leetcode：6097. Match [set record + query one by one with the same length] after replacing characters

analysis

use set Record replaceable
And then s Find length and sub same
And then we'll see if we can map In the past

ac code

class Solution:
    def matchReplacement(self, s: str, sub: str, mappings: List[List[str]]) -> bool:
        n, m = len(s), len(sub)
        
        mySet = set()
        for x, y in mappings:
            mySet.add((x, y))
            
        #print(myDict)
        for i in range(n - m + 1):
            cankao = s[i: i + m]
            #print(cankao)
            flag = True
            for j in range(m):
                if sub[j] != cankao[j] and (sub[j], cankao[j]) not in mySet:
                    flag = False
                    break
            if flag:
                return True
        
        return False
                
                

summary

Simple set