Given two jugs with the maximum capacity of m and n liters respectively. The jugs don’t have markings on them which can help us to measure smaller quantities. The task is to measure d liters of water using these two jugs. Hence our goal is to reach from initial state (m, n) to final state (0, d) or (d, 0). Examples:
Input: 4 3 2 Output: (0, 0) –> (4, 0) –> (4, 3) –> (0, 3) –> (3, 0) –> (3, 3) –> (4, 2) –> (0, 2) Input: 5 2 4 Output: (0, 0) –> (5, 0) –> (5, 2) –> (0, 2) –> (2, 0) –> (2, 2) –> (4, 0)
Approach: An approach using BFS has been discussed in the previous post. In this post an approach using memoization and recursion has been discussed. At any point, there can be a total of six possibilities:
- Empty the first jug completely
- Empty the second jug completely
- Fill the first jug
- Fill the second jug
- Fill the water from the second jug into the first jug until the first jug is full or the second jug has no water left
- Fill the water from the first jug into the second jug until the second jug is full or the first jug has no water left
Approach: Using Recursion, visit all the six possible moves one by one until one of them returns True. Since there can be repetitions of same recursive calls, hence every return value is stored using memoization to avoid calling the recursive function again and returning the stored value. Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// jug1 and jug2 contain the value// for max capacity in respective jugs// and aim is the amount of water to be measured.int jug1 = 4;int jug2 = 3; int aim = 2;// Initialize dictionary with// default value as false.int visited[10000][10000]={false};// Recursive function which prints the// intermediate steps to reach the final// solution and return boolean value// (True if solution is possible, otherwise False).// amt1 and amt2 are the amount of water present// in both jugs at a certain point of time.bool waterJugSolver(int amt1, int amt2){ // Checks for our goal and // returns true if achieved. if ((amt1 == aim && amt2 == 0) || (amt2 == aim && amt1 == 0)){ cout<<amt1<<" "<<amt2<<endl; return true; } // Checks if we have already visited the // combination or not. If not, then it proceeds further. if (visited[amt1][amt2] == false){ cout<<amt1<<" "<<amt2<<endl; // Changes the boolean value of // the combination as it is visited. visited[amt1][amt2] = true; // Check for all the 6 possibilities and // see if a solution is found in any one of them. return (waterJugSolver(0, amt2) || waterJugSolver(amt1, 0) || waterJugSolver(jug1, amt2) || waterJugSolver(amt1, jug2) || waterJugSolver(amt1 + min(amt2, (jug1-amt1)), amt2 - min(amt2, (jug1-amt1))) || waterJugSolver(amt1 - min(amt1, (jug2-amt2)), amt2 + min(amt1, (jug2-amt2)))); } // Return False if the combination is // already visited to avoid repetition otherwise // recursion will enter an infinite loop. else return false;}int main() { cout<<("Steps: ")<<endl;// Call the function and pass the// initial amount of water present in both jugs.waterJugSolver(0, 0); return 0;}// This code is contributed by Pushpesh Raj |
Java
import java.util.HashSet;public class Main { // jug1 and jug2 contain the value // for max capacity in respective jugs // and aim is the amount of water to be measured. static int jug1 = 4; static int jug2 = 3; static int aim = 2; // Initialize HashSet with // default value as false. static HashSet<String> visited = new HashSet<>(); // Recursive function which prints the // intermediate steps to reach the final // solution and return boolean value // (True if solution is possible, otherwise False). // amt1 and amt2 are the amount of water present // in both jugs at a certain point of time. static boolean waterJugSolver(int amt1, int amt2) { // Checks for our goal and // returns true if achieved. if ((amt1 == aim && amt2 == 0) || (amt2 == aim && amt1 == 0)) { System.out.println(amt1 + " " + amt2); return true; } // Checks if we have already visited the // combination or not. If not, then it proceeds // further. String state = amt1 + " " + amt2; if (!visited.contains(state)) { System.out.println(state); // Changes the boolean value of // the combination as it is visited. visited.add(state); // Check for all the 6 possibilities and // see if a solution is found in any one of // them. return ( waterJugSolver(0, amt2) || waterJugSolver(amt1, 0) || waterJugSolver(jug1, amt2) || waterJugSolver(amt1, jug2) || waterJugSolver( amt1 + Math.min(amt2, (jug1 - amt1)), amt2 - Math.min(amt2, (jug1 - amt1))) || waterJugSolver( amt1 - Math.min(amt1, (jug2 - amt2)), amt2 + Math.min(amt1, (jug2 - amt2)))); } // Return False if the combination is // already visited to avoid repetition otherwise // recursion will enter an infinite loop. else { return false; } } public static void main(String[] args) { System.out.println("Steps: "); // Call the function and pass the // initial amount of water present in both jugs. waterJugSolver(0, 0); }} |
Python3
