What is the time complexity of Water and Jug Problem?

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

What data structures are used to solve Water and Jug Problem?

This problem uses Math, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Water and Jug Problem?

Insight 1: The problem can be solved using the concept of the Greatest Common Divisor (GCD). If 'target' is divisible by GCD(x, y), a solution exists. Bézout's identity states that for integers x and y, there exist integers a and b such that ax + by = gcd(x, y). This means we can obtain multiples of gcd(x,y) through a linear combination of x and y (representing fills and empties). Insight 2: The target cannot be greater than the sum of the two jugs' capacities (x + y). If target > x + y, it's impossible to measure the target amount because you only have x and y liters jugs.

Which companies ask LeetCode 365. Water and Jug Problem in interviews?

This problem is frequently asked at Lyft, Wells Fargo.

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Water and Jug Problem - LeetCode 365 Solution

Water and Jug Problem - Complete Solution Guide

Water and Jug Problem is LeetCode problem 365, 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 jugs with capacities x liters and y liters. You have an infinite water supply. Return whether the total amount of water in both jugs may reach target using the following operations: Fill either jug completely with water. Completely empty either jug. Pour water from one jug into another until the receiving jug is full, or the transferring jug is empty. Example 1: Input: x = 3, y = 5, target = 4 Output: true Explanation: Follow these steps to reach a total of 4 liters: Fill the 5

Detailed Explanation

The Water and Jug Problem asks whether it is possible to measure exactly `target` liters of water using two jugs with capacities `x` and `y` liters, given an infinite water supply and specific operations: filling a jug, emptying a jug, and pouring water from one jug to another until the receiving jug is full or the transferring jug is empty. The problem can be visualized as a system of equations where the combination of pouring and emptying actions should result in the `target` amount. The inputs are `x`, `y`, and `target` (integers), and the output is a boolean value indicating whether it's possible to reach the `target`.

Solution Approach

The solution uses the mathematical properties of GCD to determine if the `target` amount can be reached. It leverages Bézout's identity, which is a fundamental theorem in number theory. The core idea is that if 'target' is a multiple of the GCD of `x` and `y`, then it's possible to obtain 'target' using a combination of filling, emptying, and pouring operations involving the jugs. First, it checks if target exceeds the total capacity of the jugs, in which case, it immediately returns false. Then, it calculates the GCD of x and y. Finally, it checks if the `target` is divisible by the GCD. If so, it returns true; otherwise, it returns false. The edge case of x=0 and y=0 requires checking if target is also 0.

Step-by-Step Algorithm

Step 1: Check if the `target` is greater than the sum of the capacities of the two jugs (`x + y`). If it is, return `false` because it's impossible to measure more water than the total capacity.
Step 2: Handle the special case where both `x` and `y` are 0. If both are 0, then only if `target` is also 0, return `true`; otherwise return `false`.
Step 3: Calculate the Greatest Common Divisor (GCD) of `x` and `y`.
Step 4: Check if `target` is divisible by the GCD of `x` and `y` (i.e., `target % GCD(x, y) == 0`). If it is, return `true`; otherwise, return `false`.

Key Insights

Insight 1: The problem can be solved using the concept of the Greatest Common Divisor (GCD). If 'target' is divisible by GCD(x, y), a solution exists. Bézout's identity states that for integers x and y, there exist integers a and b such that ax + by = gcd(x, y). This means we can obtain multiples of gcd(x,y) through a linear combination of x and y (representing fills and empties).
Insight 2: The target cannot be greater than the sum of the two jugs' capacities (x + y). If target > x + y, it's impossible to measure the target amount because you only have x and y liters jugs.
Insight 3: Consider the case when either x or y is 0. If both are 0, the target must also be 0 to return true. If one of x and y is 0, then the other must be equal to the target, or target must be 0 for a true return value. The edge cases of x=0 and/or y=0 must be checked to avoid errors in GCD calculation or incorrect results.

Complexity Analysis

Time Complexity: O(log(min(x, y)))

Space Complexity: O(1)

Topics

This problem involves: Math, Depth-First Search, Breadth-First Search.

Companies

Asked at: Lyft, Wells Fargo.

Water and Jug Problem | LeetCode 365

Medium

Math Depth-First Search Breadth-First Search

Time: O(log(min(x, y)))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean canMeasureWater(int x, int y, int target) {
        if (x + y < target) {
            return false;
        }
        
        if (x == 0 && y == 0) {
            return target == 0;
        }
            
        return target % gcd(x, y) == 0;
    }
    
    private int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

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Problem Info

Description

Problem 365: Water and Jug Problem - Detailed explanation and solution approaches.

