public static String solve(int a, int b, int c) {

// char* solve(int a, int b, int c) { int c1=hcf(a,b); if(c<=Math.max(a,b)&&c%c1==0) return "YES"; else return "NO";

} public static int hcf(int n1, int n2) { if (n2 != 0) return hcf(n2, n1 % n2); else return n1; } }

My Python3 solution

def solve(a, b, c):
    yes, no = "YES", "NO"
    a, b = max(a, b), min(a, b)

    b_states = {0, b}
    a_states = {0, a}
    b_state = b_states.copy()
    a_state = a_states.copy()
    if c in a_states or c in b_states:
        return yes
    search = True
    while search:
        for jug_a in a_state:
            for jug_b in b_state:
                
                a1, a2 = min(a, jug_a + jug_b), max(0, jug_a - (b - jug_b))
                b1, b2 = min(b, jug_b+jug_a), max(0, jug_b - (a - jug_a))
                
                if c in {a1, a2, b1, b2}:
                    return yes
        
                a_states.add(a1)
                a_states.add(a2)
                
                b_states.add(b1)
                b_states.add(b2)
    
        if a_states == a_state and b_states == b_state:
            break
    

        b_state = b_states.copy()
        a_state = a_states.copy()

    # print(a_states, b_states)
    return no

I first solved this problem with backtracking (which works great for games, or procedures with limited next steps available), but using the gcd was a lot simpler - I didn't think about it for a while, though. So there are a few ways to solve this problem!

I cannot understand the question,

Bruce can do the following, fill jug a with 5 gallons. a = 5, b = 0 Now, he can fill jug b with 3 gallons from jug a. a = 2, b = 3 He can empty jug b and empty 2 gallons from jug a to jug b. a = 0, b = 2 Now, he can fill jug a with 5 gallons and fill jug b with 1 gallon from jug a. This results in jug a containing exactly 4 gallons of water. a = 5, b = 2

why he made it so complex, why not just pour 4 galon at the begin in a, then it is done, a = 4, b =