS
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            else:
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                ans.append(i)
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        return ans