What is the time complexity of Minimum Number of Operations to Make X and Y Equal?

The optimal solution has a time complexity of O(log(x)).

What data structures are used to solve Minimum Number of Operations to Make X and Y Equal?

This problem uses Dynamic Programming, Breadth-First Search, Memoization. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Operations to Make X and Y Equal?

Insight 1: When x > y, dividing by 5 or 11 can get us closer to y more quickly than just incrementing or decrementing, but we need to consider the remainders when dividing. Insight 2: Dynamic Programming (Memoization) is crucial to avoid redundant calculations since the same values of 'x' may be reached through multiple paths of operations.

Which companies ask LeetCode 2998. Minimum Number of Operations to Make X and Y Equal in interviews?

This problem is frequently asked at Groww.

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Minimum Number of Operations to Make X and Y Equal - LeetCode 2998 Solution

Q: What is the space complexity of Minimum Number of Operations to Make X and Y Equal?

The space complexity is O(log(x)).

Minimum Number of Operations to Make X and Y Equal - Complete Solution Guide

Minimum Number of Operations to Make X and Y Equal is LeetCode problem 2998, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 . Divide x by 5 if x is a multiple of 5 . Decrement x by 1 . Increment x by 1 . Return the minimum number of operations required to make x and y equal. Example 1: 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

Detailed Explanation

The problem asks us to find the minimum number of operations required to transform a positive integer `x` into another positive integer `y`. The allowed operations are: dividing `x` by 11 (if divisible), dividing `x` by 5 (if divisible), decrementing `x` by 1, and incrementing `x` by 1. The input consists of two integers, `x` and `y`, and the output is a single integer representing the minimum number of operations.

Solution Approach

The solution uses dynamic programming with memoization to efficiently explore the possible operation sequences. The `solve` function recursively explores the possible transformations of `x` by dividing by 5 or 11 (if applicable) or by incrementing or decrementing by 1. The base case for the recursion is when `x` is less than or equal to `y`, in which case the number of operations is simply `y - x`. The memoization ensures that the solution for each value of `x` is computed only once, significantly improving performance.

Step-by-Step Algorithm

Step 1: Handle the base case: If x <= y, return y - x. This is the minimum number of increments required.
Step 2: If the result for the current value of x has already been computed and stored in the memo, return it directly.
Step 3: Initialize the minimum operations required as the difference between x and y (x - y). This represents only decrementing x to reach y.
Step 4: Calculate the remainder when x is divided by 5. Consider two options: dividing x by 5 after decrementing by the remainder, and dividing x by 5 after incrementing to the next multiple of 5.
Step 5: Recursively call the `solve` function for both options mentioned in Step 4, adding the number of increment/decrement operations (the remainder or 5 - remainder) and the division operation (+1) to the result.
Step 6: Calculate the remainder when x is divided by 11. Consider two options: dividing x by 11 after decrementing by the remainder, and dividing x by 11 after incrementing to the next multiple of 11.
Step 7: Recursively call the `solve` function for both options mentioned in Step 6, adding the number of increment/decrement operations (the remainder or 11 - remainder) and the division operation (+1) to the result.
Step 8: Update the minimum operations with the minimum among the decrements, the two division by 5 options, and the two division by 11 options.
Step 9: Store the computed result in the memoization table and return it.

Key Insights

Insight 1: When x > y, dividing by 5 or 11 can get us closer to y more quickly than just incrementing or decrementing, but we need to consider the remainders when dividing.
Insight 2: Dynamic Programming (Memoization) is crucial to avoid redundant calculations since the same values of 'x' may be reached through multiple paths of operations.
Insight 3: When x <= y, the optimal solution is always to simply increment x until it reaches y (i.e., y-x operations). This serves as the base case for the recursion/DP.

Complexity Analysis

Time Complexity: O(log(x))

Space Complexity: O(log(x))

Topics

This problem involves: Dynamic Programming, Breadth-First Search, Memoization.

Companies

Asked at: Groww.

