How do you solve Digit Operations to Make Two Integers Equal on LeetCode?

Digit Operations to Make Two Integers Equal (LeetCode #3377) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transformations must avoid prime numbers.

What is the time complexity of Digit Operations to Make Two Integers Equal?

The optimal solution for Digit Operations to Make Two Integers Equal runs in O(n) time and uses O(n) space. The complexity is driven by the number of transformations and the priority queue operations, which are efficient for this problem.

What is the brute force solution for Digit Operations to Make Two Integers Equal?

The brute force approach for Digit Operations to Make Two Integers Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible transformations of n by incrementing or decrementing each digit. This exhaustive search will help us find the minimum cost to reach m, but it's inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Digit Operations to Make Two Integers Equal?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice similar problems to build confidence.

What are common mistakes when solving Digit Operations to Make Two Integers Equal?

Forgetting to check primality after each transformation. Not considering all digit positions for transformations.

#3377

Digit Operations to Make Two Integers Equal

Medium

MathGraph TheoryHeap (Priority Queue)Number TheoryShortest PathGraph TraversalDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

We can model the problem as a graph where each node represents a number and edges represent valid transformations. Using a shortest path algorithm, we can efficiently find the minimum cost to transform n to m.

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Algorithm

3 steps

1Step 1: Create a priority queue to explore numbers starting from n.
2Step 2: For each number, generate valid transformations and push them into the queue if they are not prime.
3Step 3: Continue until m is reached or all possibilities are exhausted, tracking the total cost.

solution.py22 lines

1import heapq
2
3def minCost(n, m):
4    def is_prime(x):
5        if x < 2: return False
6        for i in range(2, int(x**0.5) + 1):
7            if x % i == 0: return False
8        return True
9    pq = [(n, n)]
10    visited = set()
11    while pq:
12        cost, num = heapq.heappop(pq)
13        if num == m: return cost
14        if num in visited or is_prime(num): continue
15        visited.add(num)
16        for i in range(len(str(num))):
17            digit = (num // (10**i)) % 10
18            if digit < 9:
19                heapq.heappush(pq, (cost + num + 1, num + 10**i))
20            if digit > 0:
21                heapq.heappush(pq, (cost + num + 1, num - 10**i))
22    return -1

ℹ

Complexity note: The complexity is driven by the number of transformations and the priority queue operations, which are efficient for this problem.

1Transformations must avoid prime numbers.
2Each digit can be incremented or decremented independently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.