# This function is used to initialize the # dictionary elements with a default value.from collections import defaultdict# jug1 and jug2 contain the value # for max capacity in respective jugs # and aim is the amount of water to be measured. jug1, jug2, aim = 4, 3, 2# Initialize dictionary with # default value as false.visited = defaultdict(lambda: False)# Recursive function which prints the # intermediate steps to reach the final # solution and return boolean value # (True if solution is possible, otherwise False).# amt1 and amt2 are the amount of water present # in both jugs at a certain point of time.def waterJugSolver(amt1, amt2): # Checks for our goal and # returns true if achieved. if (amt1 == aim and amt2 == 0) or (amt2 == aim and amt1 == 0): print(amt1, amt2) return True # Checks if we have already visited the # combination or not. If not, then it proceeds further. if visited[(amt1, amt2)] == False: print(amt1, amt2) # Changes the boolean value of # the combination as it is visited. visited[(amt1, amt2)] = True # Check for all the 6 possibilities and # see if a solution is found in any one of them. return (waterJugSolver(0, amt2) or waterJugSolver(amt1, 0) or waterJugSolver(jug1, amt2) or waterJugSolver(amt1, jug2) or waterJugSolver(amt1 + min(amt2, (jug1-amt1)), amt2 - min(amt2, (jug1-amt1))) or waterJugSolver(amt1 - min(amt1, (jug2-amt2)), amt2 + min(amt1, (jug2-amt2)))) # Return False if the combination is # already visited to avoid repetition otherwise # recursion will enter an infinite loop. else: return Falseprint("Steps: ")# Call the function and pass the# initial amount of water present in both jugs.waterJugSolver(0, 0) |
C#
using System;using System.Collections.Generic;public class MainClass { // jug1 and jug2 contain the value // for max capacity in respective jugs // and aim is the amount of water to be measured. static int jug1 = 4; static int jug2 = 3; static int aim = 2; // Initialize HashSet with // default value as false. static HashSet<string> visited = new HashSet<string>(); // Recursive function which prints the // intermediate steps to reach the final // solution and return boolean value // (True if solution is possible, otherwise False). // amt1 and amt2 are the amount of water present // in both jugs at a certain point of time. static bool waterJugSolver(int amt1, int amt2) { // Checks for our goal and // returns true if achieved. if ((amt1 == aim && amt2 == 0) || (amt2 == aim && amt1 == 0)) { Console.WriteLine(amt1 + " " + amt2); return true; } // Checks if we have already visited the // combination or not. If not, then it proceeds // further. string state = amt1 + " " + amt2; if (!visited.Contains(state)) { Console.WriteLine(state); // Changes the boolean value of // the combination as it is visited. visited.Add(state); // Check for all the 6 possibilities and // see if a solution is found in any one of // them. return ( waterJugSolver(0, amt2) || waterJugSolver(amt1, 0) || waterJugSolver(jug1, amt2) || waterJugSolver(amt1, jug2) || waterJugSolver( amt1 + Math.Min(amt2, (jug1 - amt1)), amt2 - Math.Min(amt2, (jug1 - amt1))) || waterJugSolver( amt1 - Math.Min(amt1, (jug2 - amt2)), amt2 + Math.Min(amt1, (jug2 - amt2)))); } // Return False if the combination is // already visited to avoid repetition otherwise // recursion will enter an infinite loop. else { return false; } } public static void Main(string[] args) { Console.WriteLine("Steps: "); // Call the function and pass the // initial amount of water present in both jugs. waterJugSolver(0, 0); }} |
Javascript
// jug1 and jug2 contain the value// for max capacity in respective jugs// and aim is the amount of water to be measured.let jug1 = 4;let jug2 = 3;let aim = 2;// Initialize dictionary with// default value as false.let visited = new Set();// Recursive function which prints the// intermediate steps to reach the final// solution and return boolean value// (True if solution is possible, otherwise False).// amt1 and amt2 are the amount of water present// in both jugs at a certain point of time.function waterJugSolver(amt1, amt2) { // Checks for our goal and // returns true if achieved. if ((amt1 == aim && amt2 == 0) || (amt2 == aim && amt1 == 0)) { console.log(amt1, amt2); return true; } let key = `${amt1}-${amt2}`; // Checks if we have already visited the // combination or not. If not, then it proceeds further. if (!visited.has(key)) { console.log(amt1, amt2); // Changes the boolean value of // the combination as it is visited. visited.add(key); // Check for all the 6 possibilities and // see if a solution is found in any one of them. return ( waterJugSolver(0, amt2) || waterJugSolver(amt1, 0) || waterJugSolver(jug1, amt2) || waterJugSolver(amt1, jug2) || waterJugSolver( amt1 + Math.min(amt2, jug1 - amt1), amt2 - Math.min(amt2, jug1 - amt1) ) || waterJugSolver( amt1 - Math.min(amt1, jug2 - amt2), amt2 + Math.min(amt1, jug2 - amt2) ) ); // Return False if the combination is // already visited to avoid repetition otherwise // recursion will enter an infinite loop. } else { return false; }}console.log("Steps:");// Call the function and pass the// initial amount of water present in both jugs.waterJugSolver(0, 0); |
Output:
Steps: 0 0 4 0 4 3 0 3 3 0 3 3 4 2 0 2
Time complexity: O(M * N) Auxiliary Space: O(M * N)
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