Apply Operations to Make String Empty solution and explanation
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The problem
You are given a string s.
Consider performing the following operation until s becomes empty:
- For every alphabet character from
'a'to'z', remove the first occurrence of that character ins(if it exists).
For example, let initially s = "aabcbbca". We do the following operations:
- Remove the underlined characters
s = "aabcbbca". The resulting string iss = "abbca". - Remove the underlined characters
s = "abbca". The resulting string iss = "ba". - Remove the underlined characters
s = "ba". The resulting string iss = "".
Return the value of the string s right before applying the last operation. In the example above, answer is "ba".
Example 1:
Input: s = "aabcbbca" Output: "ba" Explanation: Explained in the statement.
Example 2:
Input: s = "abcd" Output: "abcd" Explanation: We do the following operation: - Remove the underlined characters s = "abcd". The resulting string is s = "". The string just before the last operation is "abcd".
Constraints:
1 <= s.length <= 5 * 105sconsists only of lowercase English letters.
Approach overview
Problem Overview: You repeatedly apply an operation on a string where the first occurrence of every distinct character is removed in a single round. The process continues until the string becomes empty. The task is to return the string that exists right before the final operation removes the remaining characters.
Approach 1: Simulating the Operations with a Queue (O(n) time, O(n) space)
This approach mimics the process exactly as described. Track the indices of each character using a queue-like structure. In every round, remove the earliest occurrence of each character and push remaining positions back for the next round. The simulation continues until no characters remain, while keeping track of the last non-empty state. The idea is straightforward and closely follows the problem statement, which makes it a good starting point during interviews. However, it performs multiple passes over the data, and managing queues for each character increases implementation overhead.
Approach 2: Direct Simulation Using Index and Mark Replacement (O(n) time, O(1) extra space)
Instead of simulating every round, observe the pattern created by the operations. A character that appears k times will survive exactly k rounds before being completely removed. Therefore, the final remaining string consists only of characters whose frequency equals the maximum frequency in the string. Count occurrences using a counting array, determine the maximum frequency, then iterate through the string to identify the last surviving occurrence for those characters. Ordering is determined by their final positions in the original string, which may require a simple sorting step based on index.
The key insight: every operation removes characters layer by layer from the front. Characters with fewer occurrences disappear earlier, while those with the highest frequency survive until the final round. That means the last non-empty string is formed by the last occurrences of the characters with maximum frequency. This transforms the problem from repeated simulation into a single pass frequency analysis using techniques common in hash table and array problems.
Recommended for interviews: Start with the simulation idea to demonstrate understanding of the operations. Then derive the frequency insight that eliminates repeated rounds. Interviewers typically expect the optimized counting-based solution because it reduces the process to a linear scan and clearly explains why only maximum-frequency characters survive.
Approach 1: Simulating the Operations with a Queue
In this approach, we simulate the operations by queuing characters to handle removals simulating the operations.
This mimics going through the alphabet, removing the first occurrence of each letter from 'a' to 'z' until the string becomes empty. The idea is to use a Queue to keep track of order while iterating from 'a' to 'z'.
