What is the time complexity of Reaching Points?

The optimal solution has a time complexity of O(log(max(tx, ty))).

What data structures are used to solve Reaching Points?

This problem uses Math. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Reaching Points?

Insight 1: Working backwards from the target to the source is much easier and more efficient. We can reverse the operations to subtract instead of add. Insight 2: Using modulo operations (`%`) is crucial for efficiency because it avoids unnecessary subtractions when tx and ty are significantly larger than sx and sy.

Which companies ask LeetCode 780. Reaching Points in interviews?

This problem is frequently asked at Cloudflare, Coursera, Docusign, J.P. Morgan, KLA and 2 more companies.

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Reaching Points - LeetCode 780 Solution

Reaching Points - Complete Solution Guide

Reaching Points is LeetCode problem 780, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given four integers sx , sy , tx , and ty , return true if it is possible to convert the point (sx, sy) to the point (tx, ty) through some operations , or false otherwise . The allowed operation on some point (x, y) is to convert it to either (x, x + y) or (x + y, y) . Example 1: Input: sx = 1, sy = 1, tx = 3, ty = 5 Output: true Explanation: One series of moves that transforms the starting point to the target is: (1, 1) -> (1, 2) (1, 2) -> (3, 2) (3, 2) -> (3, 5) Example 2: Input: sx = 1, sy =

Detailed Explanation

The problem asks whether it's possible to transform a starting point (sx, sy) to a target point (tx, ty) using only two types of moves: (x, y) -> (x, x+y) or (x, y) -> (x+y, y). The inputs are four integers representing the coordinates of the starting and target points. The output is a boolean: `true` if the transformation is possible, `false` otherwise. The constraints specify that all coordinates are between 1 and 10^9, inclusive.

Solution Approach

The solution uses a backward approach. Starting from (tx, ty), we iteratively apply the reverse operations until we reach (sx, sy) or determine that it's impossible to reach. The core idea is to repeatedly subtract the smaller coordinate from the larger one until one of the coordinates equals the starting coordinate. The modulo operator is used to accelerate the process when there's a large difference between tx and ty. The base cases where tx == sx or ty == sy are specifically handled.

Step-by-Step Algorithm

Step 1: While tx is greater than or equal to sx and ty is greater than or equal to sy, repeat the following steps.
Step 2: If tx equals sx and ty equals sy, return true (base case: we have reached the starting point).
Step 3: If tx is greater than ty, then:
a. If ty is greater than sy, update tx by taking tx modulo ty (tx %= ty). This avoids unnecessary subtractions when the difference is large.
b. Otherwise, if ty equals sy, return true if the difference (tx - sx) is divisible by sy, and false otherwise. This handles the edge case when we've reached the target row.
Step 4: Else if ty is greater than tx, then:
a. If tx is greater than sx, update ty by taking ty modulo tx (ty %= tx). This avoids unnecessary subtractions when the difference is large.
b. Otherwise, if tx equals sx, return true if the difference (ty - sy) is divisible by sx, and false otherwise. This handles the edge case when we've reached the target column.
Step 5: Else, if tx equals ty, return false (because if tx and ty are equal but not equal to sx and sy, it's impossible to reach the target).
Step 6: If the loop finishes without returning true, return false (meaning it's impossible to reach the target).

Key Insights

Insight 1: Working backwards from the target to the source is much easier and more efficient. We can reverse the operations to subtract instead of add.
Insight 2: Using modulo operations (`%`) is crucial for efficiency because it avoids unnecessary subtractions when tx and ty are significantly larger than sx and sy.
Insight 3: The cases where tx == sx or ty == sy need to be handled separately after reaching a state where either tx or ty equals sx or sy, respectively. This requires checking if the difference is a multiple of the other coordinate.

Complexity Analysis

Time Complexity: O(log(max(tx, ty)))

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Cloudflare, Coursera, Docusign, J.P. Morgan, KLA, Wayfair, Workday.

Reaching Points | LeetCode 780

Hard

Math

Time: O(log(max(tx, ty)))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean reachingPoints(int sx, int sy, int tx, int ty) {
        while (tx >= sx && ty >= sy) {
            if (tx == sx && ty == sy) {
                return true;
            }
            
            if (tx > ty) {
                if (ty > sy) {
                    tx %= ty;
                } else {
                    return (tx - sx) % sy == 0;
                }
            } else if (ty > tx) {
                if (tx > sx) {
                    ty %= tx;
                } else {
                    return (ty - sy) % sx == 0;
                }
            } else { // tx == ty but not (sx, sy)
                return false;
            }
        }
        
        return false;
    }
}

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Problem Info

Description

Problem 780: Reaching Points - Detailed explanation and solution approaches.

