Skip to content

Commit de53702

Browse files
ajay-dhangarajay-dhangar
authored
Merge pull request #1347 from Hitesh4278/lexicographically-minimum-string-after-removing-stars
Added the Solution for Lexicographically Minimum String After Removing # Issue No - 1229
2 parents 36b24d1 + e3897d7 commit de53702

File tree

2 files changed

+314
-1
lines changed

2 files changed

+314
-1
lines changed

‎dsa-problems/leetcode-problems/3100-3199.md

Lines changed: 6 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -424,7 +424,12 @@ export const problems = [
424424
"leetCodeLink": "https://leetcode.com/problems/count-days-without-meetings/description/",
425425
"solutionLink": "/dsa-solutions/lc-solutions/3100-3199/count-days-without-meetings"
426426
},
427-
427+
{
428+
"problemName": "3170. Lexicographically Minimum String After Removing Stars",
429+
"difficulty": "Medium",
430+
"leetCodeLink": "https://leetcode.com/problems/lexicographically-minimum-string-after-removing-stars/description/",
431+
"solutionLink": "/dsa-solutions/lc-solutions/3100-3199/lexicographically-minimum-string-after-removing-stars"
432+
},
428433
{
429434
"problemName": "3179.Find the N-th Value After K Seconds",
430435
"difficulty": "Medium",

‎dsa-solutions/lc-solutions/3100-3199/3170-lexicographically-minimum-string-after-removing-stars.md

