Leetcode #833: Find And Replace in String
In this guide, we solve Leetcode #833 Find And Replace in String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string s that you must perform k replacement operations on. The replacement operations are given as three 0-indexed parallel arrays, indices, sources, and targets, all of length k.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, String, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "abcd", indices = [0, 2], sources = ["a", "cd"], targets = ["eee", "ffff"]
Output: "eeebffff"
Explanation:
"a" occurs at index 0 in s, so we replace it with "eee".
"cd" occurs at index 2 in s, so we replace it with "ffff".
Python Solution
class Solution:
def findReplaceString(
self, s: str, indices: List[int], sources: List[str], targets: List[str]
) -> str:
n = len(s)
d = [-1] * n
for k, (i, src) in enumerate(zip(indices, sources)):
if s.startswith(src, i):
d[i] = k
ans = []
i = 0
while i < n:
if ~d[i]:
ans.append(targets[d[i]])
i += len(sources[d[i]])
else:
ans.append(s[i])
i += 1
return "".join(ans)
Complexity
The time complexity is , and the space complexity is , where is the sum of the lengths of all strings, and is the length of the string . The space complexity is , where is the sum of the lengths of all strings, and is the length of the string .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.