Pollard's Rho Algorithm for Prime Factorization
Last Updated :
14 Aug, 2024
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 ?n
A large enough number will still mean a great deal of work. Pollard’s Rho is a prime factorization algorithm, particularly fast for a large composite number with small prime factors. The Rho algorithm’s most remarkable success was the factorization of eighth Fermat number: 1238926361552897 * 93461639715357977769163558199606896584051237541638188580280321.
The Rho algorithm was a good choice because the first prime factor is much smaller than the other one.
Concepts used in Pollard’s Rho Algorithm:
- Two numbers x and y are said to be congruent modulo n (x = y modulo n) if
- their absolute difference is an integer multiple of n, OR,
- each of them leaves the same remainder when divided by n.
- The Greatest Common Divisor is the largest number which divides evenly into each of the original numbers.
- Birthday Paradox: The probability of two persons having same birthday is unexpectedly high even for small set of people.
- Floyd's cycle-finding algorithm: If tortoise and hare start at same point and move in a cycle such that speed of hare is twice the speed of tortoise, then they must meet at some point.
Algorithm:
- Start with random x and c. Take y equal to x and f(x) = x2 + c.
- While a divisor isn't obtained
- Update x to f(x) (modulo n) [Tortoise Move]
- Update y to f(f(y)) (modulo n) [Hare Move]
- Calculate GCD of |x-y| and n
- If GCD is not unity
- If GCD is n, repeat from step 2 with another set of x, y and c
- Else GCD is our answer
Illustration:
Let us suppose n = 187 and consider different cases for different random values.
1. An Example of random values such that algorithm finds result:
y = x = 2 and c = 1, Hence, our f(x) = x2 + 1.
2. An Example of random values such that algorithm finds result faster:
y = x = 110 and ‘c’ = 183. Hence, our f(x) = x2 + 183.
3. An Example of random values such that algorithm doesn't find result:
x = y = 147 and c = 67. Hence, our f(x) = x2 + 67.
Below is implementation of above algorithm:
C++
/* C++ program to find a prime factor of composite using
Pollard's Rho algorithm */
#include<bits/stdc++.h>
using namespace std;
/* Function to calculate (base^exponent)%modulus */
long long int modular_pow(long long int base, int exponent,
long long int modulus)
{
/* initialize result */
long long int result = 1;
while (exponent > 0)
{
/* if y is odd, multiply base with result */
if (exponent & 1)
result = (result * base) % modulus;
/* exponent = exponent/2 */
exponent = exponent >> 1;
/* base = base * base */
base = (base * base) % modulus;
}
return result;
}
/* method to return prime divisor for n */
long long int PollardRho(long long int n)
{
/* initialize random seed */
srand (time(NULL));
/* no prime divisor for 1 */
if (n==1) return n;
/* even number means one of the divisors is 2 */
if (n % 2 == 0) return 2;
/* we will pick from the range [2, N) */
long long int x = (rand()%(n-2))+2;
long long int y = x;
/* the constant in f(x).
* Algorithm can be re-run with a different c
* if it throws failure for a composite. */
long long int c = (rand()%(n-1))+1;
/* Initialize candidate divisor (or result) */
long long int d = 1;
/* until the prime factor isn't obtained.
If n is prime, return n */
while (d==1)
{
/* Tortoise Move: x(i+1) = f(x(i)) */
x = (modular_pow(x, 2, n) + c + n)%n;
/* Hare Move: y(i+1) = f(f(y(i))) */
y = (modular_pow(y, 2, n) + c + n)%n;
y = (modular_pow(y, 2, n) + c + n)%n;
/* check gcd of |x-y| and n */
d = __gcd(abs(x-y), n);
/* retry if the algorithm fails to find prime factor
* with chosen x and c */
if (d==n) return PollardRho(n);
}
return d;
}
/* driver function */
int main()
{
long long int n = 10967535067;
printf("One of the divisors for %lld is %lld.",
n, PollardRho(n));
return 0;
}
/* Java program to find a prime factor of composite using
Pollard's Rho algorithm */
import java.util.*;
class GFG{
/* Function to calculate (base^exponent)%modulus */
static long modular_pow(long base, int exponent,
long modulus)
{
/* initialize result */
long result = 1;
while (exponent > 0)
{
/* if y is odd, multiply base with result */
if (exponent % 2 == 1)
result = (result * base) % modulus;
/* exponent = exponent/2 */
exponent = exponent >> 1;
/* base = base * base */
base = (base * base) % modulus;
}
return result;
}
/* method to return prime divisor for n */
static long PollardRho(long n)
{
/* initialize random seed */
Random rand = new Random();
/* no prime divisor for 1 */
if (n == 1) return n;
/* even number means one of the divisors is 2 */
if (n % 2 == 0) return 2;
/* we will pick from the range [2, N) */
long x = (long)(rand.nextLong() % (n - 2)) + 2;
long y = x;
/* the constant in f(x).
