Thread by @Iampatelajeet

"If you wanted to be rich in the 1800s you did it with Labour.
In the 1900s with Capital.
Now you do it with code."

@naval

In the series of solving @Leetcode daily challenges, we've an awesome problem today, 2 more days to finish a month, let's see how we can solve it.

@naval @LeetCode Find and Replace Pattern:
Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]

@naval @LeetCode Well, coming to solution part:

Approach 1:
-> we're given a set of strings, let's do it for one string to compare with given pattern.
-> For any string to a pattern of given 'pattern' string,
-> Both length must be equal
-> The number of unique characters must be equal

@naval @LeetCode -> Now, we come to comparison part where,
-> we map every character of given string with pattern and check a bijection matching in every iteration.
-> If it passes, means it has a same pattern as wanted.
-> Now we can generalize it to n strings.
TC: O(m*n)
AS: O(m*n)
Verdict: ✅

@naval @LeetCode Approach 2:

->As we mapped every character in the previous approach if
->>> we map them with their first occurrence index, we can get a numeric pattern.
-> Eg: "abab" = 1212 and "pqqp" = 1221
-> This way we can make patterns for every string, if they match with given pattern,

@naval @LeetCode ->We count them in our answer.
TC: O(m*n)
AS: O(m*n)
Verdict: ✅

Auxiliary space is counted from output array.
It was an easy kind of problem, just to figure out how things going on, hope you've understood.

@naval @LeetCode That's a wrap!

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