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Measure one litre using two vessels and infinite water supply

Last Updated : 11 Sep, 2023
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There are two vessels of capacities 'a' and 'b' respectively. We have infinite water supply. Give an efficient algorithm to make exactly 1 litre of water in one of the vessels. You can throw all the water from any vessel any point of time. Assume that 'a' and 'b' are Coprimes.
Following are the steps: 
Let V1 be the vessel of capacity 'a' and V2 be the vessel of capacity 'b' and 'a' is smaller than 'b'. 
1) Do following while the amount of water in V1 is not 1. 
....a) If V1 is empty, then completely fill V1 
....b) Transfer water from V1 to V2. If V2 becomes full, then keep the remaining water in V1 and empty V2 
2) V1 will have 1 litre after termination of loop in step 1. Return.
Following is C++ implementation of the above algorithm. 
 

C++ 
/* Sample run of the Algo for V1 with capacity 3 and V2 with capacity 7 1. Fill V1:                             V1 = 3, V2 = 0 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 3 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 6 3. Transfer from V1 to V2, and empty V2: V1 = 2, V2 = 0 4. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 2 5. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 5 6. Transfer from V1 to V2, and empty V2: V1 = 1, V2 = 0 7. Stop as V1 now contains 1 litre.  Note that V2 was made empty in steps 3 and 6 because it became full */  #include <iostream> using namespace std;  // A utility function to get GCD of two numbers int gcd(int a, int b) { return b? gcd(b, a % b) : a; }  // Class to represent a Vessel class Vessel {     // A vessel has capacity, and current amount of water in it     int capacity, current; public:     // Constructor: initializes capacity as given, and current as 0     Vessel(int capacity) { this->capacity = capacity; current = 0; }      // The main function to fill one litre in this vessel. Capacity of V2     // must be greater than this vessel and two capacities must be co-prime     void makeOneLitre(Vessel &V2);      // Fills vessel with given amount and returns the amount of water     // transferred to it. If the vessel becomes full, then the vessel     // is made empty.     int transfer(int amount); };  // The main function to fill one litre in this vessel. Capacity // of V2 must be greater than this vessel and two capacities // must be coprime void Vessel:: makeOneLitre(Vessel &V2) {     // solution exists if a and b are co-prime     if (gcd(capacity, V2.capacity) != 1)         return;      while (current != 1)     {         // fill A (smaller vessel)         if (current == 0)             current = capacity;          cout << "Vessel 1: " << current << " Vessel 2: "             << V2.current << endl;          // Transfer water from V1 to V2 and reduce current of V1 by         // the amount equal to transferred water         current = current - V2.transfer(current);     }      // Finally, there will be 1 litre in vessel 1     cout << "Vessel 1: " << current << " Vessel 2: "         << V2.current << endl; }  // Fills vessel with given amount and returns the amount of water // transferred to it. If the vessel becomes full, then the vessel // is made empty int Vessel::transfer(int amount) {     // If the vessel can accommodate the given amount     if (current + amount < capacity)     {         current += amount;         return amount;     }      // If the vessel cannot accommodate the given amount, then     // store the amount of water transferred     int transferred = capacity - current;      // Since the vessel becomes full, make the vessel     // empty so that it can be filled again     current = 0;      return transferred; }  // Driver program to test above function int main() {     int a = 3, b = 7; // a must be smaller than b      // Create two vessels of capacities a and b     Vessel V1(a), V2(b);      // Get 1 litre in first vessel     V1.makeOneLitre(V2);      return 0; }
C# 
/* Sample run of the Algo for V1 with capacity 3 and V2 with capacity 7 1. Fill V1:                             V1 = 3, V2 = 0 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 3 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 6 3. Transfer from V1 to V2, and empty V2: V1 = 2, V2 = 0 4. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 2 5. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 5 6. Transfer from V1 to V2, and empty V2: V1 = 1, V2 = 0 7. Stop as V1 now contains 1 litre.  Note that V2 was made empty in steps 3 and 6 because it became full */  using System;  class GFG {      // A utility function to get GCD of two numbers     static int gcd(int a, int b)      {         return b > 0 ? gcd(b, a % b) : a;     }      // Class to represent a Vessel     class Vessel      {                  // A vessel has capacity, and          // current amount of water in it         int capacity, current;          // Constructor: initializes capacity         // as given, and current as 0         public Vessel(int capacity)         {             this.capacity = capacity;             current = 0;         }          // The main function to fill one litre          // in this vessel. Capacity of V2 must be         // greater than this vessel and two capacities         // must be coprime         public void makeOneLitre(Vessel V2)         {             // solution exists if a and b are co-prime             if (gcd(capacity, V2.capacity) != 1)                 return;              while (current != 1)             {                 // fill A (smaller vessel)                 if (current == 0)                     current = capacity;                  Console.Write("Vessel 1: " + current +                              " Vessel 2: " + V2.current + "\n");                  // Transfer water from V1 to V2 and                  // reduce current of V1 by                 // the amount equal to transferred water                 current = current - V2.transfer(current);             }              // Finally, there will be 1 litre in vessel 1             Console.Write("Vessel 1: " + current +                          " Vessel 2: " + V2.current + "\n");         }          // Fills vessel with given amount and          // returns the amount of water         // transferred to it. If the vessel          // becomes full, then the vessel         // is made empty         int transfer(int amount)         {             // If the vessel can accommodate the given amount             if (current + amount < capacity)              {                 current += amount;                 return amount;             }              // If the vessel cannot accommodate             // the given amount, then store             // the amount of water transferred             int transferred = capacity - current;              // Since the vessel becomes full, make the vessel             // empty so that it can be filled again             current = 0;              return transferred;         }     }      // Driver program to test above function     public static void Main(String[] args)     {         int a = 3, b = 7; // a must be smaller than b          // Create two vessels of capacities a and b         Vessel V1 = new Vessel(a);         Vessel V2 = new Vessel(b);          // Get 1 litre in first vessel         V1.makeOneLitre(V2);     } }  // This code is contributed by Rajput-Ji
Java 
/* Sample run of the Algo for V1 with capacity 3 and V2 with capacity 7 1. Fill V1:                             V1 = 3, V2 = 0 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 3 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 6 3. Transfer from V1 to V2, and empty V2: V1 = 2, V2 = 0 4. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 2 5. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 5 6. Transfer from V1 to V2, and empty V2: V1 = 1, V2 = 0 7. Stop as V1 now contains 1 litre.  Note that V2 was made empty in steps 3 and 6 because it became full */  class GFG {      // A utility function to get GCD of two numbers     static int gcd(int a, int b)     {         return b > 0 ? gcd(b, a % b) : a;     }      // Class to represent a Vessel     static class Vessel     {                  // A vessel has capacity, and         // current amount of water in it         int capacity, current;          // Constructor: initializes capacity         // as given, and current as 0         public Vessel(int capacity)         {             this.capacity = capacity;             current = 0;         }          // The main function to fill one litre         // in this vessel. Capacity of V2 must be         // greater than this vessel and two capacities         // must be coprime         void makeOneLitre(Vessel V2)         {             // solution exists if a and b are co-prime             if (gcd(capacity, V2.capacity) != 1)                 return;              while (current != 1)             {                 // fill A (smaller vessel)                 if (current == 0)                     current = capacity;                  System.out.print("Vessel 1: " + current +                             " Vessel 2: " + V2.current + "\n");                  // Transfer water from V1 to V2 and                 // reduce current of V1 by                 // the amount equal to transferred water                 current = current - V2.transfer(current);             }              // Finally, there will be 1 litre in vessel 1             System.out.print("Vessel 1: " + current +                         " Vessel 2: " + V2.current + "\n");         }          // Fills vessel with given amount and         // returns the amount of water         // transferred to it. If the vessel         // becomes full, then the vessel         // is made empty         int transfer(int amount)         {             // If the vessel can accommodate the given amount             if (current + amount < capacity)             {                 current += amount;                 return amount;             }              // If the vessel cannot accommodate             // the given amount, then store             // the amount of water transferred             int transferred = capacity - current;              // Since the vessel becomes full, make the vessel             // empty so that it can be filled again             current = 0;              return transferred;         }     }      // Driver program to test above function     public static void main(String[] args)     {         int a = 3, b = 7; // a must be smaller than b          // Create two vessels of capacities a and b         Vessel V1 = new Vessel(a);         Vessel V2 = new Vessel(b);          // Get 1 litre in first vessel         V1.makeOneLitre(V2);     } }  // This code is contributed by 29AjayKumar
Python3 