* Algorithm can be re-run with a different c
* if it throws failure for a composite. */
long c = (long)(rand.nextLong()) % (n - 1) + 1;
/* Initialize candidate divisor (or result) */
long d = 1L;
/* until the prime factor isn't obtained.
If n is prime, return n */
while (d == 1)
{
/* Tortoise Move: x(i+1) = f(x(i)) */
x = (modular_pow(x, 2, n) + c + n) % n;
/* Hare Move: y(i+1) = f(f(y(i))) */
y = (modular_pow(y, 2, n) + c + n) % n;
y = (modular_pow(y, 2, n) + c + n) % n;
/* check gcd of |x-y| and n */
d = __gcd(Math.abs(x - y), n);
/* retry if the algorithm fails to find prime factor
* with chosen x and c */
if (d == n) return PollardRho(n);
}
return d;
}
// Recursive function to return gcd of a and b
static long __gcd(long a, long b)
{
return b == 0? a:__gcd(b, a % b);
}
/* driver function */
public static void main(String[] args)
{
long n = 10967535067L;
System.out.printf("One of the divisors for " + n + " is " +
PollardRho(n));
}
}
// This code contributed by aashish1995
# Python 3 program to find a prime factor of composite using
# Pollard's Rho algorithm
import random
import math
# Function to calculate (base^exponent)%modulus
def modular_pow(base, exponent,modulus):
# initialize result
result = 1
while (exponent > 0):
# if y is odd, multiply base with result
if (exponent & 1):
result = (result * base) % modulus
# exponent = exponent/2
exponent = exponent >> 1
# base = base * base
base = (base * base) % modulus
return result
# method to return prime divisor for n
def PollardRho( n):
# no prime divisor for 1
if (n == 1):
return n
# even number means one of the divisors is 2
if (n % 2 == 0):
return 2
# we will pick from the range [2, N)
x = (random.randint(0, 2) % (n - 2))
y = x
# the constant in f(x).
# Algorithm can be re-run with a different c
# if it throws failure for a composite.
c = (random.randint(0, 1) % (n - 1))
# Initialize candidate divisor (or result)
d = 1
# until the prime factor isn't obtained.
# If n is prime, return n
while (d == 1):
# Tortoise Move: x(i+1) = f(x(i))
x = (modular_pow(x, 2, n) + c + n)%n
# Hare Move: y(i+1) = f(f(y(i)))
y = (modular_pow(y, 2, n) + c + n)%n
y = (modular_pow(y, 2, n) + c + n)%n
# check gcd of |x-y| and n
d = math.gcd(abs(x - y), n)
# retry if the algorithm fails to find prime factor
# with chosen x and c
if (d == n):
return PollardRho(n)
return d
# Driver function
if __name__ == "__main__":
n = 10967535067
print("One of the divisors for", n , "is ",PollardRho(n))
# This code is contributed by chitranayal
/* C# program to find a prime factor of composite using
Pollard's Rho algorithm */
using System;
class GFG
{
/* Function to calculate (base^exponent)%modulus */
static long modular_pow(long _base, int exponent,
long modulus)
{
/* initialize result */
long result = 1;
while (exponent > 0)
{
/* if y is odd, multiply base with result */
if (exponent % 2 == 1)
result = (result * _base) % modulus;
/* exponent = exponent/2 */
exponent = exponent >> 1;
/* base = base * base */
_base = (_base * _base) % modulus;
}
return result;
}
/* method to return prime divisor for n */
static long PollardRho(long n)
{
/* initialize random seed */
Random rand = new Random();
/* no prime divisor for 1 */
if (n == 1) return n;
/* even number means one of the divisors is 2 */
if (n % 2 == 0) return 2;
/* we will pick from the range [2, N) */
long x = (long)(rand.Next(0, -(int)n + 1));
long y = x;
/* the constant in f(x).