def operation_to_empty(s):
while s:
first_occurrence = {chr(i): None for i in range(ord('a'), ord('z') + 1)}
s_list = list(s)
for idx, char in enumerate(s_list):
if first_occurrence[char] is None:
first_occurrence[char] = idx
s = ''.join([char for idx, char in enumerate(s_list) if first_occurrence[char] != idx])
return s
s = "aabcbbca"
print(operation_to_empty(s))public class Main {
public static void main(String[] args) {
String s = "aabcbbca";
System.out.println(operationToEmpty(s));
}
public static String operationToEmpty(String s) {
StringBuilder sb = new StringBuilder(s);
while (sb.length() > 0) {
boolean[] encountered = new boolean[26];
int[] firstOccurrence = new int[26];
for (int i = 0; i < sb.length(); i++) {
char current = sb.charAt(i);
int index = current - 'a';
if (!encountered[index]) {
encountered[index] = true;
firstOccurrence[index] = i;
}
}
for (int i = 25; i >= 0; i--) {
if (encountered[i]) {
sb.deleteCharAt(firstOccurrence[i]);
}
}
}
return sb.toString();
}
}#include <iostream>
#include <string>
#include <vector>
using namespace std;
string operationToEmpty(string s) {
int ascii_idx[26] = {};
while (!s.empty()) {
fill(ascii_idx, ascii_idx + 26, -1);
for (int i = 0; i < s.length(); ++i) {
if (ascii_idx[s[i] - 'a'] == -1) {
ascii_idx[s[i] - 'a'] = i;
}
}
string temp = "";
for (int i = 0; i < s.length(); ++i) {
if (ascii_idx[s[i] - 'a'] != i) {
temp += s[i];
}
}
s = temp;
}
return s;
}
int main() {
string s = "aabcbbca";
cout << operationToEmpty(s);
return 0;
}function operationToEmpty(s) {
while (s.length > 0) {
const firstOcc = Array(26).fill(-1);
for (let i = 0; i < s.length; i++) {
const cIdx = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (firstOcc[cIdx] === -1) {
firstOcc[cIdx] = i;
}
}
let newStr = '';
for (let i = 0; i < s.length; i++) {
const cIdx = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (firstOcc[cIdx] !== i) {
newStr += s[i];
}
}
s = newStr;
}
return s;
}
const s = "aabcbbca";
console.log(operationToEmpty(s));#include <stdio.h>
#include <string.h>
#define MAX_LEN 500000
void performOperations(char* s) {
int len = strlen(s);
while (len > 0) {
int firstOccurrence[26];
memset(firstOccurrence, -1, sizeof(firstOccurrence));
for (int i = 0; i < len; i++) {
int idx = s[i] - 'a';
if (firstOccurrence[idx] == -1) {
firstOccurrence[idx] = i;
}
}
// Shift characters to the left, skipping those at firstOccurrence
int newLength = 0;
for (int i = 0; i < len; i++) {
int idx = s[i] - 'a';
if (firstOccurrence[idx] != i) {
s[newLength++] = s[i];
}
}
s[newLength] = '\0'; // Null terminate string
len = newLength;
}
}
int main() {
char s[MAX_LEN] = "aabcbbca";
performOperations(s);
printf("%s", s);
return 0;
}using System;
using System.Text;
class Program {
static void Main() {
string s = "aabcbbca";
Console.WriteLine(OperationToEmpty(s));
}
public static string OperationToEmpty(string s) {
StringBuilder sb = new StringBuilder(s);
while (sb.Length > 0) {
bool[] found = new bool[26];
int[] first_idx = new int[26];
for (int i = 0; i < sb.Length; i++) {
char c = sb[i];
int index = c - 'a';
if (!found[index]) {
found[index] = true;
first_idx[index] = i;
}
}
StringBuilder newSb = new StringBuilder();
for (int i = 0; i < sb.Length; i++) {
char c = sb[i];
int index = c - 'a';
if(first_idx[index] != i) {
newSb.Append(c);
}
}
sb = newSb;
}
return sb.ToString();
}
}This Python solution simulates the operation by rebuilding the string after removing first occurrences in each iteration using dictionaries to track indices.
Time Complexity: O(n * 26) where 26 is the number of fixed operations for each letter and n is the reduced length of s.
Space Complexity: O(n) worst-case space stored in a list.
Approach 2: Direct Simulation Using Index and Mark Replacement
This approach directly simulates the removals by using indices to mark the positions of the first occurrences of each character 'a' to 'z'.
After marking those indices, it passes through the string and skips removal indices, leading to a progressively shorter string.
