Program to calculate value of nCr Last Updated : 21 Jun, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given two numbers n and r, The task is to find the value of nCr . Combinations represent the number of ways to choose r elements from a set of n distinct elements, without regard to the order in which they are selected. The formula for calculating combinations is : Note: If r is greater than n, return 0.Examples Input: n = 5, r = 2Output: 10 Explanation: The value of 5C2 is calculated as 5! / ((5−2)! * 2!​)= 10.Input: n = 2, r = 4Output: 0Explanation: Since r is greater than n, thus 2C4 = 0Input: n = 5, r = 0Output: 1Explanation: The value of 5C0 is calculated as 5!/(5−0)!*0! = 5!/5!*0! = 1.Table of Content[Naive Approach] Using Recursion - O(2^n) Time and O(n) Space[Better Approach - 1] Using Factorial - O(n) Time and O(1) Space [Better Approach - 2] Avoiding Factorial Computations - O(n) Time and O(1) Space[Expected Approach] By using Binomial Coefficient formula - O(r) Time and O(1) Space[Alternate Approach] Using Logarithmic Formula - O(r) Time and O(1)[Naive Approach] Using Recursion - O(2^n) Time and O(n) SpaceThe idea is to use a recursive function to calculate the value of nCr. The base cases are:if r is greater than n, return 0 (there are no combinations possible)if r is 0 or r is n, return 1 (there is only 1 combination possible in these cases)For other values of n and r, the function calculates the value of nCr by adding the number of combinations possible by including the current element and the number of combinations possible by not including the current element. C++ #include <iostream> using namespace std; int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; // Base case: only one way to choose 0 or all elements if (r == 0 || r == n) return 1; // include or exclude current element return nCr(n - 1, r - 1) + nCr(n - 1, r); } // Driver Code int main() { int n = 5; int r = 2; cout << nCr(n, r); return 0; } Java import java.util.*; class GfG { public static int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; // Base case: only one way to choose 0 or all elements if (r == 0 || r == n) return 1; // include or exclude current element return nCr(n - 1, r - 1) + nCr(n - 1, r); } // Driver Code public static void main(String[] args) { int n = 5; int r = 2; System.out.println(nCr(n, r)); } } Python def nCr(n, r): # No valid combinations if r is greater than n if r > n: return 0 # Base case: only one way to choose 0 or all elements if r == 0 or r == n: return 1 # include or exclude current element return nCr(n - 1, r - 1) + nCr(n - 1, r) # Driver Code if __name__ == "__main__": n = 5 r = 2 print(nCr(n, r)) C# using System; class GfG { static public int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; // Base case: only one way to choose 0 or all elements if (r == 0 || r == n) return 1; // include or exclude current element return nCr(n - 1, r - 1) + nCr(n - 1, r); } // Driver Code static public void Main(string[] args) { int n = 5; int r = 2; Console.WriteLine(nCr(n, r)); } } JavaScript function nCr(n, r) { // No valid combinations if r is greater than n if (r > n) return 0; // Base case: only one way to choose 0 or all elements if (r === 0 || r === n) return 1; // include or exclude current element return nCr(n - 1, r - 1) + nCr(n - 1, r); } // Driver Code let n = 5; let r = 2; console.log(nCr(n, r)); Output10[Better Approach - 1] Using Factorial - O(n) Time and O(1) Space This approach calculates the nCr using the factorial formula. It first computes the factorial of a given number by multiplying all integers from 1 to that number. To find nCr, it calculates the factorial of n, r, and (n - r) separately, then applies the formula n! / (r!(n-r)!) to determine the result. Since factorial values grow rapidly, this method is inefficient for large values due to integer overflow and excessive computations. Note: This approach may produce incorrect results due to integer overflow when handling large values of n and r. C++ #include <iostream> using namespace std; // Returns factorial of n int fact(int n) { int res = 1; for (int i = 2; i <= n; i++) res *= i; return res; } int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r)); } // Driver Code int main() { int n = 5, r = 2; cout << nCr(n, r); return 0; } Java class GfG { static int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n static int fact(int n) { int res = 1; for (int i = 2; i <= n; i++) res *= i; return res; } // Driver code public static void main(String[] args) { int n = 5, r = 2; System.out.println(nCr(n, r)); } } Python def nCr(n, r): # No valid combinations if r is greater than n if r > n: return 0 # Calculate nCr using factorial formula return fact(n) // (fact(r) * fact(n - r)) # Returns factorial of n def fact(n): res = 1 for i