* Algorithm can be re-run with a different c
* if it throws failure for a composite. */
long c = (long)(rand.Next(1, -(int)n));
/* Initialize candidate divisor (or result) */
long d = 1L;
/* until the prime factor isn't obtained.
If n is prime, return n */
while (d == 1)
{
/* Tortoise Move: x(i+1) = f(x(i)) */
x = (modular_pow(x, 2, n) + c + n) % n;
/* Hare Move: y(i+1) = f(f(y(i))) */
y = (modular_pow(y, 2, n) + c + n) % n;
y = (modular_pow(y, 2, n) + c + n) % n;
/* check gcd of |x-y| and n */
d = __gcd(Math.Abs(x - y), n);
/* retry if the algorithm fails to find prime factor
* with chosen x and c */
if (d == n) return PollardRho(n);
}
return d;
}
// Recursive function to return gcd of a and b
static long __gcd(long a, long b)
{
return b == 0 ? a:__gcd(b, a % b);
}
/* Driver code */
public static void Main(String[] args)
{
long n = 10967535067L;
Console.Write("One of the divisors for " + n + " is " +
PollardRho(n));
}
}
// This code is contributed by aashish1995
<script>
/* Javascript program to find a prime factor of composite using
Pollard's Rho algorithm */
/* Function to calculate (base^exponent)%modulus */
function modular_pow(base,exponent,modulus)
{
/* initialize result */
let result = 1;
while (exponent > 0)
{
/* if y is odd, multiply base with result */
if (exponent % 2 == 1)
result = (result * base) % modulus;
/* exponent = exponent/2 */
exponent = exponent >> 1;
/* base = base * base */
base = (base * base) % modulus;
}
return result;
}
/* method to return prime divisor for n */
function PollardRho(n)
{
/* no prime divisor for 1 */
if (n == 1)
return n;
/* even number means one of the divisors is 2 */
if (n % 2 == 0)
return 2;
/* we will pick from the range [2, N) */
let x=(Math.floor(Math.random() * (-n + 1) ));
let y = x;
/* the constant in f(x).
* Algorithm can be re-run with a different c
* if it throws failure for a composite. */
let c= (Math.floor(Math.random() * (-n + 1)));
/* Initialize candidate divisor (or result) */
let d=1;
/* until the prime factor isn't obtained.
If n is prime, return n */
while (d == 1)
{
/* Tortoise Move: x(i+1) = f(x(i)) */
x = (modular_pow(x, 2, n) + c + n) % n;
/* Hare Move: y(i+1) = f(f(y(i))) */
y = (modular_pow(y, 2, n) + c + n) % n;
y = (modular_pow(y, 2, n) + c + n) % n;
/* check gcd of |x-y| and n */
d = __gcd(Math.abs(x - y), n);
/* retry if the algorithm fails to find prime factor
* with chosen x and c */
if (d == n) return PollardRho(n);
}
return d;
}
// Recursive function to return gcd of a and b
function __gcd(a,b)
{
return b == 0? a:__gcd(b, a % b);
}
/* driver function */
let n = 10967535067;
document.write("One of the divisors for " + n + " is " +
PollardRho(n));
// This code is contributed by avanitrachhadiya2155
</script>
Output:
One of the divisors for 10967535067 is 104729
Time Complexity : O(sqrt(n)*logn)
Auxiliary Space: O(1)
How does this work?
Let n be a composite (non-prime). Since n is composite, it has a non trivial factor f < n. In fact, there is at least one f <= ?n .