Topics

Math Depth-First Search Breadth-First Search

Companies

LyftWells Fargo

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:365

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

4 steps

1
Step 1: Check if the `target` is greater than the sum of the capacities of the two jugs (`x + y`). If it is, return `false` because it's impossible to measure more water than the total capacity.
2
Step 2: Handle the special case where both `x` and `y` are 0. If both are 0, then only if `target` is also 0, return `true`; otherwise return `false`.
3
Step 3: Calculate the Greatest Common Divisor (GCD) of `x` and `y`.
4
Step 4: Check if `target` is divisible by the GCD of `x` and `y` (i.e., `target % GCD(x, y) == 0`). If it is, return `true`; otherwise, return `false`.

💡

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(min(x, y)))

🚀 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 determined by the Euclidean algorithm used to calculate the GCD, which is O(log(min(x, y))). The space complexity is O(1) because it only uses a few constant extra 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

Breadth-First Search (BFS)

Model the problem as a state-space search. Each state represents the amounts of water in the two jugs. Transitions represent the operations (fill, empty, pour). Use BFS to explore the states until the target is reached or all reachable states are visited. It explores all possible amounts in the jugs.

Time: O(x * y)Space: O(x * y)

Trade-offs: BFS provides a direct approach based on simulating operations, but its space complexity is higher because it needs to maintain a queue of visited states. The GCD-based approach is more efficient in terms of space and time complexity. BFS is useful when you need the shortest sequence of steps to reach the target if it can be achieved at all.

Depth-First Search (DFS)

Similar to BFS, DFS can explore the possible states. Keep track of visited states to avoid infinite loops. Recursively try all possible operations (fill, empty, pour).

Time: O(x * y)Space: O(x * y) in the worst case, due to the call stack.

Trade-offs: DFS offers an alternative search strategy compared to BFS. However, it might lead to stack overflow issues for large values of x and y, and it doesn't guarantee finding the shortest path like BFS.

✨

Key Insights

1Insight 1: The problem can be solved using the concept of the Greatest Common Divisor (GCD). If 'target' is divisible by GCD(x, y), a solution exists. Bézout's identity states that for integers x and y, there exist integers a and b such that ax + by = gcd(x, y). This means we can obtain multiples of gcd(x,y) through a linear combination of x and y (representing fills and empties).
2Insight 2: The target cannot be greater than the sum of the two jugs' capacities (x + y). If target > x + y, it's impossible to measure the target amount because you only have x and y liters jugs.
3Insight 3: Consider the case when either x or y is 0. If both are 0, the target must also be 0 to return true. If one of x and y is 0, then the other must be equal to the target, or target must be 0 for a true return value. The edge cases of x=0 and/or y=0 must be checked to avoid errors in GCD calculation or incorrect results.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `x = 0, y = 0, target = 0`. The code correctly handles this case by returning `true`, as it's possible to have 0 liters using two empty jugs.
Edge case 2: `x = 0, y = 5, target = 3`. The code correctly returns `false`. GCD(0,5) is 5 and 3 is not divisible by 5.
Edge case 3: `x = 3, y = 5, target = 7`. The code correctly returns `false` because `x+y = 8` and `target` cannot be greater than this
Edge case 4: x = 3, y = 5, target = 0. The code would return True, because 0 % GCD(3,5) = 0 % 1 = 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if `target` is greater than the sum of `x` and `y`. This will lead to incorrect results, as it's impossible to measure more water than the total capacity.

✗Common mistake 2: Failing to handle the edge case where both jugs have a capacity of 0. The GCD is undefined and the `target` can only be reached if it is 0. Many GCD implementations handle `x=0, y=0` incorrectly. Ensure your code addresses this.

✗Common mistake 3: Assuming the GCD is always 1. This can lead to incorrect results if the target is not divisible by the actual GCD.

✗Common mistake 4: Assuming that all combination of filling, emptying and pouring are necessary. Using GCD is sufficient, demonstrating that the complex operations are not always necessary.

✅

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 allocation. Imagine you have two containers with different capacities and need to measure a specific quantity of a resource. This problem is analogous to determining if you can obtain a certain amount of a resource by combining the contents of the two containers through filling and transferring operations.

🏦Real-world scenario 2: Cryptography. The concept of GCD is crucial in cryptography, especially in key generation algorithms like RSA. The ability to find a GCD and understand its properties is important in ensuring the security of cryptographic systems.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: June 6, 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 14

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

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Water and Jug Problem"?

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

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