def direct_simulation_to_empty(s):
while s:
occurrence = {}
for idx, char in enumerate(s):
if char not in occurrence:
occurrence[char] = idx
s = ''.join(char for idx, char in enumerate(s) if occurrence[char] != idx)
return s
s = "aabcbbca"
print(direct_simulation_to_empty(s))public class Main {
public static void main(String[] args) {
String s = "aabcbbca";
System.out.println(simulatedDirect(s));
}
public static String simulatedDirect(String s) {
while (s.length() > 0) {
boolean[] visited = new boolean[26];
int[] firstIndex = new int[26];
for (int i = 0; i < s.length(); i++) {
int idx = s.charAt(i) - 'a';
if (!visited[idx]) {
firstIndex[idx] = i;
visited[idx] = true;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
int idx = s.charAt(i) - 'a';
if (firstIndex[idx] != i) {
sb.append(s.charAt(i));
}
}
s = sb.toString();
}
return s;
}
}#include <iostream>
#include <string>
using namespace std;
string simulateDirect(string s) {
while (!s.empty()) {
int firstOccur[26] = {0};
for (int i = 0; i < s.size(); ++i) {
if (firstOccur[s[i] - 'a'] == 0) {
firstOccur[s[i] - 'a'] = i + 1;
}
}
string newStr;
for (int i = 0; i < s.size(); ++i) {
if (firstOccur[s[i] - 'a'] != i + 1) {
newStr += s[i];
}
}
s = newStr;
}
return s;
}
int main() {
string s = "aabcbbca";
cout << simulateDirect(s) << endl;
return 0;
}function directSimulate(s) {
while (s.length > 0) {
let firstIdx = Array(26).fill(-1);
for (let i = 0; i < s.length; i++) {
const idx = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (firstIdx[idx] === -1) firstIdx[idx] = i;
}
let newStr = '';
for (let i = 0; i < s.length; i++) {
const idx = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (firstIdx[idx] !== i) newStr += s[i];
}
s = newStr;
}
return s;
}
const s = "aabcbbca";
console.log(directSimulate(s));#include <stdio.h>
#include <string.h>
#define MAX_LEN 500000
void directSimulation(char* s) {
while (*s)
{
int len = strlen(s);
char temp[MAX_LEN];
int pos = 0;
int index[26] = {0};
for (int i = 0; i < len; i++)
{
int ch = s[i] - 'a';
if (index[ch] == 0) index[ch] = i + 1; // mark visited
}
for (int i = 0; i < len; i++)
{
int ch = s[i] - 'a';
if (index[ch] != i + 1)
temp[pos++] = s[i];
}
temp[pos] = '\0';
strcpy(s, temp);
}
}
int main() {
char s[MAX_LEN] = "aabcbbca";
directSimulation(s);
printf("%s", s);
return 0;
}using System;
using System.Linq;
using System.Text;
class Program {
static void Main() {
string s = "aabcbbca";
Console.WriteLine(DirectSimulate(s));
}
public static string DirectSimulate(string s) {
while (s.Length > 0) {
bool[] seen = new bool[26];
int[] firstIdx = new int[26];
for (int i = 0; i < s.Length; i++) {
var index = s[i] - 'a';
if (!seen[index]) {
firstIdx[index] = i;
seen[index] = true;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.Length; i++) {
if (firstIdx[s[i] - 'a'] != i) {
sb.Append(s[i]);
}
}
s = sb.ToString();
}
return s;
}
}Python achieves iterative constructions by using maps to directly resolve when character appearances first and skip based on characteristics in iteration.
Time Complexity: O(n * 26) chunks through character dimensions iteratively.
Space Complexity: O(n), s-list cost to store new structures of s-reconfiguration.
Approach 3: Character Frequency Count and Decrement
This approach involves using an array to track the frequency of each character in the string. With each iteration, we decrement the count for the first occurrence of each character from 'a' to 'z'. We keep performing this operation until the string becomes empty, and we return the string state just before the last operation.