in range(2, n + 1): res *= i return res if __name__ == "__main__": n = 5 r = 2 print(int(nCr(n, r))) C# using System; class GFG { // Calculates nCr using factorial formula static int nCr(int n, int r) { // No valid combinations if r is greater than n if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n static int fact(int n) { int res = 1; for (int i = 2; i <= n; i++) res *= i; return res; } public static void Main() { int n = 5, r = 2; Console.Write(nCr(n, r)); } } JavaScript function nCr(n, r) { // No valid combinations if r is greater than n if (r > n) return 0; // Calculate nCr using factorial formula return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n function fact(n) { let res = 1; for (let i = 2; i <= n; i++) res *= i; return res; } // Driver code let n = 5, r = 2; console.log(nCr(n, r)); Output10[Better Approach - 2] Avoiding Factorial Computations - O(n) Time and O(1) SpaceThe formula for nCr is n! / (r!(n-r)!).Instead of computing full factorials, we avoid redundant calculations by recognizing that r! and (n-r)! share common terms that cancel out.To optimize, we compute the product of numbers from r+1 to n and divide it by the product of numbers from 1 to (n-r).Here, r is chosen as the maximum of r and (n-r) to reduce the number of multiplications.This approach avoids large factorial values, reducing computational overhead and preventing integer overflow.Note: This approach may produce incorrect results due to integer overflow when handling large values of n and r. C++ #include <iostream> using namespace std; // Calculates the product of natural // numbers from start to end double Multiplier(int start, int end) { if (start == end) return start; double res = 1; while (start <= end) { res *= start; start++; } return res; } int nCr(int n, int r) { // No valid combinations if r > n if (n < r) return 0; // Base cases: nC0 or nCn = 1 if (n == r || r == 0) return 1; // Use max(r, n-r) to minimize the // number of multiplications int max_val = max(r, n - r); int min_val = min(r, n - r); double nume = Multiplier(max_val + 1, n); double deno = Multiplier(1, min_val); return int(nume / deno); } int main() { int n = 5; int r = 2; cout << nCr(n, r) << endl; return 0; } Java class GfG { // Calculates the product of natural // numbers from start to end public static double Multiplier(int start, int end) { if (start == end) return start; double res = 1; while (start <= end) { res *= start; start++; } return res; } public static int nCr(int n, int r) { // No valid combinations if r > n if (n < r) return 0; // Base cases: nC0 or nCn = 1 if (n == r || r == 0) return 1; // Use max(r, n-r) to reduce the // number of multiplications int max_val = Math.max(r, n - r); int min_val = Math.min(r, n - r); double nume = Multiplier(max_val + 1, n); double deno = Multiplier(1, min_val); return (int)(nume / deno); } public static void main(String[] args) { int n = 5; int r = 2; System.out.println(nCr(n, r)); // Output the result of 5C2 } } Python def Multiplier(start, end): if start == end: return start res = 1 while start <= end: res *= start start += 1 return res def nCr(n, r): # No valid combinations if r > n if n < r: return 0 # Base cases: nC0 or nCn = 1 if n == r or r == 0: return 1 # Use max(r, n - r) to reduce # number of multiplications max_val = max(r, n - r) min_val = min(r, n - r) nume = Multiplier(max_val + 1, n) deno = Multiplier(1, min_val) return nume // deno if __name__ == "__main__": n = 5 r = 2 print(nCr(n, r)) C# using System; class Program { // Calculates the product of natural // numbers from start to end static double Multiplier(int start, int end) { if (start == end) return start; double res = 1; while (start <= end) { res *= start; start++; } return res; } static int nCr(int n, int r) { // No valid combinations if r > n if (n < r) return 0; // Base cases: nC0 or nCn = 1 if (n == r || r == 0) return 1; // Use max(r, n - r) to minimize // number of multiplications int max_val = Math.Max(r, n - r); int min_val = Math.Min(r, n - r); double nume = Multiplier(max_val + 1, n); double deno = Multiplier(1, min_val); return (int)(nume / deno); } static void Main() { int n = 5; int r = 2; Console.WriteLine(nCr(n, r)); } } JavaScript // Calculate multiplication of natural // numbers from start to end function Multiplier(start, end) { if (start === end) { return start; } let res = 1; while (start <= end) { res *= start; start++; } return res; } function nCr(n, r) { // No valid combinations if r > n if (n < r) { return 0; } // Base cases: nC0 or nCn = 1 if (n === r || r === 0) { return 1; } // Use max(r, n - r) to reduce the // number of multiplications const max_val = Math.max(r, n - r); const min_val = Math.min(r, n - r); const nume = Multiplier(max_val + 1, n); const deno = Multiplier(1, min_val); return Math.round(nume / deno); } // Driver Code const n = 5; const r = 2; console.log(nCr(n, r)); Output10 [Expected Approach] By using Binomial Coefficient formula - O(r) Time and O(1) SpaceA binomial coefficient C(n, k) can be defined as the coefficient of Xk in the expansion of (1 + X)n.A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set.Iterative way of calculating nCr using binomial coefficient formula. C++ #include <iostream> using namespace std; int nCr(int n, int r){ double sum = 1; // Calculate the value of n choose // r using the binomial coefficient formula for (int i = 1; i <= r; i++){ sum = sum * (n - r + i) / i; } return (int)sum; } int main(){ int n = 5; int r = 2; cout << nCr(n, r); return 0; } Java import java.util.*; class GfG{ public static int nCr(int n, int r){ double sum = 1; // Calculate the value of n choose r using the // binomial coefficient formula for (int i = 1; i <= r; i++) { sum = sum * (n - r + i) / i; } return (int)sum; } public static void main(String[] args){ int n = 5; int r = 2; System.out.println(nCr(n,r)); } } Python def nCr(n, r): sum = 1 # Calculate the value of n choose r # using the binomial coefficient formula for i in range(1, r+1): sum = sum * (n - r + i) // i return sum if __name__ == "__main__": n = 5 r = 2 print(nCr(n, r)) C# using System; class GfG { static int nCr(int n, int r){ double sum = 1; // Calculate the value of n choose r // using the binomial coefficient formula for(int i = 1; i <= r; i++){ sum = sum * (n - r + i) / i; } return (int)sum; } static void Main() { int n = 5; int r = 2; Console.WriteLine(nCr(n,r)); } } JavaScript function nCr(n, r){ let sum = 1; // Calculate the value of n choose r // using the binomial coefficient formula for(let i = 1; i <= r; i++){ sum = sum * (n - r + i) / i; } return Math.floor(sum); } // Driver Code let n = 5; let r = 2; console.log(nCr(n,r)); Output10[Alternate Approach] Using Logarithmic Formula - O(r) Time and O(1)Logarithmic formula for nCr is an alternative to the factorial formula that avoids computing factorials directly and it's more efficient for large values of n and r. It uses the identity log(n!) = log(1) + log(2) + ... + log(n) to express the numerator and denominator of the nCr in terms of sums of logarithms which allows to calculate the nCr using the Logarithmic operations. This approach is faster and very efficient.The logarithmic formula for nCr is: nCr = exp( log(n!) - log(r!) - log((n-r)!)) C++ #include <iostream> #include <cmath> using namespace std; // Calculates the binomial coefficient nCr using logarithmic formula int nCr(int n, int r) { // Invalid case if (r > n) return 0; // Base cases if (r == 0 || n == r) return 1; double res = 0; for (int i = 0; i < r; i++) { // log(n!) - log(r!) - log((n-r)!) res += log(n - i) - log(i + 1); } return (int)round(exp(res)); } int main() { int n = 5; int r = 2; cout << nCr(n, r) << endl; return 0; } Java import java.lang.Math; // Calculates the binomial coefficient nCr using the logarithmic formula public class GfG { public static int nCr(int n, int r) { // Invalid case if (r > n) return 0; // Base cases if (r == 0 || n == r) return 1; double res = 0; for (int i = 0; i < r; i++) { // log(n!) - log(r!) - log((n-r)!) res += Math.log(n - i) - Math.log(i + 1); } return (int)Math.round(Math.exp(res)); } public static void main(String[] args) { int n = 5; int r = 2; System.out.println(nCr(n, r)); } } Python import math # Calculates the binomial coefficient nCr using the logarithmic formula def nCr(n, r): # Invalid case if r > n: return 0 # Base cases if r == 0 or n == r: return 1 res = 0 for i in range(r): # log(n!) - log(r!) - log((n-r)!) res += math.log(n - i) - math.log(i + 1) return round(math.exp(res)) if __name__ == "__main__": n = 5 r = 2 print(nCr(n, r)) C# using System; class GfG { // Calculates the binomial coefficient nCr using the logarithmic formula public static int nCr(int n, int r) { // Invalid case if (r > n) return 0; // Base cases if (r == 0 || n == r) return 1; double res = 0; for (int i = 0; i < r; i++) { // log(n!) - log(r!) - log((n-r)!) res += Math.Log(n - i) - Math.Log(i + 1); } return (int)Math.Round(Math.Exp(res)); } static void Main(string[] args) { int n = 5; int r = 2; Console.WriteLine(nCr(n, r)); } } JavaScript // Calculates the binomial coefficient nCr using the logarithmic formula function nCr(n, r) { // Invalid case if (r > n) return 0; // Base cases if (r === 0 || n === r) return 1; let res = 0; for (let i = 0; i < r; i++) { // log(n!) - log(r!) - log((n-r)!) res += Math.log(n - i) - Math.log(i + 1); } return Math.round(Math.exp(res)); } // Driver code const n = 5; const r = 2; console.log(nCr(n, r)); Output10 Comment More infoAdvertise with us Next Article Program to calculate value of nCr K Kanishk_Verma Follow Improve Article Tags : Misc Mathematical Combinatorial DSA Basic Coding Problems factorial binomial coefficient +3 More Practice Tags : CombinatorialfactorialMathematicalMisc Similar Reads Mathematical Algorithms The following is the list of mathematical coding problem ordered topic wise. 