Last updated on October 5th, 2024 at 09:08 pm

Here, We see Water and Jug Problem LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Water and Jug Problem LeetCode Solution

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:
1. Fill the 5-liter jug (0, 5).
2. Pour from the 5-liter jug into the 3-liter jug, leaving 2 liters (3, 2).
3. Empty the 3-liter jug (0, 2).
4. Transfer the 2 liters from the 5-liter jug to the 3-liter jug (2, 0).
5. Fill the 5-liter jug again (2, 5).
6. Pour from the 5-liter jug into the 3-liter jug until the 3-liter jug is full. This leaves 4 liters in the 5-liter jug (3, 4).
7. Empty the 3-liter jug. Now, you have exactly 4 liters in the 5-liter jug (0, 4).

Example 2:
Input:  x = 2, y = 6, target = 5
Output:  false

Example 3:
Input:  x = 1, y = 2, target = 3
Output:  true
Explanation: Fill both jugs. The total amount of water in both jugs is equal to 3 now.

1. Water and Jug Problem LeetCode Solution C++

class Solution {
public:
    bool canMeasureWater(int x, int y, int target) {
    return target == 0 || (target - x <= y && target % gcd(x, y) == 0);
}
private:
    int gcd(int x, int y)
    {
        return y == 0 ? x : gcd(y, x % y);        
    }
};

2. Water and Jug Problem LeetCode Solution Java

class Solution {
    public boolean canMeasureWater(int x, int y, int target) {
        if( x + y < target )
        {
            return false;
        }
        if( x == target || y == target || x + y == target )
        {
            return true;
        }
        if( target % gcd( x, y ) == 0 )
        {
            return true;
        }
        return false;
    }
    public int gcd( int a, int b ){
        if (b == 0) 
            return a;
        else 
            return gcd(b, a % b);        
    }
}

3. Water and Jug Problem LeetCode Solution JavaScript

var canMeasureWater = function(x, y, target) {
    if (target > x + y) return false;
    const terms = { '0': 1, [x]: 1, [y]: 1 };
    let xsum = x;
    let ysum = y;
    let stop = x * y;
    while (xsum < stop || ysum < stop) {
        if (xsum < ysum) {
            terms[ysum - xsum] = 1;
            xsum += x;
        } else {
            terms[xsum - ysum] = 1;
            ysum += y;
        }
    }
    for (let key in terms) {
        if (terms[target - key]) return true;
    }
    return false;
};

4. Water and Jug Problem Solution Python

class Solution(object):
    def canMeasureWater(self, x, y, target):
        a,b=x,y
        while y:
            r=x%y
            x=y
            y=r
        return bool(not target or (x and target<=a+b and not target%x))