Lexicographically Smallest String After Applying Operations

Tags : string, bfs, leetcode, cpp, medium

You are given a string s of even length consisting of digits from 0 to 9, and two integers a and b.

You can apply either of the following two operations any number of times and in any order on s:

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "0158" is lexicographically smaller than "0190" because the first position they differ is at the third letter, and '5' comes before '9'.

Examples #

Example 1:

Input: s = "5525", a = 9, b = 2
Output: "2050"
Explanation: We can apply the following operations:
Start:  "5525"
Rotate: "2555"
Add:    "2454"
Add:    "2353"
Rotate: "5323"
Add:    "5222"
​​​​​​​Add:    "5121"
​​​​​​​Rotate: "2151"
​​​​​​​Add:    "2050"​​​​​​​​​​​​
There is no way to obtain a string that is lexicographically smaller then "2050".

Example 2:

Input: s = "74", a = 5, b = 1
Output: "24"
Explanation: We can apply the following operations:
Start:  "74"
Rotate: "47"
​​​​​​​Add:    "42"
​​​​​​​Rotate: "24"​​​​​​​​​​​​
There is no way to obtain a string that is lexicographically smaller then "24".

Example 3:

Input: s = "0011", a = 4, b = 2
Output: "0011"
Explanation: There are no sequence of operations that will give us a lexicographically smaller string than "0011".

Example 4:

Input: s = "43987654", a = 7, b = 3
Output: "00553311"

Constraints #

Solutions #

class Solution {
public:
    string op_a(string s, int a){
      for(int i=1; i<s.size(); i+=2)
        s[i] = (s[i] - '0' + a) % 10 + '0';        
      return s;
    }
    string findLexSmallestString(string s, int a, int b) {
     set<string> mp;
     queue<string> q;
     q.push(s);mp.insert(s);
     string res = s;
     while(q.empty()==false){
       string temp = q.front();
       q.pop();
       res = min(res, temp);
       string one = op_a(temp, a);
       if(mp.find(one)==mp.end()) mp.insert(one), q.push(one);
       rotate(temp.begin(), temp.begin()+b, temp.end());
       if(mp.find(temp)==mp.end()) mp.insert(temp), q.push(temp);
     }
      return res;
    }
};