LeetCode 833 Find And Replace In String A Comprehensive Solution Guide

Jul 26, 2025 by Sharif Sakr 71 views

LeetCode 833 Find and Replace in String A Comprehensive Guide

Hey guys! Today, we're diving deep into LeetCode problem 833, "Find and Replace in String." This problem can be a bit tricky, but don't worry, we'll break it down step by step. We'll explore the problem statement, discuss the nuances, and then walk through a Python solution to nail this coding challenge. So, let's get started and conquer this problem together!

Understanding the Problem Statement

Before we jump into the code, let's make sure we fully grasp what the problem is asking. In LeetCode 833 Find and Replace in String, you're given a string s and a set of replacement operations. Each operation consists of an index, a source string, and a target string. The goal is to perform these replacements on the original string s. However, there's a catch! The replacements can overlap, and you need to figure out the correct order to apply them. It's like a puzzle where you need to fit the pieces (replacements) in the right spots without messing up the overall picture (the final string).

Key Components of the Problem:

Original String s: This is the string we'll be modifying.
Indices: A list of starting positions in s where replacements might occur.
Sources: A list of strings that we'll try to match in s at the given indices.
Targets: A list of strings that will replace the matched sources.

The challenge lies in applying the replacements correctly, especially when overlaps occur. Not all replacement operations may be valid, and the order matters. We need to ensure that a replacement is only made if the source string actually matches the substring in s at the given index. If there's a match, we replace the substring with the target string. If not, we skip the replacement and move on.

The Tricky Part: Overlapping Replacements

Overlapping replacements add a layer of complexity. Imagine you have two replacements:

Replace "abc" with "xyz" starting at index 0.
Replace "bcd" with "pqr" starting at index 1.

If we naively perform the first replacement, the string becomes "xyz...". Now, the second replacement won't work because "bcd" no longer exists at index 1. This is why we need a strategy to handle overlapping replacements carefully.

Clarification on the Number of Operations

The problem statement initially caused some confusion regarding the number of replacement operations. The original text mentioned performing k replacement operations, which made it sound like we had to apply all replacements, even if they weren't valid. However, the correct interpretation is that we need to perform up to k operations. This means we only apply a replacement if the source string matches the substring in s at the given index. If a replacement can't be made (due to a mismatch or overlap), we simply skip it. This clarification is crucial for understanding the problem's constraints and designing an efficient solution.

Devising a Strategy: How to Tackle the Problem

Now that we have a solid understanding of the problem, let's brainstorm a strategy to solve it. The key is to handle the replacements in a way that respects the order and avoids conflicts due to overlaps. Here's a breakdown of our approach:

Organize the Replacements: We'll start by organizing the replacement operations based on their starting indices. This will allow us to process them in the order they appear in the string s. A dictionary or a similar data structure can be used to map indices to a list of replacement pairs (source, target).
Iterate Through the String: We'll iterate through the string s, checking at each index if there are any replacement operations associated with it.
Match and Replace (if Possible): If we find a replacement operation at the current index, we'll check if the source string matches the substring in s starting at that index. If there's a match, we'll perform the replacement by adding the target string to our result. If there's no match, we'll skip the replacement and move on.
Handle Overlaps: To handle overlaps, we need to advance our index in s by the length of the replaced substring. This ensures that we don't try to apply replacements within the replaced region.
Build the Result: We'll build the final string by concatenating the replaced substrings and the characters from s that weren't part of any replacement.

Data Structures to Consider

To implement this strategy, we'll need some efficient data structures. Here are a few that come to mind:

Dictionary (or Hash Map): To map indices to replacement pairs. This will allow us to quickly look up the replacements associated with a given index.
List (or Array): To store the replacement operations and the characters of the resulting string.
String Builder (or List of Characters): To efficiently construct the final string. In Python, we can use a list of characters and then join them at the end.