Lines changed: 308 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,308 @@
1+
---
2+
id: lexicographically-minimum-string-after-removing-stars
3+
title: Lexicographically Minimum String After Removing Stars
4+
sidebar_label: 3170 . Lexicographically Minimum String After Removing Stars
5+
tags:
6+
- Hash Table
7+
- String
8+
- Stack
9+
- Greedy
10+
- Heap (Priority Queue)
11+
description: "This is a solution to the Lexicographically Minimum String After Removing Stars problem on LeetCode."
12+
---
13+
14+
## Problem Description
15+
You are given a string s. It may contain any number of '*' characters. Your task is to remove all '*' characters.
16+
17+
While there is a '*', do the following operation:
18+
19+
Delete the leftmost '*' and the smallest non-'*' character to its left. If there are several smallest characters, you can delete any of them.
20+
Return the lexicographically smallest resulting string after removing all '*' characters.
21+
22+
### Examples
23+
24+
**Example 1:**
25+
26+
```
27+
Input: s = "aaba*"
28+
29+
Output: "aab"
30+
31+
Explanation:
32+
33+
We should delete one of the 'a' characters with '*'. If we choose s[3], s becomes the lexicographically smallest.
34+
```
35+
36+
**Example 2:**
37+
```
38+
Input: s = "abc"
39+
40+
Output: "abc"
41+
42+
Explanation:
43+
44+
There is no '*' in the string.
45+
```
46+
47+
### Constraints
48+
49+
- `1 <= s.length <= 10^5`
50+
- `s consists only of lowercase English letters and '*'.`
51+
- `The input is generated such that it is possible to delete all '*' characters.`
52+
53+
## Solution for Lexicographically Minimum String After Removing Stars Problem
54+
### Approach
55+
56+
#### Initialization:
57+
58+
- We initialize two sets:
59+
- st: A set to store pairs of (character value, negative index). This helps us quickly identify and remove the leftmost character when we encounter a star.
60+
- del: A set to keep track of the indices of characters that need to be deleted (either because they are stars or they were removed by stars).
61+
#### Traversing the String:
62+
- We iterate over the string from left to right.
63+
#### For each character:
64+
- If the character is a star (*):
65+
- We identify the leftmost character in the set st by taking the smallest element (due to the ordering by character value and then by index).
66+
- We remove this leftmost character from the set st.
67+
- We record the indices of this leftmost character and the star itself in the del set.
68+
- If the character is not a star:
69+
- We add a pair of (character value, negative index) to the set st.
70+
#### Building the Result:
71+
72+
- We initialize an empty string ans to store the result.
73+
- We iterate over the original string again.
74+
- For each character:
75+
- If the index of the character is not in the del set (i.e., it hasn't been marked for deletion), we append it to ans.
76+
77+
<Tabs>
78+
<TabItem value="Solution" label="Solution">
79+
80+
#### Implementation
81+
```jsx live
82+
function Solution(arr) {
83+
var clearStars = function(s) {
84+
const mp = new Map();
85+
const n = s.length;
86+
const v = new Array(n).fill(0);
87+
88+
for (let i = 0; i < n; i++) {
89+
if (s[i] !== '*') {
90+
if (!mp.has(s[i])) {
91+
mp.set(s[i], []);
92+
}
93+
mp.get(s[i]).push(i);
94+
} else {
95+
v[i] = 1;
96+
const sortedEntries = Array.from(mp.entries()).sort((a, b) => a[0].localeCompare(b[0]));
97+
for (let [key, indices] of sortedEntries) {
98+
const m = indices.length;
99+
v[indices[m - 1]] = 1;
100+
indices.pop();
101+
if (indices.length === 0) {
102+
mp.delete(key);
103+
}
104+
break;
105+
}
106+
}
107+
}
108+
109+
let ans = "";
110+
for (let i = 0; i < n; i++) {
111+
if (v[i] !== 1) {
112+
ans += s[i];
113+
}
114+
}
115+
return ans;
116+
};
117+
const input = "aaba*"
118+
const output =clearStars(input)
119+
return (
120+
<div>
121+
<p>
122+
<b>Input: </b>
123+
{JSON.stringify(input)}
124+
</p>
125+
<p>
126+
<b>Output:</b> {output.toString()}
127+
</p>
128+
</div>
129+
);
130+
}
131+
```
132+
133+
#### Complexity Analysis
134+
135+
- Time Complexity: $ O(nlogn) $ because of Nested Loops
136+
- Space Complexity: $ O(n) $ because of prefix array
137+
138+
## Code in Different Languages
139+
<Tabs>
140+
<TabItem value="JavaScript" label="JavaScript">
141+
<SolutionAuthor name="@hiteshgahanolia"/>
142+
```javascript
143+
var clearStars = function(s) {
144+
const mp = new Map();
145+
const n = s.length;
146+
const v = new Array(n).fill(0);
147+
148+
for (let i = 0; i < n; i++) {
149+
if (s[i] !== '*') {
150+
if (!mp.has(s[i])) {
151+
mp.set(s[i], []);
152+
}
153+
mp.get(s[i]).push(i);
154+
} else {
155+
v[i] = 1;
156+
const sortedEntries = Array.from(mp.entries()).sort((a, b) => a[0].localeCompare(b[0]));
157+
for (let [key, indices] of sortedEntries) {
158+
const m = indices.length;
159+
v[indices[m - 1]] = 1;
160+
indices.pop();
161+
if (indices.length === 0) {
162+
mp.delete(key);
163+
}
164+
break;
165+
}
166+
}
167+
}
168+
169+
let ans = "";
170+
for (let i = 0; i < n; i++) {
171+
if (v[i] !== 1) {
172+
ans += s[i];
173+
}
174+
}
175+
return ans;
176+
};
177+
```
178+
179+
</TabItem>
180+
<TabItem value="TypeScript" label="TypeScript">
181+
<SolutionAuthor name="@hiteshgahanolia"/>
182+
```typescript
183+
function clearStars(s: string): string {
184+
const n = s.length;
185+
const st: Set<[number, number]> = new Set();
186+
const del: Set<number> = new Set();
187+
for (let i = 0; i < n; i++) {
188+
if (s[i] === '*') {
189+
const first = Array.from(st)[0];
190+
st.delete(first);
191+
del.add(-first[1]);
192+
del.add(i);
193+
} else {
194+
st.add([s.charCodeAt(i) - 'a'.charCodeAt(0), -i]);
195+
}
196+
}
197+
198+
let ans = '';
199+
for (let i = 0; i < n; i++) {
200+
if (!del.has(i)) ans += s[i];
201+
}
202+
return ans;
203+
}
204+
205+
```
206+
</TabItem>
207+
<TabItem value="Python" label="Python">
208+
<SolutionAuthor name="@hiteshgahanolia"/>
209+
```python
210+
def clearStars(s: str) -> str:
211+
n = len(s)
212+
st = set()
213+
del_set = set()
214+
215+
for i in range(n):
216+
if s[i] == '*':
217+
first = min(st)
218+
st.remove(first)
219+
del_set.add(-first[1])
220+
del_set.add(i)
221+
else:
222+
st.add((ord(s[i]) - ord('a'), -i))
223+
224+
ans = ''.join(s[i] for i in range(n) if i not in del_set)
225+
return ans
226+
227+
```
228+
</TabItem>
229+
<TabItem value="Java" label="Java">
230+
<SolutionAuthor name="@hiteshgahanolia"/>
231+
```java
232+
import java.util.*;
233+
234+
public class Solution {
235+
public String clearStars(String s) {
236+
int n = s.length();
237+
TreeSet<Pair> st = new TreeSet<>((a, b) -> a.first == b.first ? a.second - b.second : a.first - b.first);
238+
Set<Integer> del = new HashSet<>();
239+
240+
for (int i = 0; i < n; i++) {
241+
if (s.charAt(i) == '*') {
242+
Pair first = st.first();
243+
st.remove(first);
244+
del.add(-first.second);
245+
del.add(i);
246+
} else {
247+
st.add(new Pair(s.charAt(i) - 'a', -i));
248+
}
249+
}
250+
251+
StringBuilder ans = new StringBuilder();
252+
for (int i = 0; i < n; i++) {
253+
if (!del.contains(i)) ans.append(s.charAt(i));
254+
}
255+
return ans.toString();
256+
}
257+
258+
class Pair {
259+
int first, second;
260+
261+
Pair(int first, int second) {
262+
this.first = first;
263+
this.second = second;
264+
}
265+
}
266+
}
267+
268+
```
269+
270+
</TabItem>
271+
<TabItem value="C++" label="C++">
272+
<SolutionAuthor name="@hiteshgahanolia"/>
273+
```cpp
274+
class Solution {
275+
public:
276+
string clearStars(string s) {
277+
int n = s.size();
278+
set<pair<int,int>>st;
279+
set<int> del;
280+
for (int i = 0; i < n; i++) {
281+
if (s[i] == '*') {
282+
auto first = *st.begin();
283+
st.erase(st.begin());
284+
del.insert(-first.second);
285+
del.insert(i);
286+
} else {
287+
st.insert({s[i]-'a' , -i});
288+
}
289+
}
290+
291+
string ans = "";
292+
for (int i = 0; i < n; i++) if (!del.count(i)) ans += s[i];
293+
return ans;
294+
}
295+
};
296+
```
297+
</TabItem>
298+
</Tabs>
299+
300+
</TabItem>
301+
</Tabs>
302+
303+
## References
304+
305+
- **LeetCode Problem**: [ Lexicographically Minimum String After Removing Stars](https://leetcode.com/problems/lexicographically-minimum-string-after-removing-stars/description/)
306+
307+
- **Solution Link**: [LeetCode Solution](https://leetcode.com/problems/lexicographically-minimum-string-after-removing-stars/solutions)
308+

0 commit comments

Comments
 (0)