Codeforces 1185B. Email from Polycarp Codeforces Round #568 (Div. 2)

Problem link

It should be noted that if Polycarp press a key once, it's possible to appear one, two or more letters. So, if Polycarp press a key t$t$ times, letter will appear at least t$t$ times. Also, pressing a particular key not always prints same number of letters. So the possible correct solution is followed:

For both words s$s$ and t$t$ we should group consecutive identical letters with counting of the each group size. For ex, there 4 groups in the word "aaabbcaa": [aaa,bb,c,aa]$[aaa,bb,c,aa]$. For performance you should keep every group as the letter (char) and the group size (int).

Then, if the number of groups in word s$s$ isn't equal to the number of groups in word t$t$, then t$t$ can not be printed during typing of s$s$. Let's go through array with such groups and compare the i$i$-th in the word s$s$with the i$i$-th in the word t$t$. If letters in groups are different, answer is NO. If the group in s$s$ are greater than group in t$t$, answer is NO. Answer is YES in all other cases.

Every string can be splitted into such groups by one loop. So, the total time complexity is ∑|s|+∑|t|

A solution in c++