Distance-based problems in algorithmic challenges often revolve around measuring proximity or calculating the minimum steps to reach a goal. The “Shortest Distance to a Character” problem on Leetcode encapsulates this theme, with a twist that involves textual data. This article will explore the problem, dissect its intricacies, present viable solution paths, and conclude with an implementation in Python.

### Table of Contents

- Problem Statement
- Intuitive Approach
- Solution Strategies
- Python Implementation
- Testing and Analysis
- Enhanced Approaches and Variations
- Conclusion

### 1. Problem Statement

Given a string `S`

and a character `C`

, return an array representing the shortest distance from the character `C`

in the string for each character in `S`

.

### 2. Intuitive Approach

To understand the problem better, consider the string `S = "loveleetcode"`

and character `C = 'e'`

. The resulting shortest distances for each character would be `[3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]`

.

At a glance, the problem asks us to calculate the distance of each character in `S`

from its nearest occurrence of character `C`

.

### 3. Solution Strategies

One could think of multiple approaches to solve the problem:

**Brute Force Approach**: For each character in`S`

, search to the left and right to find the closest occurrence of`C`

and take the minimum of the two distances.**Two-Pass Approach**: Make two passes over the string — first from left to right and then from right to left — to calculate distances.

The brute force approach is simple but can be inefficient, especially for longer strings. The two-pass approach is more efficient and is detailed further below.

### 4. Python Implementation

Let’s break down the Two-Pass Approach:

- In the first pass (left to right), initialize a variable
`prev`

to a large negative value. For each character, if it’s an occurrence of`C`

, set`prev`

to its index. Otherwise, set the distance for the character as its index minus`prev`

. - In the second pass (right to left), reset
`prev`

but this time to a large positive value. Repeat the above process, but this time, also consider the minimum of the existing distance and the new distance.

Here’s the Python code for the described approach:

```
def shortestToChar(S, C):
prev = float('-inf')
res = [0] * len(S)
# First pass: left to right
for i in range(len(S)):
if S[i] == C:
prev = i
res[i] = i - prev
# Second pass: right to left
prev = float('inf')
for i in range(len(S) - 1, -1, -1):
if S[i] == C:
prev = i
res[i] = min(res[i], prev - i)
return res
```

### 5. Testing and Analysis

Testing the function using our example:

```
S = "loveleetcode"
C = 'e'
print(shortestToChar(S, C)) # Expected output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
```

**Time Complexity**: O(n), where n is the length of string `S`

. We traverse the string twice, but the operations are constant time during these traversals.

**Space Complexity**: O(n), since we store the result in an array of length n.

### 6. Enhanced Approaches and Variations

**Sparse Distance Arrays**: If`C`

occurs infrequently, one could think of creating a list of indices where`C`

occurs and then using this list to determine distances.**Real-world Use Case**: In applications like text editors, finding the distance to a particular character might be useful for features like “jump to next occurrence”.

### 7. Conclusion

The “Shortest Distance to a Character” problem on Leetcode provides an interesting foray into the realm of textual data processing combined with distance calculations. The two-pass approach offers an elegant and efficient solution to the problem. As always, understanding the problem, identifying patterns, and implementing methodically are crucial. Challenges like these not only sharpen one’s algorithmic thinking but also provide insights into real-world application scenarios.