LeetCode 1625: Lexicographically Smallest String After Applying Operations Solution

Master LeetCode problem 1625 (Lexicographically Smallest String After Applying Operations), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1625. Lexicographically Smallest String After Applying Operations

Medium

String Depth-First Search Breadth-First Search Enumeration

Problem Explanation

Explanation

To solve this problem, we can try all possible combinations of applying the two operations: rotating the string and adding 'a' to odd indices. We can use a breadth-first search (BFS) approach to explore all possible states of the string and then return the lexicographically smallest string obtained.

Start with the initial string 's' and add it to a queue.
For each string in the queue, apply both operations (rotate and add) to generate new strings.
Keep track of visited strings to avoid loops and also maintain the lexicographically smallest string found.
Continue this process until all possible combinations are explored or until we reach the lexicographically smallest string.

Time Complexity: O(n^2 * m^2), where n is the length of the string 's' and m is the number of unique states that can be generated by rotating or adding 'a'. Space Complexity: O(n^2 * m^2) for the queue and visited set.

Solution Code

class Solution {
    public String findLexSmallestString(String s, int a, int b) {
        Set<String> visited = new HashSet<>();
        Queue<String> queue = new LinkedList<>();
        String smallest = s;
        queue.offer(s);

        while (!queue.isEmpty()) {
            String current = queue.poll();
            if (smallest.compareTo(current) > 0) {
                smallest = current;
            }

            String rotated = rotate(current, b);
            if (!visited.contains(rotated)) {
                visited.add(rotated);
                queue.offer(rotated);
            }

            String added = add(current, a);
            if (!visited.contains(added)) {
                visited.add(added);
                queue.offer(added);
            }
        }

        return smallest;
    }

    private String rotate(String s, int b) {
        return s.substring(s.length() - b) + s.substring(0, s.length() - b);
    }

    private String add(String s, int a) {
        char[] chars = s.toCharArray();
        for (int i = 1; i < s.length(); i += 2) {
            int num = Character.getNumericValue(chars[i]) + a;
            chars[i] = (char) ('0' + num % 10);
        }
        return new String(chars);
    }
}

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