Algorithm in Plain English

Before we dive into the code, let's outline the algorithm in plain English:

Create a dictionary to store replacement operations, mapping each index to a list of (source, target) pairs.
Iterate through the input lists (indices, sources, targets) and populate the dictionary.
Initialize an empty list to store the characters of the result.
Iterate through the string s:
- If the current index has replacement operations:
  - For each (source, target) pair at that index:
    - If the source matches the substring in s starting at the current index:
      - Append the target to the result.
      - Advance the index by the length of the source.
      - Break the inner loop (we've made a replacement).
  - If no replacement was made at this index:
    - Append the character at the current index to the result.
    - Advance the index by 1.
- If the current index has no replacement operations:
  - Append the character at the current index to the result.
  - Advance the index by 1.
Join the characters in the result list to form the final string.

With this strategy and algorithm in mind, we're well-equipped to write the code. Let's move on to the next section and implement our solution in Python.

Python Implementation: Bringing the Strategy to Life

Alright, guys, let's get our hands dirty and translate our strategy into Python code. We'll walk through the code step by step, explaining each part and how it contributes to the overall solution. We'll also highlight some Pythonic tricks and best practices to make our code clean and efficient.

from collections import defaultdict
from typing import List

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        ret = []
        index_to_pair = defaultdict(list)
        for i in range(len(sources)):
            index_to_pair[indices[i]].append((sources[i], targets[i]))
        
        i = 0
        while i < len(s):
            if i in index_to_pair:
                found = False
                for to_find, replacement in index_to_pair[i]:
                    if s[i:i+len(to_find)] == to_find:
                        ret.append(replacement)
                        i += len(to_find)
                        found = True
                        break
                if found:
                    continue
            ret.append(s[i])
            i += 1
        return "".join(ret)

Code Breakdown:

Imports: We start by importing the necessary modules.
- collections.defaultdict: This is a handy data structure that allows us to create a dictionary where, if a key is not found, it automatically creates a default value for it. In our case, we'll use it to create a dictionary that maps indices to lists of replacement pairs.
- typing.List: This is used for type hinting, which helps make our code more readable and maintainable.
findReplaceString Method: This is the main method that takes the input string s, the list of indices, the list of source strings, and the list of target strings as arguments and returns the modified string.
Initialization:
- ret = []: We initialize an empty list called ret. This list will store the characters and replaced substrings that will form our final string. We use a list here because string concatenation in Python can be inefficient for large strings. Appending to a list is generally faster, and we can join the list elements into a string at the end.
- index_to_pair = defaultdict(list): We create a defaultdict called index_to_pair. This dictionary will map each index to a list of (source, target) pairs. The defaultdict is initialized with list as the default factory, meaning that if we try to access a key that doesn't exist, it will automatically create an empty list for that key.
Organizing Replacements:
```
for i in range(len(sources)):
    index_to_pair[indices[i]].append((sources[i], targets[i]))
```
This loop iterates through the input lists (indices, sources, targets) and populates the index_to_pair dictionary. For each index, it appends the corresponding (source, target) pair to the list associated with that index. This step organizes the replacement operations by their starting indices, which will be crucial for processing them in the correct order.
Iterating Through the String:
```
i = 0
while i < len(s):
    ...
```
We initialize a pointer i to 0 and start iterating through the string s using a while loop. This loop will continue until we've processed the entire string.
Checking for Replacements:
```
if i in index_to_pair:
    ...
```
Inside the loop, we first check if the current index i exists as a key in the index_to_pair dictionary. This tells us whether there are any replacement operations associated with this index.
Matching and Replacing:
```
found = False
for to_find, replacement in index_to_pair[i]:
    if s[i:i+len(to_find)] == to_find:
        ret.append(replacement)
        i += len(to_find)
        found = True
        break
if found:
    continue
```
If we find replacement operations at the current index, we iterate through the list of (source, target) pairs. For each pair, we check if the source string (to_find) matches the substring in s starting at the current index. We use string slicing (s[i:i+len(to_find)]) to extract the substring and compare it with the source string.