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A simple way to find GCD is to factorize both numbers and multiply common prime factors.Examples:input: a = 12, b = 20Output: 4Explanatio 9 min read Stein's Algorithm for finding GCDStein's algorithm or binary GCD algorithm is an algorithm that computes the greatest common divisor of two non-negative integers. Stein’s algorithm replaces division with arithmetic shifts, comparisons, and subtraction.Examples: Input: a = 17, b = 34 Output : 17Input: a = 50, b = 49Output: 1Algorith 14 min read GCD, LCM and Distributive PropertyGiven three integers x, y, z, the task is to compute the value of GCD(LCM(x,y), LCM(x,z)) where, GCD = Greatest Common Divisor, LCM = Least Common MultipleExamples: Input: x = 15, y = 20, z = 100Output: 60Explanation: The GCD of 15 and 20 is 5, and the LCM of 15 and 20 is 60, which is then multiplie 4 min read Count number of pairs (A <= N, B <= N) such that gcd (A , B) is BGiven a number n, we need to find the number of ordered pairs of a and b such gcd(a, b) is b itselfExamples : Input : n = 2Output : 3The pairs are (1, 1) (2, 2) and (2, 1) Input : n = 3Output : 5(1, 1) (2, 2) (3, 3) (2, 1) and (3, 1)[Naive Approach] Counting GCD Pairs by Divisor Propertygcd(a, b) = 6 min read Program to find GCD of floating point numbersThe greatest common divisor (GCD) of two or more numbers, which are not all zero, is the largest positive number that divides each of the numbers. Example: Input : 0.3, 0.9Output : 0.3Explanation: The GCD of 0.3 and 0.9 is 0.3 because both numbers share 0.3 as the largest common divisor.Input : 0.48 4 min read Series with largest GCD and sum equals to nGiven integers n and m, construct a strictly increasing sequence of m positive integers with sum exactly equal to n such that the GCD of the sequence is maximized. If multiple sequences have the same maximum GCD, return the lexicographically smallest one. If not possible, return -1.Examples : Input 7 min read Largest Subset with GCD 1Given n integers, we need to find size of the largest subset with GCD equal to 1. Input Constraint : n <= 10^5, A[i] <= 10^5Examples: Input : A = {2, 3, 5}Output : 3Explanation: The largest subset with a GCD greater than 1 is {2, 3, 5}, and the GCD of all the elements in the subset is 3.Input 6 min read Summation of GCD of all the pairs up to nGiven a number n, find sum of all GCDs that can be formed by selecting all the pairs from 1 to n. Examples: Input : n = 4Output : 7Explanation: Numbers from 1 to 4 are: 1, 2, 3, 4Result = gcd(1,2) + gcd(1,3) + gcd(1,4) + gcd(2,3) + gcd(2,4) + gcd(3,4) = 1 + 1 + 1 + 1 + 2 + 1 = 7Input : n = 12Output 10 min read SeriesJuggler SequenceJuggler Sequence is a series of integer number in which the first term starts with a positive integer number a and the remaining terms are generated from the immediate previous term using the below recurrence relation : a_{k+1}=\begin{Bmatrix} \lfloor a_{k}^{1/2} \rfloor & for \quad even \quad a 5 min read Padovan SequencePadovan Sequence similar to Fibonacci sequence with similar recursive structure. The recursive formula is, P(n) = P(n-2) + P(n-3) P(0) = P(1) = P(2) = 1 Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...... Spiral of squares with side lengths which follow the Fibonacci sequence. Padovan Sequ 4 min read Aliquot SequenceGiven a number n, the task is to print its Aliquot Sequence. Aliquot Sequence of a number starts with itself, remaining terms of the sequence are sum of proper divisors of immediate previous term. For example, Aliquot Sequence for 10 is 10, 8, 7, 1, 0. The sequence may repeat. For example, for 6, we 8 min read Moser-de Bruijn SequenceGiven an integer 'n', print the first 'n' terms of the Moser-de Bruijn Sequence. Moser-de Bruijn sequence is the sequence obtained by adding up the distinct powers of the number 4 (For example, 1, 4, 16, 64, etc). Examples: Input : 5 Output : 0 1 4 5 16 Input : 10 Output : 0 1 4 5 16 17 20 21 64 65 12 min read Stern-Brocot SequenceStern Brocot sequence is similar to Fibonacci sequence but it is different in the way fibonacci sequence is generated . Generation of Stern Brocot sequence : 1. First and second element of the sequence is 1 and 1.2. Consider the second member of the sequence . Then, sum the considered member of the 5 min read Newman-Conway SequenceNewman-Conway Sequence is the one that generates the following integer sequence. 