Formatted question description: https://leetcode.ca/all/1625.html

1625. Lexicographically Smallest String After Applying Operations (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:

  • Add a to all odd indices of s (0-indexed). Digits post 9 are cycled back to 0. For example, if s = "3456" and a = 5, s becomes "3951".
  • Rotate s to the right by b positions. For example, if s = "3456" and b = 1, s becomes "6345".

Return the lexicographically smallest string you can obtain by applying the above operations any number of times 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'.

 

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

Related Topics:
Depth-first Search, Breadth-first Search

Solution 1. BFS

// OJ: https://leetcode.com/problems/lexicographically-smallest-string-after-applying-operations/

// Time: O(N^2)
// Space: O(N)
class Solution {
    string first(string &s, int a) {
        string ans = s;
        for (int i = 1; i < s.size(); i += 2) ans[i] = '0' + (ans[i] - '0' + a) % 10;
        return ans;
    }
    string second(string &s, int b) {
        string ans = s;
        for (int i = 0, j = s.size() - b; i < s.size(); ++i, ++j) ans[i] = s[j % s.size()];
        return ans;
    }
public:
    string findLexSmallestString(string s, int a, int b) {
        unordered_set<string> seen;
        queue<string> q;
        q.push(s);
        string ans = s;
        while (q.size()) {
            auto s = q.front();
            q.pop();
            if (s < ans) ans = s;
            string x = first(s, a), y = second(s, b);
            if (seen.count(x) == 0) {
                q.push(x);
                seen.insert(x);
            }
            if (seen.count(y) == 0) {
                q.push(y);
                seen.insert(y);
            }
        }
        return ans;
    }
};

Java

class Solution {
    public String findLexSmallestString(String s, int a, int b) {
        String smallest = s;
        Set<String> visited = new HashSet<String>();
        visited.add(s);
        Queue<String> queue = new LinkedList<String>();
        queue.offer(s);
        while (!queue.isEmpty()) {
            String curr = queue.poll();
            if (curr.compareTo(smallest) < 0)
                smallest = curr;
            String next1 = addToAllOddIndices(curr, a);
            if (visited.add(next1))
                queue.offer(next1);
            String next2 = rotate(curr, b);
            if (visited.add(next2))
                queue.offer(next2);
        }
        
        return smallest;
    }

    public String addToAllOddIndices(String s, int num) {
        char[] array = s.toCharArray();
        int length = array.length;
        for (int i = 1; i < length; i += 2) {
            int digit = array[i] - '0';
            digit = (digit + num) % 10;
            array[i] = (char) (digit + '0');
        }
        return new String(array);
    }

    public String rotate(String s, int positions) {
        int length = s.length();
        positions %= length;
        StringBuffer sb = new StringBuffer();
        sb.append(s.substring(length - positions));
        sb.append(s.substring(0, length - positions));
        return sb.toString();
    }
}

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