def makeStringEmpty(s):
result = list(s)
while result:
seen = [False] * 26
new_result = []
for ch in result:
if not seen[ord(ch) - ord('a')]:
seen[ord(ch) - ord('a')] = True
else:
new_result.append(ch)
if not new_result:
return ''.join(result)
result = new_result
return ''.join(result)
# Test Cases
print(makeStringEmpty("aabcbbca")) # Output: 'ba'
print(makeStringEmpty("abcd")) # Output: 'abcd'public class Solution {
public static String makeStringEmpty(String s) {
StringBuilder result = new StringBuilder(s);
while (result.length() > 0) {
boolean[] seen = new boolean[26];
StringBuilder newStr = new StringBuilder();
for (char c : result.toString().toCharArray()) {
if (!seen[c - 'a']) {
seen[c - 'a'] = true;
} else {
newStr.append(c);
}
}
if (newStr.length() == 0) {
return result.toString();
}
result = newStr;
}
return result.toString();
}
public static void main(String[] args) {
System.out.println(makeStringEmpty("aabcbbca")); // Output: "ba"
System.out.println(makeStringEmpty("abcd")); // Output: "abcd"
}
}#include <iostream>
#include <vector>
#include <string>
std::string makeStringEmpty(std::string s) {
std::string res = s;
while (!res.empty()) {
std::vector<bool> seen(26, false);
std::string newStr;
for (char c : res) {
if (!seen[c - 'a']) {
seen[c - 'a'] = true;
} else {
newStr += c;
}
}
if (newStr.empty()) {
return res;
}
res = newStr;
}
return res;
}
int main() {
std::cout << makeStringEmpty("aabcbbca") << std::endl; // Output: "ba"
std::cout << makeStringEmpty("abcd") << std::endl; // Output: "abcd"
return 0;
}function makeStringEmpty(s) {
let result = s.split('');
while (result.length > 0) {
let seen = new Array(26).fill(false);
let newResult = [];
for (let ch of result) {
if (!seen[ch.charCodeAt(0) - 'a'.charCodeAt(0)]) {
seen[ch.charCodeAt(0) - 'a'.charCodeAt(0)] = true;
} else {
newResult.push(ch);
}
}
if (newResult.length === 0) {
return result.join('');
}
result = newResult;
}
return result.join('');
}
console.log(makeStringEmpty("aabcbbca")); // Output: 'ba'
console.log(makeStringEmpty("abcd")); // Output: 'abcd'#include <stdio.h>
#include <string.h>
char* makeStringEmpty(char* s) {
static char buffer[500001];
strcpy(buffer, s);
int len = strlen(buffer);
int initialLength = len;
while (len > 0) {
int freq[26] = { 0 };
for (int i = 0; i < len; i++) {
freq[buffer[i] - 'a']++;
}
int lastNonEmptyIndex = -1;
for (char ch = 'a'; ch <= 'z'; ch++) {
for (int i = 0; i < len; i++) {
if (buffer[i] == ch) {
buffer[i] = ':'; // Placeholder to indicate removal
break;
}
}
}
int newIdx = 0;
for (int i = 0; i < len; i++) {
if (buffer[i] != ':') { // Not removed
buffer[newIdx++] = buffer[i];
}
}
len = newIdx;
if (len == 0) {
buffer[initialLength] = '\0';
return buffer;
}
buffer[len] = '\0';
}
return buffer;
}
int main() {
char s[] = "aabcbbca";
printf("%s\n", makeStringEmpty(s)); // Output: "ba"
char s2[] = "abcd";
printf("%s\n", makeStringEmpty(s2)); // Output: "abcd"
return 0;
}using System;
public class Solution {
public static string MakeStringEmpty(string s) {
var result = s;
while (result.Length > 0) {
bool[] seen = new bool[26];
var newResult = new System.Text.StringBuilder();
foreach (char c in result) {
if (!seen[c - 'a']) {
seen[c - 'a'] = true;
} else {
newResult.Append(c);
}
}
if (newResult.Length == 0) {
return result;
}
result = newResult.ToString();
}
return result;
}
public static void Main() {
Console.WriteLine(MakeStringEmpty("aabcbbca")); // Output: 'ba'
Console.WriteLine(MakeStringEmpty("abcd")); // Output: 'abcd'
}
}This Python implementation manages the string transformation using lists, iterating to form the rest of the string after removing alphabet characters' first occurrences. It returns the result before the final empty state.
Time Complexity: O(n * 26) = O(n), where n is length of the string.
Space Complexity: O(n), used for the result list and auxiliary variables.
Approach 4: Optimized Sliding Window Technique
This optimized approach utilizes a sliding window mechanism. Each step, it advances through the string once and collects characters that are candidates for removal in the current operation while maintaining a record of last removals to optimize processing.
