LeetCode #3377 — MEDIUM

Digit Operations to Make Two Integers Equal

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given two integers n and m that consist of the same number of digits.

You can perform the following operations any number of times:

  • Choose any digit from n that is not 9 and increase it by 1.
  • Choose any digit from n that is not 0 and decrease it by 1.

The integer n must not be a prime number at any point, including its original value and after each operation.

The cost of a transformation is the sum of all values that n takes throughout the operations performed.

Return the minimum cost to transform n into m. If it is impossible, return -1.

Example 1:

Input: n = 10, m = 12

Output: 85

Explanation:

We perform the following operations:

  • Increase the first digit, now n = 20.
  • Increase the second digit, now n = 21.
  • Increase the second digit, now n = 22.
  • Decrease the first digit, now n = 12.

Example 2:

Input: n = 4, m = 8

Output: -1

Explanation:

It is impossible to make n equal to m.

Example 3:

Input: n = 6, m = 2

Output: -1

Explanation: 

Since 2 is already a prime, we can't make n equal to m.

Constraints:

  • 1 <= n, m < 104
  • n and m consist of the same number of digits.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given two integers n and m that consist of the same number of digits. You can perform the following operations any number of times: Choose any digit from n that is not 9 and increase it by 1. Choose any digit from n that is not 0 and decrease it by 1. The integer n must not be a prime number at any point, including its original value and after each operation. The cost of a transformation is the sum of all values that n takes throughout the operations performed. Return the minimum cost to transform n into m. If it is impossible, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

10
12

Example 2

4
8

Example 3

6
2
Step 02

Core Insight

What unlocks the optimal approach

  • Consider a directed, weighted graph where an edge exists from a node <code>x</code> to a node <code>y</code> if and only if <code>x</code> can be transformed into <code>y</code> through a single operation.
  • Apply a shortest path algorithm on this graph to find the shortest path from <code>n</code> to <code>m</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3377: Digit Operations to Make Two Integers Equal
class Solution {
    private boolean[] sieve;

    private void runSieve() {
        sieve = new boolean[100000];
        Arrays.fill(sieve, true);
        sieve[0] = false;
        sieve[1] = false;
        for (int i = 2; i < 100000; i++) {
            if (sieve[i]) {
                for (int j = 2 * i; j < 100000; j += i) {
                    sieve[j] = false;
                }
            }
        }
    }

    private int solve(int n, int m) {
        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        pq.add(new int[] {n, n});
        Set<Integer> visited = new HashSet<>();

        while (!pq.isEmpty()) {
            int[] top = pq.poll();
            int sum = top[0], cur = top[1];

            if (visited.contains(cur)) {
                continue;
            }
            visited.add(cur);

            if (cur == m) {
                return sum;
            }

            char[] s = String.valueOf(cur).toCharArray();
            for (int i = 0; i < s.length; i++) {
                char c = s[i];

                if (s[i] < '9') {
                    s[i] = (char) (s[i] + 1);
                    int next = Integer.parseInt(new String(s));
                    if (!sieve[next] && !visited.contains(next)) {
                        pq.add(new int[] {sum + next, next});
                    }
                    s[i] = c;
                }

                if (s[i] > '0' && !(i == 0 && s[i] == '1')) {
                    s[i] = (char) (s[i] - 1);
                    int next = Integer.parseInt(new String(s));
                    if (!sieve[next] && !visited.contains(next)) {
                        pq.add(new int[] {sum + next, next});
                    }
                    s[i] = c;
                }
            }
        }

        return -1;
    }

    public int minOperations(int n, int m) {
        runSieve();
        if (sieve[n] || sieve[m]) {
            return -1;
        }
        return solve(n, m);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

ITERATIVE
O(n) time
O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT
O(log n) time
O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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