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 p^x (p raised to power x) divides n!.

Examples:

Input: n = 7, p = 3
Output: x = 2
Explanation: 3² divides 7! and 2 is the largest such power of 3.

Input: n = 10, p = 3
Output: x = 4
Explanation: 3⁴ divides 10! and 4 is the largest such power of 3.

Using Legendre's Formula - Iterative Approach

We know that n! is the product of all the integers from 1 to n. For each multiple of p in the range [1, n], we know that we get at least one factor of p. Therefore in n!, there are at least floor(n!/p) integers divisible by p. Furthermore, for each multiple of p², we get one more factor of p. Similarly, for each multiple of p³, we get another extra factor of p. So, the largest value of x is the sum of total number of factors, that is floor(n/p) + floor(n/p²) + floor(n/p³) ... and so on till the floor value reaches 0.

Legendre's Formula:
Largest exponent of p in n! = Sum of (floor(n/(pⁱ)), where i = 1, 2, 3, 4..... and so on.

4

Time complexity: O(log_pn), as in each iteration we are reducing n to n/p.
Auxiliary Space: O(1)

Using Legendre's Formula - Recursive Approach

Since we are repeatedly counting the factors of p and keep reducing the value of n, we can also use recursion to solve the problem. The recurrence relation will be:

largestPower(n, p) = n/p + largestPower(n/p, p)

The base case will be when n = 0, we can return the count of factors as 0.

4

Time complexity: O(log_pn), as in each recursive call we are reducing n to n/p.
Auxiliary Space: O(log_pn), as we are using recursive call stack.