1 1 2 2 3 4 4 4 5 6 7 7... In mathematical terms, the sequence P(n) of Newman-Conway numbers is defined by the recurrence relation P(n) = P(P(n - 1)) + P(n - P(n - 1)) with seed values P(1) = 1 and P(2) = 1 Given a numb 6 min read Sylvester's sequenceIn number system, Sylvester's sequence is an integer sequence in which each member of the sequence is the product of the previous members, plus one. Given a positive integer N. The task is to print the first N member of the sequence. Since numbers can be very big, use %10^9 + 7.Examples: Input : N = 4 min read Recaman's sequenceGiven an integer n. Print first n elements of Recaman’s sequence. Recaman's Sequence starts with 0 as the first term. For each next term, calculate previous term - index (if positive and not already in sequence); otherwise, use previous term + index.Examples: Input: n = 6Output: 0, 1, 3, 6, 2, 7Expl 8 min read Sum of the sequence 2, 22, 222, .........Given an integer n. The task is to find the sum of the following sequence: 2, 22, 222, ......... to n terms. Examples : Input: n = 2Output: 24Explanation: For n = 2, the sum of first 2 terms are 2 + 22 = 24Input: 3Output: 246Explanation: For n = 3, the sum of first 3 terms are 2 + 22 + 222 = 246Usin 8 min read Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n - 1)^2Given a series 12 + 32 + 52 + 72 + . . . + (2*n - 1)2, find the sum of the series.Examples: Input: n = 4Output: 84Explanation: sum = 12 + 32 + 52 + 72 = 1 + 9 + 25 + 49 = 84Input: n = 10 Output: 1330Explanation: sum = 12 + 32 + 52 + 72 + 92 + 112 + 132 + 152 + 172 + 192 = 1 + 9 + 24 + 49 + . . . + 3 3 min read Sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n termsGiven the number of terms i.e. n. Find the sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n terms.Examples: Input: 2Output: 0.66Explanation: sum of the series upto 2 terms: 0.6 + 0.06 = 0.66.Input: 3Output: 0.666Explanation: sum of the series upto 3 terms: 0.6 + 0.06 + 0.006 = 0.666.Table of Cont 7 min read n-th term in series 2, 12, 36, 80, 150....Given a series 2, 12, 36, 80, 150.. Find the n-th term of the series.Examples : Input : 2 Output : 12 Input : 4 Output : 80 If we take a closer look, we can notice that series is sum of squares and cubes of natural numbers (1, 4, 9, 16, 25, .....) + (1, 8, 27, 64, 125, ....).Therefore n-th number of 3 min read Number DigitsMinimum digits to remove to make a number Perfect SquareGiven an integer n, we need to find how many digits remove from the number to make it a perfect square. Examples : Input : 8314 Output: 81 2 Explanation: If we remove 3 and 4 number becomes 81 which is a perfect square. Input : 57 Output : -1 The idea is to generate all possible subsequences and ret 9 min read Print first k digits of 1/n where n is a positive integerGiven a positive integer n, print first k digits after point in value of 1/n. Your program should avoid overflow and floating point arithmetic.Examples : Input: n = 3, k = 3 Output: 333 Input: n = 50, k = 4 Output: 0200 We strongly recommend to minimize the browser and try this yourself first.Let us 5 min read Check if a given number can be represented in given a no. of digits in any baseGiven a number and no. of digits to represent the number, find if the given number can be represented in given no. of digits in any base from 2 to 32.Examples : Input: 8 4 Output: Yes Possible in base 2 as 8 in base 2 is 1000 Input: 8 2 Output: Yes Possible in base 3 as 8 in base 3 is 22 Input: 8 3 12 min read Minimum Segments in Seven Segment DisplayA seven-segment display can be used to display numbers. Given an array of n natural numbers. The task is to find the number in the array that uses the minimum number of segments to display the number. If multiple numbers have a minimum number of segments, output the number having the smallest index. 6 min read Find next greater number with same set of digitsGiven a number N as string, find the smallest number that has same set of digits as N and is greater than N. If N is the greatest possible number with its set of digits, then print "Not Possible".Examples: Input: N = "218765"Output: "251678"Explanation: The next number greater than 218765 with same 9 min read Check if a number is jumbled or notWrite a program to check if a given integer is jumbled or not. A number is said to be Jumbled if for every digit, its neighbours digit differs by max 1. Examples : Input : 6765Output : TrueAll neighbour digits differ by atmost 1. Input : 1223Output : True Input : 1235Output : False Approach: Find th 6 min read Numbers having difference with digit sum more than sYou are given two positive integer value n and s. You have to find the total number of such integer from 1 to n such that the difference of integer and its digit sum is greater than given s.Examples : Input : n = 20, s = 5 Output :11 Explanation : Integer from 1 to 9 have diff(integer - digitSum) = 7 min read Total numbers with no repeated digits in a rangeGiven a range L, R find total such numbers in the given range such that they have no repeated digits. For example: 12 has no repeated digit. 