# Sample run of the Algo for V1 with capacity 3 and V2 with capacity 7 # 1. Fill V1:                             V1 = 3, V2 = 0 # 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 3 # 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 6 # 3. Transfer from V1 to V2, and empty V2: V1 = 2, V2 = 0 # 4. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 2 # 5. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 5 # 6. Transfer from V1 to V2, and empty V2: V1 = 1, V2 = 0 # 7. Stop as V1 now contains 1 litre.  # Note that V2 was made empty in steps 3 and 6 because it became full       # A utility function to get GCD of two numbers def gcd(a,b):     return gcd(b, a % b) if(b > 0) else a           # Class to represent a Vessel class Vessel:      # Constructor: initializes capacity     # as given, and current as 0     def __init__(self,capacity):          self.capacity = capacity         self.current = 0      # The main function to fill one litre     # in this vessel. Capacity of V2 must be     # greater than this vessel and two capacities     # must be coprime     def makeOneLitre(self,V2):          # solution exists if a and b are co-prime         if (gcd(self.capacity, V2.capacity) != 1):             return          while (self.current != 1):                      # fill A (smaller vessel)             if (self.current == 0):                 self.current = self.capacity              print(f"Vessel 1: {self.current} Vessel 2: {V2.current}")              # Transfer water from V1 to V2 and             # reduce current of V1 by             # the amount equal to transferred water             self.current = self.current - V2.transfer(self.current)                      # Finally, there will be 1 litre in vessel 1         print(f"Vessel 1: {self.current} Vessel 2: {V2.current}")      # Fills vessel with given amount and     # returns the amount of water     # transferred to it. If the vessel     # becomes full, then the vessel     # is made empty     def transfer(self,amount):          # If the vessel can accommodate the given amount         if (self.current + amount < self.capacity):             self.current += amount             return amount                  # If the vessel cannot accommodate         # the given amount, then store         # the amount of water transferred         transferred = self.capacity - self.current          # Since the vessel becomes full, make the vessel         # empty so that it can be filled again         self.current = 0          return transferred  # Driver program to test above function a,b = 3,7 # a must be smaller than b  # Create two vessels of capacities a and b V1 = Vessel(a) V2 = Vessel(b)  # Get 1 litre in first vessel V1.makeOneLitre(V2)  # This code is contributed by shinjanpatra
JavaScript 
<script> /* Sample run of the Algo for V1 with capacity 3 and V2 with capacity 7 1. Fill V1:                             V1 = 3, V2 = 0 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 3 2. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 6 3. Transfer from V1 to V2, and empty V2: V1 = 2, V2 = 0 4. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 2 5. Transfer from V1 to V2, and fill V1: V1 = 3, V2 = 5 6. Transfer from V1 to V2, and empty V2: V1 = 1, V2 = 0 7. Stop as V1 now contains 1 litre.    Note that V2 was made empty in steps 3 and 6 because it became full */          // A utility function to get GCD of two numbers     function gcd(a,b)     {         return b > 0 ? gcd(b, a % b) : a;     }       // Class to represent a Vessel     class Vessel  {     // Constructor: initializes capacity         // as given, and current as 0     constructor(capacity)     {         this.capacity = capacity;         this.current = 0;     }          // The main function to fill one litre          // in this vessel. Capacity of V2 must be         // greater than this vessel and two capacities         // must be coprime     makeOneLitre(V2)     {         // solution exists if a and b are co-prime             if (gcd(this.capacity, V2.capacity) != 1)                 return;                while (this.current != 1)             {                 // fill A (smaller vessel)                 if (this.current == 0)                     this.current = this.capacity;                    document.write("Vessel 1: " + this.current +                              " Vessel 2: " + V2.current + "<br>");                    // Transfer water from V1 to V2 and                  // reduce current of V1 by                 // the amount equal to transferred water                 this.current = this.current - V2.transfer(this.current);             }                // Finally, there will be 1 litre in vessel 1             document.write("Vessel 1: " + this.current +                          " Vessel 2: " + V2.current + "<br>");     }          // Fills vessel with given amount and          // returns the amount of water         // transferred to it. If the vessel          // becomes full, then the vessel         // is made empty     transfer(amount)     {         // If the vessel can accommodate the given amount             if (this.current + amount < this.capacity)              {                 this.current += amount;                 return amount;             }                // If the vessel cannot accommodate             // the given amount, then store             // the amount of water transferred             let transferred = this.capacity - this.current;                // Since the vessel becomes full, make the vessel             // empty so that it can be filled again             this.current = 0;                return transferred;     } }  // Driver program to test above function let a = 3, b = 7; // a must be smaller than b    // Create two vessels of capacities a and b let V1 = new Vessel(a); let V2 = new Vessel(b);  // Get 1 litre in first vessel V1.makeOneLitre(V2);   // This code is contributed by rag2127 </script>

