Approach Summary
Optimal solution for Remove All Occurrences of a Substring with O(n*m) time complexity and O(n+m) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n*m) | O(n+m) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public String removeOccurrences(String s, String part) {
StringBuilder result = new StringBuilder(); // using this as a stack
for(int i=0; i<s.length(); i++){
result.append(s.charAt(i));
if(result.length() >= part.length() && result.substring(result.length() - part.length()).equals(part)){
int n = part.length();
while(n > 0){
result.deleteCharAt(result.length()-1);
n = n - 1;
}
}
}
return result.toString();
}
}class Solution {
public:
string removeOccurrences(string s, string part) {
string result; // using this as a stack
for(int i=0; i<s.size(); i++){
result.push_back(s[i]);
if(result.size() >= part.size() && result.substr(result.size() - part.size()) == part){
int n = part.size();
while(n > 0){
result.pop_back();
n = n - 1;
}
}
}
return result;
}
};class Solution:
def remove_all_occurrences_of_a_substring(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Remove All Occurrences of a Substring
*/
function removeAllOccurrencesOfASubstring(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n*m)
Space: O(n+m)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public String removeOccurrences(String s, String part) {
StringBuilder result = new StringBuilder(); // using this as a stack
for(int i=0; i<s.length(); i++){
result.append(s.charAt(i));
if(result.length() >= part.length() && result.substring(result.length() - part.length()).equals(part)){
int n = part.length();
while(n > 0){
result.deleteCharAt(result.length()-1);
n = n - 1;
}
}
}
return result.toString();
}
}class Solution {
public:
string removeOccurrences(string s, string part) {
string result; // using this as a stack
for(int i=0; i<s.size(); i++){
result.push_back(s[i]);
if(result.size() >= part.size() && result.substr(result.size() - part.size()) == part){
int n = part.size();
while(n > 0){
result.pop_back();
n = n - 1;
}
}
}
return result;
}
};class Solution:
def remove_all_occurrences_of_a_substring(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Remove All Occurrences of a Substring
*/
function removeAllOccurrencesOfASubstring(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Stack Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Stack pattern.
How to Recognize This Pattern
- Common Stack interview pattern
- Optimal time: O(n*m)
Complexity Analysis
Time Complexity
O(n*m)
Space Complexity
O(n+m)
Tags
Stack
Frequently Asked Questions
What is the optimal time and space complexity for Remove All Occurrences of a Substring?
The optimal solution for Remove All Occurrences of a Substring achieves O(n*m) runtime complexity and requires O(n+m) auxiliary space. Time Complexity is O(n*m) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(n+m) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Remove All Occurrences of a Substring requires the Stack pattern?
In technical interviews, look for key problem signals: Common Stack interview pattern; Optimal time: O(n*m). Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Stack strategy.
What critical edge cases must be handled in Remove All Occurrences of a Substring?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Remove All Occurrences of a Substring?
Remove All Occurrences of a Substring is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.