How do you solve Find And Replace in String on LeetCode?

Find And Replace in String (LeetCode #833) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Replacements should not overlap, allowing for simpler logic.

What is the time complexity of Find And Replace in String?

The optimal solution for Find And Replace in String runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the storage of the result.

What is the brute force solution for Find And Replace in String?

The brute force approach for Find And Replace in String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each replacement operation one by one. For each operation, it verifies if the source string exists at the specified index and performs the replacement if it does. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find And Replace in String?

Find And Replace in String uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find And Replace in String?

Clarify the problem constraints and edge cases. Discuss your thought process and approach before coding. Test your code with different inputs after writing.

What are common mistakes when solving Find And Replace in String?

Not checking if the source string matches before replacing. Forgetting to handle the remaining part of the string after replacements.

#833

Find And Replace in String

Medium

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a list to collect parts of the string that need to be replaced and then constructs the final string in one go. This avoids repeated string concatenation, which is inefficient.

⚙️

Algorithm

6 steps

1Step 1: Create a list 'result' to hold parts of the final string.
2Step 2: Initialize a variable 'lastIndex' to track the end of the last replacement.
3Step 3: Loop through each index in 'indices'.
4Step 4: For each index, check if 'sources[i]' matches the substring of 's' starting at 'indices[i]'.
5Step 5: If it matches, add the substring from 'lastIndex' to 'indices[i]' and the corresponding 'targets[i]' to 'result'. Update 'lastIndex' to the end of the current replacement.
6Step 6: After the loop, add the remaining part of 's' from 'lastIndex' to 'result'. Join the list into a final string and return it.

solution.py15 lines

1s = 'abcd'
2indices = [0, 2]
3sources = ['a', 'cd']
4targets = ['eee', 'ffff']
5
6result = []
7lastIndex = 0
8for i in range(len(indices)):
9    if s[indices[i]:indices[i]+len(sources[i])] == sources[i]:
10        result.append(s[lastIndex:indices[i]])
11        result.append(targets[i])
12        lastIndex = indices[i] + len(sources[i])
13result.append(s[lastIndex:])
14final_result = ''.join(result)
15print(final_result)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the storage of the result.

1Replacements should not overlap, allowing for simpler logic.
2Using a list to build the result is more efficient than repeated string concatenation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.