LeetCode 3377: Digit Operations to Make Two Integers Equal Solution

Master LeetCode problem 3377 (Digit Operations to Make Two Integers Equal), 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.

3377. Digit Operations to Make Two Integers Equal

Medium

Math Graph Heap (Priority Queue)Number Theory Shortest Path

Problem Explanation

Explanation

To solve this problem, we can use a breadth-first search (BFS) approach. We start with the initial number n and keep track of the visited numbers to avoid revisiting them. We generate all possible transformations from n by incrementing or decrementing each digit (except 0 and 9) and check if the resulting number is not prime. We continue this process until we reach the target number m or exhaust all possibilities. Finally, we return the minimum cost if m is reachable; otherwise, we return -1.

Time complexity: O(10^d) where d is the number of digits in the input number
Space complexity: O(10^d) for storing visited numbers

Solution Code

import java.util.*;

class Solution {
    public int minCost(int n, int m) {
        Queue<Integer> queue = new LinkedList<>();
        Set<Integer> visited = new HashSet<>();
        queue.add(n);
        visited.add(n);
        int cost = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();

            for (int i = 0; i < size; i++) {
                int curr = queue.poll();

                if (curr == m) {
                    return cost;
                }

                for (int j = 0; j < String.valueOf(curr).length(); j++) {
                    int digit = Character.getNumericValue(String.valueOf(curr).charAt(j));

                    if (digit > 0) {
                        int nextNum = curr - (int) Math.pow(10, String.valueOf(curr).length() - j - 1);
                        if (!visited.contains(nextNum) && !isPrime(nextNum)) {
                            queue.add(nextNum);
                            visited.add(nextNum);
                        }
                    }

                    if (digit < 9) {
                        int nextNum = curr + (int) Math.pow(10, String.valueOf(curr).length() - j - 1);
                        if (!visited.contains(nextNum) && !isPrime(nextNum)) {
                            queue.add(nextNum);
                            visited.add(nextNum);
                        }
                    }
                }
            }
            cost++;
        }

        return -1;
    }

    private boolean isPrime(int num) {
        if (num <= 1) return false;
        for (int i = 2; i <= Math.sqrt(num); i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
}

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