If there's a match, we append the target string (replacement) to the ret list, advance the index i by the length of the source string (to handle overlaps), set the found flag to True, and break out of the inner loop (since we've made a replacement at this index).

If the found flag is True after the inner loop, it means we've made a replacement at this index, so we use the continue statement to skip to the next iteration of the outer loop.
Handling No Replacements:
```
ret.append(s[i])
i += 1
```
If we reach this part of the code, it means either there were no replacement operations at the current index, or none of the source strings matched the substring in s. In this case, we simply append the character at the current index (s[i]) to the ret list and advance the index i by 1.
Building the Final String:
```
return "".join(ret)
```
After processing the entire string, we join the characters and replaced substrings in the ret list using the "".join(ret) method. This efficiently creates the final modified string, which is then returned.

Key Pythonic Elements:

defaultdict: Using defaultdict simplifies the code by automatically creating a list for each index if it doesn't already exist.
List Comprehension (Implicit): Building the ret list character by character is more efficient than repeated string concatenation.
String Slicing: Python's string slicing (s[i:i+len(to_find)]) makes it easy to extract substrings for comparison.
continue Statement: The continue statement helps us skip to the next iteration of the outer loop when we've made a replacement, making the code cleaner and more readable.

With this detailed explanation, you should have a clear understanding of the Python implementation. We've covered everything from the overall structure to the nitty-gritty details. Now, let's move on to the next section and discuss some potential optimizations and alternative approaches.

Optimizations and Alternative Approaches

While our current solution works well, it's always a good practice to think about potential optimizations and alternative approaches. This not only improves our problem-solving skills but also helps us write more efficient and robust code. Let's explore some ways we can optimize our solution and discuss alternative methods to tackle the "Find and Replace in String" problem.

Optimization 1: Sorting Replacements by Index

In our current solution, we iterate through the index_to_pair dictionary, which is essentially a hash map. While hash map lookups are generally fast (O(1) on average), the order in which we process the indices is not guaranteed. This might lead to unnecessary comparisons and replacements if we encounter overlapping replacements. A potential optimization is to sort the replacement operations by their indices before processing them.

How Sorting Helps:

By sorting the replacements by index, we ensure that we process them in the order they appear in the string s. This can help us avoid conflicts due to overlaps. For example, if we have two replacements:

Replace "abc" with "xyz" starting at index 0.
Replace "bcd" with "pqr" starting at index 1.

If we process the replacement at index 0 first, we can correctly replace "abc" with "xyz". Then, when we reach index 1, we'll see that "bcd" no longer exists, and we can skip the replacement. If we didn't sort, we might have tried to replace "bcd" first, which would have led to an incorrect result.

Implementation:

To implement this optimization, we can sort the keys of the index_to_pair dictionary before iterating through them. Here's how we can modify our code:

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        ret = []
        index_to_pair = defaultdict(list)
        for i in range(len(sources)):
            index_to_pair[indices[i]].append((sources[i], targets[i]))
        
        # Sort the indices
        sorted_indices = sorted(index_to_pair.keys())
        
        i = 0
        while i < len(s):
            if i in index_to_pair:
                found = False
                for to_find, replacement in index_to_pair[i]:
                    if s[i:i+len(to_find)] == to_find:
                        ret.append(replacement)
                        i += len(to_find)
                        found = True
                        break
                if found:
                    continue
            ret.append(s[i])
            i += 1
        return "".join(ret)

We've added a line sorted_indices = sorted(index_to_pair.keys()) to sort the indices. Then, we can iterate through the sorted_indices list instead of directly iterating through the keys of the index_to_pair dictionary.

Optimization 2: Pre-compute Match Results

Another optimization is to pre-compute the match results for each replacement operation. In our current solution, we check for a match every time we encounter a potential replacement. We can avoid redundant checks by pre-computing the match results and storing them in a separate data structure.