Output: 

Vessel 1: 3   Vessel 2: 0  Vessel 1: 3   Vessel 2: 3  Vessel 1: 3   Vessel 2: 6  Vessel 1: 2   Vessel 2: 0  Vessel 1: 3   Vessel 2: 2  Vessel 1: 3   Vessel 2: 5  Vessel 1: 1   Vessel 2: 0


How does this work? 
To prove that the algorithm works, we need to proof that after certain number of iterations in the while loop, we will get 1 litre in V1. 
Let 'a' be the capacity of vessel V1 and 'b' be the capacity of V2. Since we repeatedly transfer water from V1 to V2 until V2 becomes full, we will have 'a - b (mod a)' water in V1 when V2 becomes full first time . Once V2 becomes full, it is emptied. We will have 'a - 2b (mod a)' water in V1 when V2 is full second time. We repeat the above steps, and get 'a - nb (mod a)' water in V1 after the vessel V2 is filled and emptied 'n' times. We need to prove that the value of 'a - nb (mod a)' will be 1 for a finite integer 'n'. To prove this, let us consider the following property of coprime numbers. 
For any two coprime integers 'a' and 'b', the integer 'b' has a multiplicative inverse modulo 'a'. In other words, there exists an integer 'y' such that 'b*y ? 1(mod)a' (See 3rd point here). After '(a - 1)*y' iterations, we will have 'a - [(a-1)*y*b (mod a)]' water in V1, the value of this expression is 'a - [(a - 1) * 1] mod a' which is 1. So the algorithm converges and we get 1 litre in V1.
This article is compiled by Aashish Barnwal.
 


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    Given three positive integers a, b, and M, the objective is to find (a/b) % M i.e., find the value of (a × b-1 ) % M, where b-1 is the modular inverse of b modulo M.Examples: Input: a = 10, b = 2, M = 13Output: 5Explanation: The modular inverse of 2 modulo 13 is 7, so (10 / 2) % 13 = (10 × 7) % 13 =
    13 min read
    Cyclic Redundancy Check and Modulo-2 Division
    Cyclic Redundancy Check or CRC is a method of detecting accidental changes/errors in the communication channel. CRC uses Generator Polynomial which is available on both sender and receiver side. An example generator polynomial is of the form like x3 + x + 1. This generator polynomial represents key
    15+ min read
    Primitive root of a prime number n modulo n
    Given a prime number n, the task is to find its primitive root under modulo n. The primitive root of a prime number n is an integer r between[1, n-1] such that the values of r^x(mod n) where x is in the range[0, n-2] are different. Return -1 if n is a non-prime number. Examples: Input : 7 Output : S
    15 min read
    Euler's criterion (Check if square root under modulo p exists)
    Given a number 'n' and a prime p, find if square root of n under modulo p exists or not. A number x is square root of n under modulo p if (x*x)%p = n%p. Examples : Input: n = 2, p = 5 Output: false There doesn't exist a number x such that (x*x)%5 is 2 Input: n = 2, p = 7 Output: true There exists a
    11 min read
    Using Chinese Remainder Theorem to Combine Modular equations
    Given N modular equations: A ? x1mod(m1) . . A ? xnmod(mn) Find x in the equation A ? xmod(m1*m2*m3..*mn) where mi is prime, or a power of a prime, and i takes values from 1 to n. The input is given as two arrays, the first being an array containing values of each xi, and the second array containing
    12 min read
    Multiply large integers under large modulo
    Given an integer a, b, m. Find (a * b ) mod m, where a, b may be large and their direct multiplication may cause overflow. However, they are smaller than half of the maximum allowed long long int value. Examples: Input: a = 426, b = 964, m = 235Output: 119Explanation: (426 * 964) % 235 = 410664 % 23
    7 min read
    Compute n! under modulo p
    Given a large number n and a prime p, how to efficiently compute n! % p?Examples : Input: n = 5, p = 13 Output: 3 5! = 120 and 120 % 13 = 3 Input: n = 6, p = 11 Output: 5 6! = 720 and 720 % 11 = 5 A Naive Solution is to first compute n!, then compute n! % p. This solution works fine when the value o
    15+ min read
    Wilson's Theorem
    Wilson's Theorem is a fundamental result in number theory that provides a necessary and sufficient condition for determining whether a given number is prime. It states that a natural number p > 1 is a prime number if and only if:(p - 1)! ≡ −1 (mod p)This means that the factorial of p - 1 (the pro
    2 min read