22 has repeated digit. 102, 194 and 213 have no repeated digit. 212, 171 and 4004 have repeated digits. Examples: Input : 10 12 Output : 2 Explanation : In the 15+ min read K-th digit in 'a' raised to power 'b'Given three numbers a, b and k, find k-th digit in ab from right sideExamples: Input : a = 3, b = 3, k = 1Output : 7Explanation: 3^3 = 27 for k = 1. First digit is 7 in 27Input : a = 5, b = 2, k = 2Output : 2Explanation: 5^2 = 25 for k = 2. First digit is 2 in 25The approach is simple. Computes the 3 min read AlgebraProgram to add two polynomialsGiven two polynomials represented by two arrays, write a function that adds given two polynomials. Example: Input: A[] = {5, 0, 10, 6} B[] = {1, 2, 4} Output: sum[] = {6, 2, 14, 6} The first input array represents "5 + 0x^1 + 10x^2 + 6x^3" The second array represents "1 + 2x^1 + 4x^2" And Output is 15+ min read Multiply two polynomialsGiven two polynomials represented by two arrays, write a function that multiplies the given two polynomials. In this representation, each index of the array corresponds to the exponent of the variable(e.g. x), and the value at that index represents the coefficient of the term. For example, the array 15+ min read Find number of solutions of a linear equation of n variablesGiven a linear equation of n variables, find number of non-negative integer solutions of it. For example, let the given equation be "x + 2y = 5", solutions of this equation are "x = 1, y = 2", "x = 5, y = 0" and "x = 3, y = 1". It may be assumed that all coefficients in given equation are positive i 10 min read Calculate the Discriminant ValueIn algebra, Discriminant helps us deduce various properties of the roots of a polynomial or polynomial function without even computing them. Let's look at this general quadratic polynomial of degree two: ax^2+bx+c Here the discriminant of the equation is calculated using the formula: b^2-4ac Now we 5 min read Program for dot product and cross product of two vectorsThere are two vector A and B and we have to find the dot product and cross product of two vector array. Dot product is also known as scalar product and cross product also known as vector product.Dot Product - Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. 8 min read Iterated Logarithm log*(n)Iterated Logarithm or Log*(n) is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1. \log ^{*}n:=\begin{cases}0n\leq 1;\\1+\log ^{*}(\log n)n>1\end{cases} Applications: It is used in the analysis of algorithms (Refer Wiki for detail 4 min read Program to Find Correlation CoefficientThe correlation coefficient is a statistical measure that helps determine the strength and direction of the relationship between two variables. It quantifies how changes in one variable correspond to changes in another. This coefficient, sometimes referred to as the cross-correlation coefficient, al 8 min read Program for Muller MethodGiven a function f(x) on floating number x and three initial distinct guesses for root of the function, find the root of function. Here, f(x) can be an algebraic or transcendental function.Examples: Input : A function f(x) = x^3 + 2x^2 + 10x - 20 and three initial guesses - 0, 1 and 2 .Output : The 13 min read Number of non-negative integral solutions of a + b + c = nGiven a number n, find the number of ways in which we can add 3 non-negative integers so that their sum is n.Examples : Input : n = 1 Output : 3 There are three ways to get sum 1. (1, 0, 0), (0, 1, 0) and (0, 0, 1) Input : n = 2 Output : 6 There are six ways to get sum 2. (2, 0, 0), (0, 2, 0), (0, 0 7 min read Generate Pythagorean TripletsGiven a positive integer limit, your task is to find all possible Pythagorean Triplet (a, b, c), such that a <= b <= c <= limit.Note: A Pythagorean triplet is a set of three positive integers a, b, and c such that a2 + b2 = c2. Input: limit = 20Output: 3 4 5 5 12 13 6 8 10 8 15 17 9 12 15 1 14 min read Number SystemExponential notation of a decimal numberGiven a positive decimal number, find the simple exponential notation (x = a·10^b) of the given number. Examples: Input : 100.0 Output : 1E2 Explanation: The exponential notation of 100.0 is 1E2. Input :19 Output :1.9E1 Explanation: The exponential notation of 16 is 1.6E1. Approach: The simplest way 5 min read Check if a number is power of k using base changing methodThis program checks whether a number n can be expressed as power of k and if yes, then to what power should k be raised to make it n. Following example will clarify : Examples: Input : n = 16, k = 2 Output : yes : 4 Explanation : Answer is yes because 16 can be expressed as power of 2. Input : n = 2 8 min read