How Pre-computation Helps:

By pre-computing the match results, we can reduce the number of string comparisons we need to perform. This can be especially beneficial if we have a large number of replacement operations or if the source strings are long. Instead of repeatedly checking if a source string matches the substring in s, we can simply look up the pre-computed result.

Implementation:

To implement this optimization, we can create a list or a dictionary to store the match results. Here's how we can modify our code:

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        ret = []
        index_to_pair = defaultdict(list)
        matches = [False] * len(indices)  # Pre-compute match results
        
        for i in range(len(sources)):
            if s[indices[i]:indices[i]+len(sources[i])] == sources[i]:
                matches[i] = True
            index_to_pair[indices[i]].append((i, sources[i], targets[i]))
        
        i = 0
        while i < len(s):
            if i in index_to_pair:
                found = False
                for idx, to_find, replacement in index_to_pair[i]:
                    if matches[idx]:  # Use pre-computed match result
                        ret.append(replacement)
                        i += len(to_find)
                        found = True
                        break
                if found:
                    continue
            ret.append(s[i])
            i += 1
        return "".join(ret)

We've added a list called matches to store the match results. We pre-compute the match results in the first loop and store them in the matches list. Then, in the inner loop, we use the pre-computed match result (matches[idx]) instead of performing the string comparison again.

Alternative Approach: Using Regular Expressions

Another approach to solve this problem is to use regular expressions. Regular expressions provide a powerful way to search and replace patterns in strings. We can use the re.sub() function in Python to perform the replacements. However, using regular expressions for this problem can be tricky due to the overlapping replacements. We need to carefully construct the regular expression pattern and handle the replacements in the correct order.

Challenges with Regular Expressions:

Overlapping Replacements: Regular expressions typically process the string from left to right and perform replacements greedily. This can lead to incorrect results if we have overlapping replacements.
Constructing the Pattern: Building the regular expression pattern can be complex, especially if we have a large number of replacement operations.

When to Consider Regular Expressions:

Regular expressions might be a viable option if the number of replacement operations is small and the patterns are relatively simple. However, for complex scenarios with overlapping replacements, it's generally better to stick with the iterative approach we discussed earlier.

Choosing the Right Approach:

When deciding which approach to use, consider the following factors:

Number of Replacement Operations: If the number of operations is large, optimizations like sorting and pre-computation can significantly improve performance.
Complexity of Patterns: If the patterns are simple, regular expressions might be an option. However, for complex patterns, the iterative approach is generally more robust.
Overlapping Replacements: For scenarios with overlapping replacements, the iterative approach with sorting is usually the best choice.

In most cases, the iterative approach with sorting and pre-computation provides a good balance between performance and code complexity. It's also easier to understand and maintain compared to the regular expression approach.

Common Pitfalls and How to Avoid Them

Solving the "Find and Replace in String" problem involves several nuances, and it's easy to fall into common pitfalls if you're not careful. Let's discuss some of these pitfalls and how to avoid them. By being aware of these potential issues, you can write more robust and bug-free code.

Pitfall 1: Incorrectly Handling Overlapping Replacements

As we've discussed earlier, overlapping replacements are the trickiest part of this problem. If you don't handle them correctly, you might end up with incorrect results. The key is to ensure that you process the replacements in the correct order and that you don't try to apply replacements within a replaced region.

How to Avoid It:

Sort Replacements by Index: Sorting the replacements by their starting indices ensures that you process them in the order they appear in the string s. This helps avoid conflicts due to overlaps.
Advance Index Correctly: When you make a replacement, advance the index i by the length of the source string. This ensures that you don't try to apply replacements within the replaced region.

Pitfall 2: Modifying the String In-Place

Another common mistake is trying to modify the string s in-place. Strings in Python are immutable, which means you can't directly modify them. If you try to modify a string in-place, you'll end up creating a new string, which can be inefficient.