    Number Theory

    Introduction to Primality Test and School Method
    Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples of the first few prime numbers are {2, 3, 5, ...}Examples : Input: n = 11Output: trueInput: n = 15Output: falseInput: n = 1Output:
    10 min read
    Fermat Method of Primality Test
    Given a number n, check if it is prime or not. We have introduced and discussed the School method for primality testing in Set 1.Introduction to Primality Test and School MethodIn this post, Fermat's method is discussed. This method is a probabilistic method and is based on Fermat's Little Theorem.
    10 min read
    Primality Test | Set 3 (Miller–Rabin)
    Given a number n, check if it is prime or not. We have introduced and discussed School and Fermat methods for primality testing.Primality Test | Set 1 (Introduction and School Method) Primality Test | Set 2 (Fermat Method)In this post, the Miller-Rabin method is discussed. This method is a probabili
    15+ min read
    Solovay-Strassen method of Primality Test
    We have already been introduced to primality testing in the previous articles in this series. Introduction to Primality Test and School MethodFermat Method of Primality TestPrimality Test | Set 3 (Miller–Rabin)The Solovay–Strassen test is a probabilistic algorithm used to check if a number is prime
    13 min read
    Legendre's formula - Largest power of a prime p in n!
    Given an integer n and a prime number p, the task is to find the largest x such that px (p raised to power x) divides n!.Examples: Input: n = 7, p = 3Output: x = 2Explanation: 32 divides 7! and 2 is the largest such power of 3.Input: n = 10, p = 3Output: x = 4Explanation: 34 divides 10! and 4 is the
    6 min read
    Carmichael Numbers
    A number n is said to be a Carmichael number if it satisfies the following modular arithmetic condition: power(b, n-1) MOD n = 1, for all b ranging from 1 to n such that b and n are relatively prime, i.e, gcd(b, n) = 1 Given a positive integer n, find if it is a Carmichael number. These numbers have
    8 min read
    Number Theory | Generators of finite cyclic group under addition
    Given a number n, find all generators of cyclic additive group under modulo n. Generator of a set {0, 1, ... n-1} is an element x such that x is smaller than n, and using x (and addition operation), we can generate all elements of the set.Examples: Input : 10 Output : 1 3 7 9 The set to be generated
    5 min read
    Sum of divisors of factorial of a number
    Given a number n, we need to calculate the sum of divisors of factorial of the number. Examples: Input : 4 Output : 60 Factorial of 4 is 24. Divisors of 24 are 1 2 3 4 6 8 12 24, sum of these is 60. Input : 6 Output : 2418 A Simple Solution is to first compute the factorial of the given number, then
    14 min read
    GFact | 2x + 1(where x > 0) is prime if and only if x is a power of 2
    A number of the form 2x + 1 (where x > 0) is prime if and only if x is a power of 2, i.e., x = 2n. So overall number becomes 22n + 1. Such numbers are called Fermat Number (Numbers of form 22n + 1). The first few Fermat numbers are 3, 5, 17, 257, 65537, 4294967297, .... An important thing to note
    1 min read
    Sieve of Eratosthenes
    Given a number n, find all prime numbers less than or equal to n.Examples:Input: n = 10Output: [2, 3, 5, 7]Explanation: The prime numbers up to 10 obtained by Sieve of Eratosthenes are [2, 3, 5, 7].Input: n = 35Output: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]Explanation: The prime numbers up to 35 o
    5 min read
    Program for Goldbach’s Conjecture (Two Primes with given Sum)
    Goldbach's conjecture is one of the oldest and best-known unsolved problems in the number theory of mathematics. Every even integer greater than 2 can be expressed as the sum of two primes. Examples: Input : n = 44 Output : 3 + 41 (both are primes) Input : n = 56 Output : 3 + 53 (both are primes) Re
    12 min read
    Pollard's Rho Algorithm for Prime Factorization
    Given a positive integer n, and that it is composite, find a divisor of it.Example:Input: n = 12;Output: 2 [OR 3 OR 4]Input: n = 187;Output: 11 [OR 17]Brute approach: Test all integers less than n until a divisor is found. Improvisation: Test all integers less than ?nA large enough number will still
    14 min read