Minimum Number of Operations to Make X and Y Equal | LeetCode 2998

Medium

Dynamic Programming Breadth-First Search Memoization

Time: O(log(x))Space: O(log(x))

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    private Map<Integer, Integer> memo = new HashMap<>();

    public int minimumOperationsToMakeEqual(int x, int y) {
        if (x <= y) {
            return y - x;
        }

        return solve(x, y);
    }

    private int solve(int current_x, int y) {
        if (memo.containsKey(current_x)) {
            return memo.get(current_x);
        }

        if (current_x <= y) {
            return y - current_x;
        }

        int res = current_x - y;

        int rem5 = current_x % 5;
        res = Math.min(res, solve(current_x / 5, y) + rem5 + 1);
        res = Math.min(res, solve((current_x / 5) + 1, y) + (5 - rem5) + 1);

        int rem11 = current_x % 11;
        res = Math.min(res, solve(current_x / 11, y) + rem11 + 1);
        res = Math.min(res, solve((current_x / 11) + 1, y) + (11 - rem11) + 1);

        memo.put(current_x, res);
        return res;
    }
}

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Problem Info

Description

Problem 2998: Minimum Number of Operations to Make X and Y Equal - Detailed explanation and solution approaches.

Topics

Dynamic Programming Breadth-First Search Memoization

Companies

Groww

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2998

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Handle the base case: If x <= y, return y - x. This is the minimum number of increments required.
2
Step 2: If the result for the current value of x has already been computed and stored in the memo, return it directly.
3
Step 3: Initialize the minimum operations required as the difference between x and y (x - y). This represents only decrementing x to reach y.
4
Step 4: Calculate the remainder when x is divided by 5. Consider two options: dividing x by 5 after decrementing by the remainder, and dividing x by 5 after incrementing to the next multiple of 5.
5
Step 5: Recursively call the `solve` function for both options mentioned in Step 4, adding the number of increment/decrement operations (the remainder or 5 - remainder) and the division operation (+1) to the result.
6
Step 6: Calculate the remainder when x is divided by 11. Consider two options: dividing x by 11 after decrementing by the remainder, and dividing x by 11 after incrementing to the next multiple of 11.
7
Step 7: Recursively call the `solve` function for both options mentioned in Step 6, adding the number of increment/decrement operations (the remainder or 11 - remainder) and the division operation (+1) to the result.
8
Step 8: Update the minimum operations with the minimum among the decrements, the two division by 5 options, and the two division by 11 options.
9
Step 9: Store the computed result in the memoization table and return it.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(x))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(log(x))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is O(log(x)) because in the worst case, x is repeatedly divided by 5 or 11 until it becomes less than or equal to y. The number of recursive calls is proportional to the number of times x can be divided by 5 or 11. The space complexity is O(log(x)) due to the memoization table storing the results of the recursive calls. In the worst case, the memoization table will store results for a logarithmic number of different values of x.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

BFS can also be used to solve this problem by treating each possible value of x as a node in a graph, and the operations as edges. The goal is to find the shortest path from x to y.

Time: O(x)Space: O(x)

Trade-offs: BFS might be easier to conceptualize, but the space complexity can be higher than the DP approach. In general, the memoized DP solution is preferred. BFS also needs to avoid cycles, which can make implementation trickier. For the given constraint range BFS could pass but DP is more efficient.

✨

Key Insights

1Insight 1: When x > y, dividing by 5 or 11 can get us closer to y more quickly than just incrementing or decrementing, but we need to consider the remainders when dividing.
2Insight 2: Dynamic Programming (Memoization) is crucial to avoid redundant calculations since the same values of 'x' may be reached through multiple paths of operations.
3Insight 3: When x <= y, the optimal solution is always to simply increment x until it reaches y (i.e., y-x operations). This serves as the base case for the recursion/DP.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: x <= y: The algorithm correctly handles this case by simply returning y - x. This is essential for both correctness and for defining the base case of the recursion.
Edge case 2: x and y are very close to each other. The algorithm avoids unnecessary divisions by 5 or 11 and considers direct increment/decrement operations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization: Without memoization, the recursive solution will have exponential time complexity and will likely time out for larger inputs.

✗Common mistake 2: Not considering both incrementing and decrementing before dividing by 5 or 11: To find the *minimum* number of operations, it's important to check both options before dividing.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource optimization: Determining the fewest steps or operations to achieve a desired configuration or target state, such as minimizing the cost of production by adjusting machine parameters.

🏦Real-world scenario 2: Game playing: Finding the shortest sequence of moves to win a game, where the operations are the possible moves and the goal is to reach a winning state.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Operations to Make X and Y Equal"?

For this medium-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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