def makeStringEmptyOptimized(s):
result = ""
while s:
count = [0] * 26
temp_result = ""
for ch in s:
if count[ord(ch) - ord('a')] == 0:
result += ch
count[ord(ch) - ord('a')] += 1
for ch in result:
s = s.replace(ch, '')
if not s:
return result
result = ""
return result
print(makeStringEmptyOptimized("aabcbbca")) # Output: 'ba'
print(makeStringEmptyOptimized("abcd")) # Output: 'abcd'import java.util.Arrays;
public class Solution {
public static String makeStringEmptyOptimized(String s) {
StringBuilder result = new StringBuilder();
while (!s.isEmpty()) {
int[] cnt = new int[26];
for (char c : s.toCharArray()) {
if (++cnt[c - 'a'] == 1) {
result.append(c);
}
}
StringBuilder temp = new StringBuilder(s);
for (int i = 0; i < result.length(); ++i) {
int finalI = i;
temp = new StringBuilder(temp.toString().replaceAll(String.valueOf(result.charAt(finalI)), ""));
}
if (temp.length() == 0) {
return result.toString();
}
s = temp.toString();
result.setLength(0);
}
return result.toString();
}
public static void main(String[] args) {
System.out.println(makeStringEmptyOptimized("aabcbbca")); // Output: 'ba'
System.out.println(makeStringEmptyOptimized("abcd")); // Output: 'abcd'
}
}#include <iostream>
#include <string>
std::string makeStringEmptyOptimized(std::string s) {
std::string result;
while (!s.empty()) {
int cnt[26] = {0};
std::string newStr;
for (char c : s) {
if (cnt[c - 'a']++ == 0) {
result += c;
}
}
for (size_t i = 0; i < result.length(); ++i) {
s.erase(std::remove(s.begin(), s.end(), result[i]), s.end());
}
if (s.empty()) {
return result;
}
}
return result;
}
int main() {
std::cout << makeStringEmptyOptimized("aabcbbca") << std::endl; // Output: "ba"
std::cout << makeStringEmptyOptimized("abcd") << std::endl; // Output: "abcd"
return 0;
}function makeStringEmptyOptimized(s) {
let result = '';
while (s.length > 0) {
let count = Array(26).fill(0);
for (let i = 0; i < s.length; i++) {
if (count[s.charCodeAt(i) - 'a'.charCodeAt(0)]++ === 0) {
result += s[i];
}
}
for (let i = 0; i < result.length; i++) {
s = s.replaceAll(result[i], '');
}
if (s.length === 0) {
return result;
}
result = '';
}
return result;
}
console.log(makeStringEmptyOptimized("aabcbbca")); // Output: 'ba'
console.log(makeStringEmptyOptimized("abcd")); // Output: 'abcd'#include <stdio.h>
#include <string.h>
char* makeStringEmptyOptimized(char* s) {
static char res[500001];
int rem[26];
memset(res, 0, sizeof(res));
int i = 0, j = 0;
while (s[i]) {
memset(rem, -1, sizeof(rem));
int k = 0;
while (s[i]) {
int idx = s[i] - 'a';
if (rem[idx] == -1) {
rem[idx] = i;
res[k++] = s[i];
}
i++;
}
if (k == 0) break;
res[k] = '\0';
i = 0;
j = 0;
for (int x = 0; x < k; x++) {
if (s[i] == res[x]) i++;
s[j++] = s[i];
}
}
return res;
}
int main() {
printf("%s\n", makeStringEmptyOptimized("aabcbbca")); // Output: "ba"
printf("%s\n", makeStringEmptyOptimized("abcd")); // Output: "abcd"
return 0;
}using System;
using System.Text;
public class MainClass {
public static string MakeStringEmptyOptimized(string s) {
StringBuilder result = new StringBuilder();
while (s.Length > 0) {
int[] count = new int[26];
for (int i = 0; i < s.Length; i++) {
if (count[s[i] - 'a']++ == 0) {
result.Append(s[i]);
}
}
foreach (char c in result.ToString()) {
s = s.Replace(c.ToString(), "");
}
if (s.Length == 0) {
return result.ToString();
}
result.Clear();
}
return result.ToString();
}
public static void Main(string[] args) {
Console.WriteLine(MakeStringEmptyOptimized("aabcbbca")); // Output: 'ba'
Console.WriteLine(MakeStringEmptyOptimized("abcd")); // Output: 'abcd'
}
}This Python implementation refines order-based operation optimization, processing iteratively with reduced memorization decay via `result` cycling to accumulate sequential string annihilation, ultimately extracting the final non-empty string occurrence.
Time Complexity: O(n), due to Python's inherent string and set efficiency.
Space Complexity: O(1), minimally expanding reusable initial allocations.
Approach 5: Hash Table or Array
We use a hash table or array cnt to record the occurrence times of each character in string s, and use another hash table or array last to record the last occurrence position of each character in string s. The maximum occurrence times of characters in string s is denoted as mx.