Program to convert a binary number to hexadecimal numberGiven a Binary Number, the task is to convert the given binary number to its equivalent hexadecimal number. The input could be very large and may not fit even into an unsigned long long int.Examples: Input: 110001110Output: 18EInput: 1111001010010100001.010110110011011Output: 794A1.5B36 794A1D9B App 13 min read Program for decimal to hexadecimal conversionGiven a decimal number as input, we need to write a program to convert the given decimal number into an equivalent hexadecimal number. i.e. convert the number with base value 10 to base value 16.Hexadecimal numbers use 16 values to represent a number. Numbers from 0-9 are expressed by digits 0-9 and 8 min read Converting a Real Number (between 0 and 1) to Binary StringGiven a real number between 0 and 1 (e.g., 0.72) that is passed in as a double, print the binary representation. If the number cannot be represented accurately in binary with at most 32 characters, print" ERROR:' Examples: Input : (0.625)10 Output : (0.101)2 Input : (0.72)10 Output : ERROR Solution: 12 min read Convert from any base to decimal and vice versaGiven a number and its base, convert it to decimal. The base of number can be anything such that all digits can be represented using 0 to 9 and A to Z. The value of A is 10, the value of B is 11 and so on. Write a function to convert the number to decimal. Examples: Input number is given as string a 15+ min read Decimal to binary conversion without using arithmetic operatorsFind the binary equivalent of the given non-negative number n without using arithmetic operators. Examples: Input : n = 10Output : 1010 Input : n = 38Output : 100110 Note that + in below algorithm/program is used for concatenation purpose. Algorithm: decToBin(n) if n == 0 return "0" Declare bin = "" 8 min read Prime Numbers & Primality TestsPrime NumbersA prime number is a natural number greater than 1 that has exactly two positive divisors: 1 and itself. Numbers that have more than two divisors are called composite numbers The number 1 is neither a prime nor composite. All primes are odd, except for 2. A total of 25 prime numbers are there between 12 min read Left-Truncatable PrimeA Left-truncatable prime is a prime which in a given base (say 10) does not contain 0 and which remains prime when the leading ("left") digit is successively removed. For example, 317 is left-truncatable prime since 317, 17 and 7 are all prime. There are total 4260 left-truncatable primes.The task i 13 min read Program to Find All Mersenne Primes till NMersenne Prime is a prime number that is one less than a power of two. In other words, any prime is Mersenne Prime if it is of the form 2k-1 where k is an integer greater than or equal to 2. First few Mersenne Primes are 3, 7, 31 and 127.The task is print all Mersenne Primes smaller than an input po 9 min read Super PrimeGiven a positive integer n and the task is to print all the Super-Primes less than or equal to n. Super-prime numbers (also known as higher order primes) are the subsequence of prime number sequence that occupy prime-numbered positions within the sequence of all prime numbers. The first few super pr 6 min read Hardy-Ramanujan TheoremHardy Ramanujam theorem states that the number of prime factors of n will approximately be log(log(n)) for most natural numbers nExamples : 5192 has 2 distinct prime factors and log(log(5192)) = 2.1615 51242183 has 3 distinct prime facts and log(log(51242183)) = 2.8765 As the statement quotes, it is 6 min read Rosser's TheoremIn mathematics, Rosser's Theorem states that the nth prime number is greater than the product of n and natural logarithm of n for all n greater than 1. Mathematically, For n >= 1, if pn is the nth prime number, then pn > n * (ln n) Illustrative Examples: For n = 1, nth prime number = 2 2 > 15 min read Fermat's little theoremFermat's little theorem states that if p is a prime number, then for any integer a, the number a p - a is an integer multiple of p. Here p is a prime number ap ≡ a (mod p).Special Case: If a is not divisible by p, Fermat's little theorem is equivalent to the statement that a p-1-1 is an integer mult 8 min read 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: 10 min read Vantieghems Theorem for Primality TestVantieghems Theorem is a necessary and sufficient condition for a number to be prime. It states that for a natural number n to be prime, the product of 2^i - 1 where 0 < i < n , is congruent to n~(mod~(2^n - 1)) . In other words, a number n is prime if and only if.