    Game Theory

    Minimax Algorithm in Game Theory | Set 1 (Introduction)
    Minimax is a kind of backtracking algorithm that is used in decision making and game theory to find the optimal move for a player, assuming that your opponent also plays optimally. It is widely used in two player turn-based games such as Tic-Tac-Toe, Backgammon, Mancala, Chess, etc.In Minimax the tw
    9 min read
    Combinatorial Game Theory | Set 2 (Game of Nim)
    We strongly recommend to refer below article as a prerequisite of this. Combinatorial Game Theory | Set 1 (Introduction) In this post, Game of Nim is discussed. The Game of Nim is described by the following rules- “ Given a number of piles in which each pile contains some numbers of stones/coins. In
    15+ min read
    Combinatorial Game Theory | Set 4 (Sprague - Grundy Theorem)
    Prerequisites : Grundy Numbers/Numbers and MexWe have already seen in Set 2 (https://www.geeksforgeeks.org/combinatorial-game-theory-set-2-game-nim/), that we can find who wins in a game of Nim without actually playing the game.Suppose we change the classic Nim game a bit. This time each player can
    15 min read

    Practice Problems

    Rabin-Karp Algorithm for Pattern Searching
    Given two strings text and pattern string, your task is to find all starting positions where the pattern appears as a substring within the text. The strings will only contain lowercase English alphabets.While reporting the results, use 1-based indexing (i.e., the first character of the text is at po
    12 min read
    Measure one litre using two vessels and infinite water supply
    There are two vessels of capacities 'a' and 'b' respectively. We have infinite water supply. Give an efficient algorithm to make exactly 1 litre of water in one of the vessels. You can throw all the water from any vessel any point of time. Assume that 'a' and 'b' are Coprimes.Following are the steps
    15 min read
    Program to find last digit of n'th Fibonacci Number
    Given a number 'n', write a function that prints the last digit of n'th ('n' can also be a large number) Fibonacci number. Examples : Input : n = 0 Output : 0 Input: n = 2 Output : 1 Input : n = 7 Output : 3 Recommended PracticeThe Nth FibonnaciTry It! Method 1 : (Naive Method) Simple approach is to
    13 min read
    GCD of two numbers when one of them can be very large
    Given two numbers 'a' and 'b' such that (0 <= a <= 10^12 and b <= b < 10^250). Find the GCD of two given numbers.Examples : Input: a = 978 b = 89798763754892653453379597352537489494736 Output: 6 Input: a = 1221 b = 1234567891011121314151617181920212223242526272829 Output: 3 Solution : In
    9 min read
    Find Last Digit of a^b for Large Numbers
    You are given two integer numbers, the base a (number of digits d, such that 1 <= d <= 1000) and the index b (0 <= b <= 922*10^15). You have to find the last digit of a^b.Examples: Input : 3 10Output : 9Input : 6 2Output : 6Input : 150 53Output : 0 After taking few examples, we can notic
    9 min read
    Remainder with 7 for large numbers
    Given a large number as a string, find the remainder of number when divided by 7. Examples : Input : num = 1234 Output : 2 Input : num = 1232 Output : 0 Input : num = 12345 Output : 4Recommended PracticeRemainder with 7Try It! Simple Approach is to convert a string into number and perform the mod op
    8 min read
    Find (a^b)%m where 'a' is very large
    Given three numbers a, b and m where 1<=b,m<=10^6 and 'a' may be very large and contains upto 10^6 digits. The task is to find (a^b)%m. Examples: Input : a = 3, b = 2, m = 4 Output : 1 Explanation : (3^2)%4 = 9%4 = 1 Input : a = 987584345091051645734583954832576, b = 3, m = 11 Output: 10Recomm
    15+ min read
    Find sum of modulo K of first N natural number
    Given two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ..... + N%K. Examples : Input : N = 10 and K = 2. Output : 5 Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 + 7%2 + 8%2 + 9%2 + 10%2 = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 = 5.Recommended PracticeReve
