Given two numbers N and M(M > 1), the task is to find the highest power of M that divides N.
Examples:
Input: N = 12, M = 2
Output: 2
Explanation: The powers of 2 which divide 12 are 1 and 2 (21 = 2 and 22 = 4 which both divide 12).
The higher power is 2, hence consider 2.Input: N = 500, M = 5
Output: 3.
Naive and Bit Manipulation Approach: The naive approach and bit manipulation approach is already mentioned in the Set 1 of this problem.
Efficient Approach: The task can be solved using a binary search technique over the range [1, logB(A)]. For each value x in the range, check if Mx divides N. Finally, return the largest value possible
Follow the below steps to solve the problem:
- Find the value of logM(N)
- Run binary search in range [1, logM(N)] .
- For each value x, check if Mx divides N, and find the largest such value possible.
Below is the implementation of the above approach.
// C++ program to find the Highest // Power of M that divides N #include <bits/stdc++.h> using namespace std;
// Function to find any log(N) base M int log_a_to_base_b( int a, int b)
{ return log (a) / log (b);
} // Function to find the Highest Power // of M which divides N int HighestPower( int N, int M)
{ int start = 0, end = log_a_to_base_b(N, M);
int ans = 0;
while (start <= end) {
int mid = start + (end - start) / 2;
int temp = ( int )( pow (M, mid));
if (N % temp == 0) {
ans = mid;
start = mid + 1;
}
else {
end = mid - 1;
}
}
return ans;
} // Driver code int main()
{ int N = 12;
int M = 2;
cout << HighestPower(N, M) << endl;
return 0;
} |
// Java program to find the Highest // Power of M that divides N import java.util.*;
public class GFG
{ // Function to find any log(N) base M
static int log_a_to_base_b( int a, int b)
{
return ( int )(Math.log(a) / Math.log(b));
}
// Function to find the Highest Power
// of M which divides N
static int HighestPower( int N, int M)
{
int start = 0 , end = log_a_to_base_b(N, M);
int ans = 0 ;
while (start <= end) {
int mid = start + (end - start) / 2 ;
int temp = ( int )(Math.pow(M, mid));
if (N % temp == 0 ) {
ans = mid;
start = mid + 1 ;
}
else {
end = mid - 1 ;
}
}
return ans;
}
// Driver code
public static void main(String args[])
{
int N = 12 ;
int M = 2 ;
System.out.println(HighestPower(N, M));
}
} // This code is contributed by Samim Hossain Mondal. |
# Python program to find the Highest # Power of M that divides N import math
# Function to find any log(N) base M def log_a_to_base_b(a, b):
return math.log(a) / math.log(b)
# Function to find the Highest Power # of M which divides N def HighestPower(N, M):
start = 0
end = log_a_to_base_b(N, M)
ans = 0
while (start < = end):
mid = math.floor(start + (end - start) / 2 )
temp = math. pow (M, mid)
if (N % temp = = 0 ):
ans = mid
start = mid + 1
else :
end = mid - 1
return ans
# Driver code N = 12
M = 2
print (HighestPower(N, M))
# This code is contributed by Samim Hossain Mondal. |
// C# program for the above approach using System;
class GFG
{ // Function to find any log(N) base M
static int log_a_to_base_b( int a, int b)
{
return ( int )(Math.Log(a) / Math.Log(b));
}
// Function to find the Highest Power
// of M which divides N
static int HighestPower( int N, int M)
{
int start = 0, end = log_a_to_base_b(N, M);
int ans = 0;
while (start <= end) {
int mid = start + (end - start) / 2;
int temp = ( int )(Math.Pow(M, mid));
if (N % temp == 0) {
ans = mid;
start = mid + 1;
}
else {
end = mid - 1;
}
}
return ans;
}
// Driver code
public static void Main()
{
int N = 12;
int M = 2;
// Function call
Console.Write(HighestPower(N, M));
}
} // This code is contributed by Samim Hossain Mondal. |
<script> // JavaScript code for the above approach
// Function to find any log(N) base M
function log_a_to_base_b(a, b) {
return Math.log(a) / Math.log(b);
}
// Function to find the Highest Power
// of M which divides N
function HighestPower(N, M)
{
let start = 0, end = log_a_to_base_b(N, M);
let ans = 0;
while (start <= end) {
let mid = start + Math.floor((end - start) / 2);
let temp = (Math.pow(M, mid));
if (N % temp == 0) {
ans = mid;
start = mid + 1;
}
else {
end = mid - 1;
}
}
return ans;
}
// Driver code
let N = 12;
let M = 2;
document.write(HighestPower(N, M) + '<br>' );
// This code is contributed by Potta Lokesh
</script>
|
2
Time Complexity: O(log(logM(N)))
Auxiliary Space: O(1)