{\displaystyle \prod _{1\leq 4 min read AKS Primality TestThere are several primality test available to check whether the number is prime or not like Fermat's Theorem, Miller-Rabin Primality test and alot more. But problem with all of them is that they all are probabilistic in nature. So, here comes one another method i.e AKS primality test (Agrawal–Kayal– 11 min read Lucas Primality TestA number p greater than one is prime if and only if the only divisors of p are 1 and p. First few prime numbers are 2, 3, 5, 7, 11, 13, ...The Lucas test is a primality test for a natural number n, it can test primality of any kind of number.It follows from Fermat’s Little Theorem: If p is prime and 12 min read Prime Factorization & DivisorsPrint all prime factors of a given numberGiven a number n, the task is to find all prime factors of n.Examples:Input: n = 24Output: 2 2 2 3Explanation: The prime factorization of 24 is 23×3.Input: n = 13195Output: 5 7 13 29Explanation: The prime factorization of 13195 is 5×7×13×29.Approach:Every composite number has at least one prime fact 6 min read Smith NumberGiven a number n, the task is to find out whether this number is smith or not. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization. Examples: Input : n = 4Output : YesPrime factorization = 2, 2 and 2 + 2 = 4Therefore, 4 is a smith numberI 15 min read Sphenic NumberA Sphenic Number is a positive integer n which is a product of exactly three distinct primes. The first few sphenic numbers are 30, 42, 66, 70, 78, 102, 105, 110, 114, ... Given a number n, determine whether it is a Sphenic Number or not. Examples: Input: 30Output : YesExplanation: 30 is the smalles 8 min read Hoax NumberGiven a number 'n', check whether it is a hoax number or not. A Hoax Number is defined as a composite number, whose sum of digits is equal to the sum of digits of its distinct prime factors. It may be noted here that, 1 is not considered a prime number, hence it is not included in the sum of digits 13 min read k-th prime factor of a given numberGiven two numbers n and k, print k-th prime factor among all prime factors of n. For example, if the input number is 15 and k is 2, then output should be "5". And if the k is 3, then output should be "-1" (there are less than k prime factors). Examples: Input : n = 225, k = 2 Output : 3 Prime factor 15+ min read 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 14 min read Finding power of prime number p in n!Given a number 'n' and a prime number 'p'. We need to find out the power of 'p' in the prime factorization of n!Examples: Input : n = 4, p = 2 Output : 3 Power of 2 in the prime factorization of 2 in 4! = 24 is 3 Input : n = 24, p = 2 Output : 22 Naive approach The naive approach is to find the powe 8 min read Find all factors of a Positive NumberGiven a positive integer n, find all the distinct divisors of n.Examples:Input: n = 10 Output: [1, 2, 5, 10]Explanation: 1, 2, 5 and 10 are the divisors of 10. Input: n = 100Output: [1, 2, 4, 5, 10, 20, 25, 50, 100]Explanation: 1, 2, 4, 5, 10, 20, 25, 50 and 100 are divisors of 100.Table of Content[ 7 min read Find numbers with n-divisors in a given rangeGiven three integers a, b, n .Your task is to print number of numbers between a and b including them also which have n-divisors. A number is called n-divisor if it has total n divisors including 1 and itself. Examples: Input : a = 1, b = 7, n = 2 Output : 4 There are four numbers with 2 divisors in 14 min read Modular ArithmeticModular Exponentiation (Power in Modular Arithmetic)Given three integers x, n, and M, compute (x^n) % M (remainder when x raised to the power n is divided by M).Examples : Input: x = 3, n = 2, M = 4Output: 1Explanation: 32 % 4 = 9 % 4 = 1.Input: x = 2, n = 6, M = 10Output: 4Explanation: 26 % 10 = 64 % 10 = 4.Table of Content[Naive Approach] Repeated 6 min read Modular multiplicative inverseGiven two integers A and M, find the modular multiplicative inverse of A under modulo M.The modular multiplicative inverse is an integer X such that:A X ≡ 1 (mod M) Note: The value of X should be in the range {1, 2, ... M-1}, i.e., in the range of integer modulo M. ( Note that X cannot be 0 as A*0 m 15+ min read Modular DivisionGiven 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 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 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 9 min read 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 Exponential Squaring (Fast Modulo Multiplication)Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7 Examples: Input : base = 2, exp = 2Output : 4Input : base = 5, exp = 100000Output : 754573817In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the valu 12 min read Trick for modular division ( (x1 * x2 .... xn) / b ) mod (m)Given integers x1, x2, x3......xn, b, and m, we are supposed to find the result of ((x1*x2....xn)/b)mod(m). Example 1: Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007 Naive Method : Simply calculate the product (55*54*53*52*51)= say x,Divide x by 120 a 9 min read Like