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Leetcode #2998: Minimum Number of Operations to Make X and Y Equal

In this guide, we solve Leetcode #2998 Minimum Number of Operations to Make X and Y Equal in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given two positive integers x and y. In one operation, you can do one of the four following operations: Divide x by 11 if x is a multiple of 11.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Breadth-First Search, Memoization, Dynamic Programming

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

  • Choose a DP state definition.
  • Write the recurrence and base cases.
  • Compute states in the correct order.

Example

Input: x = 26, y = 1 Output: 3 Explanation: We can make 26 equal to 1 by applying the following operations: 1. Decrement x by 1 2. Divide x by 5 3. Divide x by 5 It can be shown that 3 is the minimum number of operations required to make 26 equal to 1.

Python Solution

class Solution: def minimumOperationsToMakeEqual(self, x: int, y: int) -> int: @cache def dfs(x: int) -> int: if y >= x: return y - x ans = x - y ans = min(ans, x % 5 + 1 + dfs(x // 5)) ans = min(ans, 5 - x % 5 + 1 + dfs(x // 5 + 1)) ans = min(ans, x % 11 + 1 + dfs(x // 11)) ans = min(ans, 11 - x % 11 + 1 + dfs(x // 11 + 1)) return ans return dfs(x)

Complexity

The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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