    9 min read
    Count sub-arrays whose product is divisible by k
    Given an integer K and an array arr[], the task is to count all the sub-arrays whose product is divisible by K.Examples: Input: arr[] = {6, 2, 8}, K = 4 Output: 4 Required sub-arrays are {6, 2}, {6, 2, 8}, {2, 8}and {8}.Input: arr[] = {9, 1, 14}, K = 6 Output: 1 Naive approach: Run nested loops and
    15+ min read
    Partition a number into two divisible parts
    Given a number (as string) and two integers a and b, divide the string in two non-empty parts such that the first part is divisible by a and the second part is divisible by b. If the string can not be divided into two non-empty parts, output "NO", else print "YES" with the two parts. Examples: Input
    15+ min read
    Find power of power under mod of a prime
    Given four numbers A, B, C and M, where M is prime number. Our task is to compute A raised to power (B raised to power C) modulo M. Example: Input : A = 2, B = 4, C = 3, M = 23Output : 643 = 64 so,2^64(mod 23) = 6 A Naive Approach is to calculate res = BC and then calculate Ares % M by modular expon
    7 min read
    Rearrange an array in maximum minimum form in O(1) extra space
    Given a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on. Examples:Input: arr[] = {1, 2, 3, 4, 5, 6, 7} Output: arr[] = {7, 1, 6, 2, 5, 3, 4}Explanation: First 7 is th
    8 min read
    Subset with no pair sum divisible by K
    Given an array of integer numbers, we need to find maximum size of a subset such that sum of each pair of this subset is not divisible by K. Examples : Input : arr[] = [3, 7, 2, 9, 1] K = 3 Output : 3 Maximum size subset whose each pair sum is not divisible by K is [3, 7, 1] because, 3+7 = 10, 3+1 =
    7 min read
    Number of substrings divisible by 6 in a string of integers
    Given a string consisting of integers 0 to 9. The task is to count the number of substrings which when convert into integer are divisible by 6. Substring does not contain leading zeroes. Examples: Input : s = "606". Output : 5 Substrings "6", "0", "6", "60", "606" are divisible by 6. Input : s = "48
    9 min read

    Miscellaneous Practice Problems

    How to compute mod of a big number?
    Given a big number 'num' represented as string and an integer x, find value of "num % a" or "num mod a". Output is expected as an integer. Examples : Input: num = "12316767678678", a = 10 Output: num (mod a) ? 8 The idea is to process all digits one by one and use the property that xy (mod a) ? ((x
    4 min read
    BigInteger Class in Java
    BigInteger class is used for the mathematical operation which involves very big integer calculations that are outside the limit of all available primitive data types. In this way, BigInteger class is very handy to use because of its large method library and it is also used a lot in competitive progr
    6 min read
    Modulo 10^9+7 (1000000007)
    In most programming competitions, we are required to answer the result in 10^9+7 modulo. The reason behind this is, if problem constraints are large integers, only efficient algorithms can solve them in an allowed limited time.What is modulo operation: The remainder obtained after the division opera
    10 min read
    How to avoid overflow in modular multiplication?
    Consider below simple method to multiply two numbers. C++ #include <iostream> using namespace std; #define ll long long // Function to multiply two numbers modulo mod ll multiply(ll a, ll b, ll mod) { // Multiply two numbers and then take modulo to prevent // overflow return ((a % mod) * (b %
    6 min read
    RSA Algorithm in Cryptography
    RSA(Rivest-Shamir-Adleman) Algorithm is an asymmetric or public-key cryptography algorithm which means it works on two different keys: Public Key and Private Key. The Public Key is used for encryption and is known to everyone, while the Private Key is used for decryption and must be kept secret by t
    13 min read
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