## Solution

#### Approach 1: Group into Blocks

Intuition and Algorithm

For a string like S = 'aabbbbccc', we can group it into blocks groupify(S) = [('a', 2), ('b', 4), ('c', 3)], that consist of a key 'abc' and a count [2, 4, 3].

Then, the necessary and sufficient condition for typed to be a long-pressed version of name is that the keys are the same, and each entry of the count of typed is at least the entry for the count of name.

For example, 'aaleex' is a long-pressed version of 'alex': because when considering the groups [('a', 2), ('l', 1), ('e', 2), ('x', 1)] and [('a', 1), ('l', 1), ('e', 1), ('x', 1)], they both have the key 'alex', and the count [2,1,2,1] is at least [1,1,1,1] when making an element-by-element comparison (2 >= 1, 1 >= 1, 2 >= 1, 1 >= 1).

Complexity Analysis

• Time Complexity: , where are the lengths of name and typed.

• Space Complexity: .

#### Approach 2: Two Pointer

Intuition

As in Approach 1, we want to check the key and the count. We can do this on the fly.

Suppose we read through the characters name, and eventually it doesn't match typed.

There are some cases for when we are allowed to skip characters of typed. Let's use a tuple to denote the case (name, typed):

• In a case like ('aab', 'aaaaab'), we can skip the 3rd, 4th, and 5th 'a' in typed because we have already processed an 'a' in this block.

• In a case like ('a', 'b'), we can't skip the 1st 'b' in typed because we haven't processed anything in the current block yet.

Algorithm

This leads to the following algorithm:

• For each character in name, if there's a mismatch with the next character in typed:
• If it's the first character of the block in typed, the answer is False.
• Else, discard all similar characers of typed coming up. The next (different) character coming must match.

Also, we'll keep track on the side of whether we are at the first character of the block.

Complexity Analysis

• Time Complexity: , where are the lengths of name and typed.

• Space Complexity: in additional space complexity. (In Java, .toCharArray makes this , but this can be easily remedied.)