# Leetcode – Long Pressed Name Solution in Python

Leetcode frequently presents problems that test a programmer’s understanding of string manipulations and pattern recognition. Among these, the “Long Pressed Name” problem is notable due to its real-world implications related to touchscreen typing errors. This article will dive deep into the problem, elucidate its underlying logic, provide Python-based solutions, and conclude with a complexity analysis.

1. Understanding the Problem Statement
2. Problem Analysis
3. Two-Pointer Technique
4. Python Implementation
5. Complexity Analysis
6. Alternate Approaches
7. Conclusion

## 1. Understanding the Problem Statement

### Problem:

Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long-pressed, and the character will be typed 1 or more times. Given two strings name and typed, return True if typed can be produced by long pressing some characters (possibly zero characters) of name, and False otherwise.

Example:

Input: name = "alex", typed = "aaleex"
Output: True
Explanation: 'a' and 'e' in 'alex' were long pressed.

## 2. Problem Analysis

The main challenge is to determine if typed is a valid long-press variation of name. The problem can be tackled by comparing characters from both strings and checking the frequency of characters in typed.

## 3. Two-Pointer Technique

The two-pointer technique shines in problems that involve comparisons within two sequences. For this problem:

• i: pointer for traversing the name string
• j: pointer for traversing the typed string

We’ll move through both strings, comparing the characters at pointers i and j.

1. If they match, move both pointers ahead.
2. If they don’t match, check if typed[j] is the same as typed[j-1]. If true, it indicates a long press, so only move j ahead.
3. If neither of the above conditions is met, it means the characters don’t match, and there’s no evidence of a long press. Hence, return False.

## 4. Python Implementation

Here’s a Python solution using the two-pointer technique:

def isLongPressedName(name, typed):
i, j = 0, 0

while j < len(typed):
if i < len(name) and name[i] == typed[j]:
i += 1
j += 1
elif j > 0 and typed[j] == typed[j - 1]:
j += 1
else:
return False

return i == len(name)

## 5. Complexity Analysis

Time Complexity: O(n+m)
We might traverse both the name and typed strings completely, leading to a linear time complexity where n and m are the lengths of name and typed respectively.

Space Complexity: O(1)
We use only a constant amount of space, regardless of the input size.

## 6. Alternate Approaches

1. Grouping: One could also solve this by grouping characters and comparing the groups. Use the itertools.groupby() function in Python to group consecutive similar characters in both name and typed. Then, compare the character and length of groups from both strings.
2. Regular Expressions: Create a regex pattern from the name string where each character is followed by a ‘+’. This indicates one or more of the character. Then, use this pattern to match the typed string.

## 7. Conclusion

The “Long Pressed Name” problem is a fantastic example of a scenario where string manipulations meet real-world applications. It mimics the challenges faced in predictive text inputs and touch typing errors. By offering a concrete use-case scenario, this problem emphasizes the importance of mastering string manipulation.