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 po

findLexSmallestString.js

/**
 * @param {string} s
 * @param {number} a
 * @param {number} b
 * @return {string}
 */
var findLexSmallestString = function(s, a, b) {
    // Set to keep track of visited strings to avoid revisiting the same state
    const visited = new Set();

    // Queue for BFS traversal, starting with the original string
    const queue = [s];

    // Initialize result with the original string
    let result = s;

    // BFS loop: explore all reachable strings
    while (queue.length > 0) {
        // Get the next string from the queue
        const curr = queue.shift();

        // Skip if we've already visited this string
        if (visited.has(curr)) continue;

        // Mark this string as visited
        visited.add(curr);

        // Update result if current string is lexicographically smaller
        if (curr < result) result = curr;

        // --- Operation 1: Add 'a' to all digits at odd indices ---
        let chars = curr.split('');
        for (let i = 1; i < chars.length; i += 2) {
            // Add 'a' to digit and wrap around using modulo 10
            chars[i] = (parseInt(chars[i]) + a) % 10;
        }
        const added = chars.join('');
        // If this new string hasn't been visited, add it to the queue
        if (!visited.has(added)) queue.push(added);

        // --- Operation 2: Rotate the string to the right by 'b' positions ---
        const rotated = curr.slice(-b) + curr.slice(0, curr.length - b);
        // If this rotated string hasn't been visited, add it to the queue
        if (!visited.has(rotated)) queue.push(rotated);
    }

    // Return the smallest string found
    return result;
};