Now suppose we have to pick two numbers x and y from the range [0, n-1]. The only time we get x = y modulo n is when x and y are identical. However, since f < ?n, there is a good chance x = y modulo f even when x and y are not identical (Birthday Paradox).
We begin by randomly selecting x with replacement from the set {0, 1, …, n-1} to form a sequence s1, s2, s3 … Defining śi = si mod f, our sequence now has each śi belonging to {0, 1, …, f-1}. Because both the sets are finite, eventually there will be a repeated integer in both. We expect to achieve the repeat earlier in śi, since f < n.
Now, say śi = śj for i ? j, then, si = sj modulo d. And hence, |si - sj| will be a multiple of f. As per assumed above, n is also a multiple of f. Together this means that GCD of |si - sj| and n will be positive integral multiple of f, and also our candidate divisor d! The catch here is that we just knew there had to be some divisor of n, and we didn’t even care of its value. Once we hit si and sj (our final x and y) then each element in the sequence starting with si will be congruent modulo f to the corresponding element in the sequence starting with sj, and hence, a cycle. If we graph the sequence si, we will observe the shape of Greek letter Rho (?).
At the heart of Rho algorithm is picking up random values and evaluating GCDs. To decrease the costly GCD calculations, we can implement the Pollard’s Rho with Floyd’s cycle detection (which can be understood with the tortoise-hare analogy where the tortoise moves through each element one at a time in order, and the hare starts at the same point but moves twice as fast as the tortoise). We shall have some polynomial f(x) for the same, start with random x0, y0 = x0, and compute xi+1 = f(xi) and yi+1 = f(f(yi)). Since we don’t know much about d, a typical choice for the polynomial is f(x) = x2 + c (modulo n) (Yes, ‘c’ is also be chosen randomly).
Note:
- Algorithm will run indefinitely for prime numbers.
- The algorithm may not find the factors and return a failure for composite n. In that case, we use a different set of x, y and c and try again.
- The above algorithm only finds a divisor. To find a prime factor, we may recursively factorize the divisor d, run algorithm for d and n/d. The cycle length is typically of the order ?d.
Time Complexity analysis:
The algorithm offers a trade-off between its running time and the probability that it finds a factor. A prime divisor can be achieved with a probability around 0.5, in O(?d) <= O(n1/4) iterations. This is a heuristic claim, and rigorous analysis of the algorithm remains open.
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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
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Using Chinese Remainder Theorem to Combine Modular equationsGiven 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
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Multiply large integers under large moduloGiven 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
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Compute n! under modulo pGiven 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
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Wilson's TheoremWilson'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
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Number Theory
Introduction to Primality Test and School MethodGiven 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:
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Fermat Method of Primality TestGiven 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.
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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
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Solovay-Strassen method of Primality TestWe 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
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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
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Carmichael NumbersA 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
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Number Theory | Generators of finite cyclic group under additionGiven 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
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Sum of divisors of factorial of a numberGiven 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
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GFact | 2x + 1(where x > 0) is prime if and only if x is a power of 2A 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
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Sieve of EratosthenesGiven 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
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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
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Pollard's Rho Algorithm for Prime FactorizationGiven 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
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Game Theory
Practice Problems
Rabin-Karp Algorithm for Pattern SearchingGiven 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
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Measure one litre using two vessels and infinite water supplyThere 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
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Program to find last digit of n'th Fibonacci NumberGiven 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
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GCD of two numbers when one of them can be very largeGiven 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
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Find Last Digit of a^b for Large NumbersYou 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
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Remainder with 7 for large numbersGiven 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
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Find (a^b)%m where 'a' is very largeGiven 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
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Find sum of modulo K of first N natural numberGiven 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
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Count sub-arrays whose product is divisible by kGiven 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
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Partition a number into two divisible partsGiven 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
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Find power of power under mod of a primeGiven 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
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Rearrange an array in maximum minimum form in O(1) extra spaceGiven 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
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Subset with no pair sum divisible by KGiven 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 =
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Number of substrings divisible by 6 in a string of integersGiven 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
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Miscellaneous Practice Problems