How to Avoid It:

Use a List to Build the Result: Instead of modifying the string s in-place, use a list to store the characters and replaced substrings. This allows you to efficiently build the final string. Once you've processed all the replacements, you can join the elements of the list to form the final string.

Pitfall 3: Incorrectly Checking for Matches

It's crucial to correctly check if the source string matches the substring in s at the given index. If you make a mistake in the matching logic, you might end up making incorrect replacements or skipping valid replacements.

How to Avoid It:

Use String Slicing: Python's string slicing (s[i:i+len(to_find)]) makes it easy to extract substrings for comparison. Make sure you use the correct indices when slicing the string.
Double-Check the Comparison: Carefully double-check your comparison logic to ensure that you're comparing the correct substrings.

Pitfall 4: Not Handling Edge Cases

Failing to handle edge cases can lead to unexpected behavior and incorrect results. It's important to consider edge cases when designing your solution.

Common Edge Cases:

Empty String: What happens if the input string s is empty?
Empty Replacement Lists: What happens if the indices, sources, and targets lists are empty?
Invalid Indices: What happens if the indices are out of bounds?
Empty Source Strings: What happens if the source strings are empty?

How to Avoid It:

Add Explicit Checks: Add explicit checks for edge cases at the beginning of your function. This can help you handle these cases gracefully.
Test with Edge Cases: Always test your solution with edge cases to ensure that it behaves correctly.

Pitfall 5: Inefficient Code

Writing inefficient code can lead to performance issues, especially for large input sizes. It's important to consider the time and space complexity of your solution and to optimize it if necessary.

How to Avoid It:

Use Efficient Data Structures: Choose the right data structures for your problem. For example, using a defaultdict can simplify the code and improve performance.
Avoid Redundant Computations: Avoid performing redundant computations. Pre-compute match results or sort the replacements if necessary.
Optimize Loops: Make sure your loops are efficient. Avoid unnecessary iterations or comparisons.

By being aware of these common pitfalls and how to avoid them, you can write more robust, efficient, and bug-free code for the "Find and Replace in String" problem. Always take the time to carefully analyze your solution and test it thoroughly to ensure that it behaves correctly in all cases.

Conclusion

Alright, guys! We've reached the end of our comprehensive guide to LeetCode problem 833, "Find and Replace in String." We've covered a lot of ground, from understanding the problem statement to implementing an efficient Python solution. We've also discussed potential optimizations, alternative approaches, and common pitfalls to avoid. By now, you should have a solid understanding of how to tackle this challenging problem.

Key Takeaways:

Understanding the Problem: The key to solving any coding problem is to first understand it thoroughly. Make sure you grasp the problem statement, the constraints, and the edge cases.
Devising a Strategy: Before you start coding, take the time to devise a strategy. Break the problem down into smaller subproblems and think about how you'll solve each one.
Implementing the Solution: Translate your strategy into code. Use appropriate data structures and algorithms to implement your solution efficiently.
Optimizing the Code: Once you have a working solution, think about potential optimizations. Can you improve the time or space complexity of your code?
Avoiding Pitfalls: Be aware of common pitfalls and how to avoid them. This will help you write more robust and bug-free code.
Testing Thoroughly: Always test your solution thoroughly. Use a variety of test cases, including edge cases, to ensure that your code behaves correctly.

Final Thoughts:

The "Find and Replace in String" problem is a great example of a coding challenge that requires careful analysis and a well-thought-out strategy. It's not just about writing code; it's about problem-solving, algorithm design, and attention to detail. By mastering problems like this, you'll not only improve your coding skills but also your overall problem-solving abilities.

So, keep practicing, keep learning, and keep challenging yourself. The world of coding is full of exciting problems to solve, and with the right approach, you can conquer them all. Thanks for joining me on this journey, and I'll see you in the next coding adventure! Keep coding, guys!