1625. Lexicographically Smallest String After Applying Operations

问题：

给定一个由0～9构成的字符串s

对其有两种操作：

• 偶数位置+a（超过9，则从0开始循环）
  • 例：if s = "3456" and a = 5, s becomes "3951".
• 字符串向后移动 b位（后面的字符循环到字符串前部）
  • 例：if s = "3456" and b = 1, s becomes "6345".

求进行多次操作后，能获得的数值最小的字符串是什么。

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:
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

　　

解法：BFS

本题的操作会出现字符串循环出现。

• 状态：
  • 字符串s，
  • visited标记已经访问过的字符串。
• 选择：
  • add
  • rotate

 

对于每次出队字符串cur，和结果res取最小

最终遍历完成，得到res，即为最小。

 

代码参考：

class Solution {
public:
    int n;
    string add(string s, int a) {
        for(int i=1; i<n; i+=2) {
            s[i] = ((s[i]-'0'+a)%10) + '0';
        }
        return s;
    }
    string rotate(string s, int b) {
        string s_back = s.substr(n-b);
        string s_front = s.substr(0,n-b);
        return s_back+s_front;
    }
    string findLexSmallestString(string s, int a, int b) {
        n = s.length();
        string res = s;
        queue<string> q;
        unordered_set<string> visited;
        q.push(s);
        while(!q.empty()) {
            string cur = q.front();
            q.pop();
            res = min(res, cur);
            string adds = add(cur, a);
            string rtts = rotate(cur, b);
            if(visited.insert(adds).second) {
                q.push(adds);
            }
            if(visited.insert(rtts).second) {
                q.push(rtts);
            }
        }
        return res;
    }
};