Leetcode 1625: Lexicographically Smallest String After Applying Operations
Given a string s of even length consisting of digits from 0 to 9, and two integers a and b, return the lexicographically smallest string that can be obtained by applying two operations any number of times: adding a to all odd indices of s (with wrapping) and rotating s to the right by b positions.
Problem
Approach
Steps
Complexity
Input: You are given a string s of even length, containing digits from 0 to 9, and two integers a and b.
Example: s = "3637", a = 7, b = 2
Constraints:
• 2 <= s.length <= 100
• s.length is even.
• s consists of digits from 0 to 9 only.
• 1 <= a <= 9
• 1 <= b <= s.length - 1
Output: Return the lexicographically smallest string that can be obtained by applying the two operations any number of times, in any order.
Example: Output: "3036"
Constraints:
• The output should be a string of the same length as s.
Goal: The goal is to find the lexicographically smallest string after applying the operations any number of times.
Steps:
• Apply rotations to s in all possible ways to generate different versions of s.
• For each version, apply the addition operation on odd indices of s.
• Track the lexicographically smallest result by comparing the modified strings.
Goal: The length of the string is between 2 and 100, and the string contains only digits from 0 to 9.
Steps:
• 2 <= s.length <= 100
• s.length is even.
• s consists of digits from 0 to 9 only.
• 1 <= a <= 9
• 1 <= b <= s.length - 1
Assumptions:
• The input string s will always have even length and consist of digits from 0 to 9.
• Input: s = "3637", a = 7, b = 2
• Explanation: By rotating the string twice and adding 7 to odd positions, we get the smallest possible string, which is "3036".
• Input: s = "9783", a = 3, b = 1
• Explanation: After rotating the string and adding 3 to odd indices, we get the smallest possible string, which is "3789".
• Input: s = "1567", a = 4, b = 2
• Explanation: In this case, the string is already the lexicographically smallest, and no further operations can make it smaller.
Approach: To solve this problem, we apply the operations in various combinations, keeping track of the lexicographically smallest string generated.
Observations:
• The operations can be applied multiple times, and we need to explore all possibilities to find the smallest string.
• We will rotate the string and apply the addition operation to each resulting string, comparing to find the smallest possible string.
Steps:
• Use depth-first search (DFS) to explore all possible rotations and additions.
• For each new string generated, check if it is lexicographically smaller than the current smallest string.
• Keep track of visited strings to avoid redundant calculations.
Empty Inputs:
• The string is guaranteed to have a length of at least 2, so we do not need to handle empty inputs.
Large Inputs:
• Ensure the solution works for strings of length up to 100.
Special Values:
• Handle cases where no further operations can produce a smaller string.
Constraints:
• Ensure all operations are applied efficiently, especially when dealing with large strings.
class Solution {
string ans;
int a, b;
int n;
unordered_set<string> seen;
bool isVisited(string s) {
return seen.find(s) != seen.end();
}
void visit(string s) {
ans = min(ans, s);
seen.insert(s);
}
string rotate(string s, int x) {
reverse(s.begin(), s.end());
reverse(s.begin(), s.begin()+x);
reverse(s.begin()+x, s.end());
return s;
}
string add(string s, int x) {
for (int i=1; i<n; i += 2) {
char &c = s[i];
c = '0' + (c-'0'+x)%10;
}
return s;
}
public:
string findLexSmallestString(string s, int a, int b) {
ans = s;
this->a = a;
this->b = b;
n = s.size();
dfs(s);
return ans;
}
void dfs(string s){
if(isVisited(s)) return;
visit(s);
dfs(rotate(s, b));
dfs(add(s, a));
}
1 : Class Definition
class Solution {
Class definition starts. This class encapsulates the solution for the problem.
2 : Variable Initialization
string ans;
String variable 'ans' is used to store the lexicographically smallest string found.
3 : Variable Initialization
int a, b;
Integers 'a' and 'b' are parameters that control the transformation of the string.
4 : Variable Initialization
int n;
Integer 'n' stores the size of the string 's'.
5 : Set Initialization
unordered_set<string> seen;
An unordered_set 'seen' is used to track visited strings to avoid cycles.
6 : Method Definition
bool isVisited(string s) {
Method to check if a string 's' has already been visited.
7 : Set Operations
return seen.find(s) != seen.end();
Returns true if the string 's' has already been visited by checking the 'seen' set.
8 : Method Definition
void visit(string s) {
Method to process and record a visited string.
9 : String Manipulations
ans = min(ans, s);
Update 'ans' to be the lexicographically smallest string between the current 'ans' and 's'.
10 : Set Operations
seen.insert(s);
Insert the string 's' into the 'seen' set to mark it as visited.
11 : Method Definition
string rotate(string s, int x) {
Method to rotate the string 's' by 'x' positions.
12 : String Manipulations
reverse(s.begin(), s.end());
Reverse the entire string 's'.
13 : String Manipulations
reverse(s.begin(), s.begin()+x);
Reverse the first 'x' characters of the string 's'.
14 : String Manipulations
reverse(s.begin()+x, s.end());
Reverse the remaining part of the string from index 'x' to the end.
15 : Return Statement
return s;
Return the rotated string 's'.
16 : Method Definition
string add(string s, int x) {
Method to add 'x' to every alternate digit of the string 's'.
17 : Loop Constructs
for (int i=1; i<n; i += 2) {
Loop through the string 's', starting at index 1 and skipping every other character.
18 : String Manipulations
char &c = s[i];
Get a reference to the character 'c' at the current position 'i' in the string 's'.
19 : Mathematical Operations
c = '0' + (c-'0'+x)%10;
Add 'x' to the character 'c', mod 10 to keep it a single digit, and update 'c'.
20 : Return Statement
return s;
Return the modified string 's'.
21 : Access Modifiers
public:
Start of the public section, where the main function is defined.
22 : Method Definition
string findLexSmallestString(string s, int a, int b) {
Main method to find the lexicographically smallest string using the transformations.
23 : Variable Initialization
ans = s;
Initialize 'ans' with the original string 's'.
24 : Variable Initialization
this->a = a;
Assign the value of 'a' to the class variable 'a'.
25 : Variable Initialization
this->b = b;
Assign the value of 'b' to the class variable 'b'.
26 : Variable Initialization
n = s.size();
Set 'n' to the size of the string 's'.
27 : Recursive Calls
dfs(s);
Call the depth-first search (DFS) to explore all possible transformations.
28 : Return Statement
return ans;
Return the lexicographically smallest string found.
29 : Method Definition
void dfs(string s){
Method to perform depth-first search on the string 's'.
30 : Conditional Statements
if(isVisited(s)) return;
If the string 's' has already been visited, return immediately to avoid cycles.
31 : Method Calls
visit(s);
Mark the current string 's' as visited and possibly update 'ans'.
32 : Recursive Calls
dfs(rotate(s, b));
Recursively explore the rotated version of 's'.
33 : Recursive Calls
dfs(add(s, a));
Recursively explore the version of 's' after applying the 'add' transformation.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The worst case occurs when all rotations and additions must be explored.
Best Case: O(n)
Worst Case: O(n)
Description: Space complexity is O(n) due to the storage of strings in the recursion stack and visited set.
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