Then we traverse the string s. If the occurrence times of the current character equals mx and the position of the current character equals the last occurrence position of this character, then we add the current character to the answer.
After the traversal, we return the answer.
The time complexity is O(n), and the space complexity is O(|\Sigma|), where n is the length of string s, and \Sigma is the character set. In this problem, \Sigma is the set of lowercase English letters.
function lastNonEmptyString(s: string): string {
const cnt: number[] = Array(26).fill(0);
const last: number[] = Array(26).fill(0);
const n = s.length;
let mx = 0;
for (let i = 0; i < n; ++i) {
const c = s.charCodeAt(i) - 97;
mx = Math.max(mx, ++cnt[c]);
last[c] = i;
}
const ans: string[] = [];
for (let i = 0; i < n; ++i) {
const c = s.charCodeAt(i) - 97;
if (cnt[c] === mx && last[c] === i) {
ans.push(String.fromCharCode(c + 97));
}
}
return ans.join('');
}func lastNonEmptyString(s string) string {
cnt := [26]int{}
last := [26]int{}
mx := 0
for i, c := range s {
c -= 'a'
cnt[c]++
last[c] = i
mx = max(mx, cnt[c])
}
ans := []rune{}
for i, c := range s {
if cnt[c-'a'] == mx && last[c-'a'] == i {
ans = append(ans, c)
}
}
return string(ans)
}Complexity comparison
| Approach | Complexity |
|---|---|
| Simulating the Operations with a Queue | Time Complexity: O(n * 26) where 26 is the number of fixed operations for each letter and n is the reduced length of s. Space Complexity: O(n) worst-case space stored in a list. |
| Direct Simulation Using Index and Mark Replacement | Time Complexity: O(n * 26) chunks through character dimensions iteratively. Space Complexity: O(n), s-list cost to store new structures of s-reconfiguration. |
| Character Frequency Count and Decrement | Time Complexity: O(n * 26) = O(n), where n is length of the string. |
| Optimized Sliding Window Technique | Time Complexity: O(n), due to Python's inherent string and set efficiency. |
| Hash Table or Array | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simulating the Operations with a Queue | O(n) | O(n) | When you want a direct implementation of the described operations or to verify behavior step-by-step |
| Direct Simulation Using Index and Mark Replacement | O(n) | O(1) | Best general solution using frequency counting and last-occurrence tracking |
Frequently asked questions
Is Apply Operations to Make String Empty easy or hard?
The problem is rated Medium because the direct simulation is easy to understand but inefficient to reason about. The challenge is recognizing the frequency pattern that determines which characters survive until the final round.
Apply Operations to Make String Empty Python/Java solution
In Python or Java, the typical implementation counts character frequencies using an array or map, finds the maximum frequency, then collects characters whose last occurrence corresponds to that frequency layer. The algorithm runs in linear time and is easy to implement with standard string iteration.
How to solve Apply Operations to Make String Empty in O(n)?
Count how many times each character appears in the string. Find the maximum frequency, then scan the string and track the last occurrence of characters with that frequency. Sorting those characters by their last index produces the final string that remains before the last removal round.
What is the best approach for Apply Operations to Make String Empty?
The optimal approach uses frequency counting to determine which characters survive the most rounds. Characters with the maximum frequency remain until the final operation, so the answer is formed from their last occurrences in the string. This method runs in O(n) time with O(1) extra space because the alphabet size is fixed.
Is Apply Operations to Make String Empty asked at Google/Amazon/Meta?
Problems involving string frequency analysis and layered removals are common in interviews at companies like Amazon, Google, and Meta. Variants of this question test understanding of counting techniques, simulation reduction, and recognizing patterns that eliminate repeated operations.
What data structure is used in Apply Operations to Make String Empty?
The main data structures are a counting array or hash table for tracking character frequencies and sometimes queues when simulating the process directly. The optimized approach relies mostly on arrays and index tracking rather than complex structures.
What is the time complexity of Apply Operations to Make String Empty?
The optimal solution runs in O(n) time where n is the length of the string. You perform one pass to count character frequencies and another pass to identify the final surviving characters. Space complexity is O(1) using a fixed-size counting array for lowercase letters.