Vpropel VIT | POD | 24/05/2020 | Water and Jugs | 34

Question(water_and_jugs).txt

Water and Jugs
You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

Fill any of the jugs completely with water.

Empty any of the jugs.

Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example:-

Input: x = 3, y = 5, z = 4

Output: Yes

Example:-

Input: x = 2, y = 6, z = 5

Output: No

Input format:-

X

Y

Z

Output format:-

Yes/No

water_and_jugs.cpp

#include<iostream>
#include<algorithm>
using namespace std;
/*class Solution 
{
public:
    int hcf(int a, int b)
    {
        if(b==0) return a;
        return hcf(b,a%b);
    }
    bool canMeasureWater(int x, int y, int z) 
    {
        if(x+y<z) return false;
        
        if(z==0) return true;
        if(x==0&&y==0&&z!=0) return false;
        if(x==0&&y!=0&&y!=z) return false;
        if(x==0&&y!=0&&y==z) return true;
        if(x!=0&&y==0&&x!=z) return false;
        if(x!=0&&y==0&&x==z) return true;
        if(x<y) swap(x,y);
        if(z%hcf(x,y)==0) return true;
        return false;
    }
};
*/

class Solution
{
public:
    bool canMeasureWater(int x, int y, int z) 
    {
        return (z == 0 || x + y >= z && z % __gcd(x, y) == 0);
    }
};
int main()
{
    int x,y,z;
    cin>>x>>y>>z;
    Solution s;
    if(s.canMeasureWater(x,y,z)) cout<<"Yes";
    else cout<<"No";
    return 0;
}

water_and_jugs2.cpp

// Arithmatic Approach (to represent with process). Not the solution. 
#include<iostream>
#include<algorithm>

int gcd(int a,int b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
int pour(int fromCap,int toCap,int d)
{
    int from=fromCap;
    int to=0;
    std::cout<<"("<<from<<','<<to<<")\n";
    int step=1;
    while(from!=d && to!=d)
    {
        int temp=std::min(from,toCap-to);
        to+=temp;
        from-=temp;
        std::cout<<"("<<from<<','<<to<<")\n";
        step++;
        if(from==d || to==d)    break;
        if(from==0)
        {
            from=fromCap;
            std::cout<<"("<<from<<','<<to<<")\n";
            step++;
        }
        if(to==toCap)
        {
            to=0;
            std::cout<<"("<<from<<','<<to<<")\n";
            step++;
        }
    }
    std::cout<<"\n\n\n";
    return step;

}
int minSteps(int m,int n,int d)
{
    if(m<n) std::swap(m,n);

    if(d>n) return -1;

    if(d%gcd(n,m)!=0)  return -1;

    return std::min(pour(n,m,d),pour(m,n,d));
}
int main()
{
    int n,m,d;
    std::cin>>n>>m>>d;
    std::cout<<minSteps(m,n,d);
    return 0;
}

https://www.geeksforgeeks.org/water-jug-problem-using-memoization/
https://www.geeksforgeeks.org/water-jug-problem-using-bfs/
https://www.geeksforgeeks.org/two-water-jug-puzzle/