Topics

Math

Companies

CloudflareCourseraDocusignJ.P. MorganKLAWayfairWorkday

Stats

Difficulty:Hard

Languages:4

Companies:7

Problem #:780

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

10 steps

1
Step 1: While tx is greater than or equal to sx and ty is greater than or equal to sy, repeat the following steps.
2
Step 2: If tx equals sx and ty equals sy, return true (base case: we have reached the starting point).
3
Step 3: If tx is greater than ty, then:
4
a. If ty is greater than sy, update tx by taking tx modulo ty (tx %= ty). This avoids unnecessary subtractions when the difference is large.
5
b. Otherwise, if ty equals sy, return true if the difference (tx - sx) is divisible by sy, and false otherwise. This handles the edge case when we've reached the target row.
6
Step 4: Else if ty is greater than tx, then:
7
a. If tx is greater than sx, update ty by taking ty modulo tx (ty %= tx). This avoids unnecessary subtractions when the difference is large.
8
b. Otherwise, if tx equals sx, return true if the difference (ty - sy) is divisible by sx, and false otherwise. This handles the edge case when we've reached the target column.
9
Step 5: Else, if tx equals ty, return false (because if tx and ty are equal but not equal to sx and sy, it's impossible to reach the target).
10
Step 6: If the loop finishes without returning true, return false (meaning it's impossible to reach the target).

💡

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(max(tx, ty)))

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

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because the modulo operation (tx %= ty or ty %= tx) reduces the size of the larger number significantly in each step. The loop continues as long as tx >= sx and ty >= sy. The space complexity is constant because the algorithm only uses a fixed number of variables regardless of the input size.

📖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

Depth-First Search (DFS)

A recursive approach where we try to move from (sx, sy) to (tx, ty) by exploring both possible moves (x+y, y) and (x, x+y).

Time: O(2^(tx+ty)), where tx and ty are target coordinates. Exponential time complexity due to exploring all possible paths.Space: O(tx+ty). The maximum depth of the recursion tree can be at most tx + ty.

Trade-offs: DFS is very inefficient due to its exponential time complexity and is highly susceptible to stack overflow errors, especially for larger inputs. The backwards approach using modulo operations is much more efficient and should be preferred.

✨

Key Insights

1Insight 1: Working backwards from the target to the source is much easier and more efficient. We can reverse the operations to subtract instead of add.
2Insight 2: Using modulo operations (`%`) is crucial for efficiency because it avoids unnecessary subtractions when tx and ty are significantly larger than sx and sy.
3Insight 3: The cases where tx == sx or ty == sy need to be handled separately after reaching a state where either tx or ty equals sx or sy, respectively. This requires checking if the difference is a multiple of the other coordinate.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: sx = tx and sy = ty. The solution correctly returns true immediately.
Edge case 2: Cases where tx is much larger than ty, or vice versa. The modulo operator handles these efficiently by quickly reducing the larger value.
Edge case 3: Cases where either tx or ty have reached sx or sy, respectively. These cases need special handling using the modulo operator to check if the remaining difference is divisible by the other coordinate.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to solve the problem by moving forward from (sx, sy). This approach is much harder because it's difficult to determine which move to take at each step, and it's easy to get lost in irrelevant paths.

✗Common mistake 2: Not handling the base cases where tx == sx or ty == sy correctly. Missing the modulo check will lead to incorrect answers.

✗Common mistake 3: Integer overflow. While the problem statement implies that sx, sy, tx, and ty are integers, repeatedly adding them together in the forward direction could lead to integer overflow if intermediate results exceed the maximum integer value.

✅

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: This problem demonstrates a pattern of reversing operations to simplify a problem. This pattern can be used in path-finding problems where it is difficult or inefficient to search in the forward direction.

🏦Real-world scenario 2: The use of the modulo operator to optimize calculations with large numbers is a technique commonly applied in cryptography and other areas of computer science that deal with large numbers and arithmetic operations.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: July 31, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Reaching Points"?

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

🤖 Need help with this problem?