Fifth root of a number

Given a number, print floor of 5’th root of the number.
Examples:

Input  : n = 32
Output : 2
2 raise to power 5 is 32

Input  : n = 250
Output : 3
Fifth square root of 250 is between 3 and 4
So floor value is 3.

Method 1 (Simple)
A simple solution is initialize result as 0, keep incrementing result while result⁵ is smaller than or equal to n. Finally return result – 1. 

C++14

// A C++ program to find floor of 5th root
#include<bits/stdc++.h>
using namespace std;

// Returns floor of 5th root of n
int floorRoot5(int n)
{
    // Base cases
    if (n == 0 || n == 1)
        return n;

    // Initialize result
    int res = 0;

    // Keep incrementing res while res^5 is
    // smaller than or equal to n
    while (res*res*res*res*res <= n)
        res++;

    // Return floor of 5'th root
    return res-1;
}

// Driver program
int main()
{
    int n = 250;
    cout << "Floor of 5'th root is "
         << floorRoot5(n);
    return 0;
}

Java

// Java program to find floor of 5th root

class GFG {

// Returns floor of 5th root of n
static int floorRoot5(int n)
{

    // Base cases
    if (n == 0 || n == 1)
        return n;

    // Initialize result
    int res = 0;

    // Keep incrementing res while res^5
    // is smaller than or equal to n
    while (res * res * res * res * res <= n)
        res++;

    // Return floor of 5'th root
    return res-1;
}

    // Driver Code
    public static void main(String []args)
    {
        int n = 250;
        System.out.println("Floor of 5'th root is "
                            + floorRoot5(n));
    }
}

// This code is contributed by Anshul Aggarwal.

Python3

# A Python3 program to find the floor
# of the 5th root

# Returns floor of 5th root of n
def floorRoot5(n):

    # Base cases
    if n == 0 and n == 1:
        return n

    # Initialize result
    res = 0

    # Keep incrementing res while res^5
    # is smaller than or equal to n
    while res * res * res * res * res <= n:
        res += 1

    # Return floor of 5'th root
    return res-1

# Driver Code
if __name__ == "__main__":

    n = 250
    print("Floor of 5'th root is",
                    floorRoot5(n))

# This code is contributed by Rituraj Jain

C#

// C# program to find floor of 5th root
using System;

class GFG {

// Returns floor of 5th root of n
static int floorRoot5(int n)
{

    // Base cases
    if (n == 0 || n == 1)
        return n;

    // Initialize result
    int res = 0;

    // Keep incrementing res while res^5
    // is smaller than or equal to n
    while (res * res * res * res * res <= n)
        res++;

    // Return floor of 5'th root
    return res-1;
}

    // Driver Code
    public static void Main()
    {
        int n = 250;
        Console.Write("Floor of 5'th root is "
                       + floorRoot5(n));
    }
}

// This code is contributed by Sumit Sudhakar.

PHP

<?php
// PHP program to find
// floor of 5th root

// Returns floor of
// 5th root of n
function floorRoot5($n)
{

    // Base cases
    if ($n == 0 || $n == 1)
        return $n;

    // Initialize result
    $res = 0;

    // Keep incrementing res while
    // res^5 is smaller than or
    // equal to n
    while ($res * $res * $res *
           $res * $res <= $n)
        $res++;

    // Return floor
    // of 5'th root
    return $res - 1;
}

    // Driver Code
    $n = 250;
    echo "Floor of 5'th root is "
                , floorRoot5($n);

// This code is contributed by nitin mittal.
?>

Javascript

<script>

// JavaScript program to find floor of 5th root

// Returns floor of 5th root of n
function floorRoot5(n)
{

    // Base cases
    if (n == 0 || n == 1)
        return n;

    // Initialize result
    let res = 0;

    // Keep incrementing res while res^5
    // is smaller than or equal to n
    while (res * res * res * res * res <= n)
        res++;

    // Return floor of 5'th root
    return res-1;
}

// Driver Code

        let n = 250;
       document.write("Floor of 5'th root is "
                            + floorRoot5(n));

</script>

Output:

Floor of 5'th root is 3

Time complexity: O(n^1/5)

Auxiliary Space: O(1)

We can do better. See below solution. 
Method 2 (Binary Search)
The idea is to do Binary Search. We start from n/2 and if its 5’th power is more than n, we recur for interval from n/2+1 to n. Else if power is less, we recur for interval 0 to n/2-1 

C++

// A C++ program to find floor of 5'th root
#include<bits/stdc++.h>
using namespace std;

// Returns floor of 5'th root of n
int floorRoot5(int n)
{
    // Base cases
    if (n == 0 || n == 1)
       return n;

    // Do Binary Search for floor of 5th square root
    int low = 1, high = n, ans = 0;
    while (low <= high)
    {
        // Find the middle point and its power 5
        int mid = (low + high) / 2;
        long int mid5 = mid*mid*mid*mid*mid;

        // If mid is the required root
        if (mid5 == n)
            return mid;

        // Since we need floor, we update answer when
        // mid5 is smaller than n, and move closer to
        // 5'th root
        if (mid5 < n)
        {
            low = mid + 1;
            ans = mid;
        }
        else // If mid^5 is greater than n
            high = mid - 1;
    }
    return ans;
}

// Driver program
int main()
{
    int n = 250;
    cout << "Floor of 5'th root is "
         << floorRoot5(n);
    return 0;
}

Java

// A Java program to find
// floor of 5'th root

class GFG {

    // Returns floor of 5'th
    // root of n
    static int floorRoot5(int n)
    {

        // Base cases
        if (n == 0 || n == 1)
        return n;

        // Do Binary Search for
        // floor of 5th square root
        int low = 1, high = n, ans = 0;
        while (low <= high)
        {

            // Find the middle point
            // and its power 5
            int mid = (low + high) / 2;
            long mid5 = mid * mid * mid *
                            mid * mid;

            // If mid is the required root
            if (mid5 == n)
                return mid;

            // Since we need floor,
            // we update answer when
            // mid5 is smaller than n,
            // and move closer to
            // 5'th root
            if (mid5 < n)
            {
                low = mid + 1;
                ans = mid;
            }

            // If mid^5 is greater
            // than n
            else
                high = mid - 1;
        }
        return ans;
    }

    // Driver Code
    public static void main(String []args)
    {
        int n = 250;
        System.out.println("Floor of 5'th root is " +
                                     floorRoot5(n));
    }
}

// This code is contributed by Anshul Aggarwal.

Python3

# A Python3 program to find the floor
# of 5'th root

# Returns floor of 5'th root of n
def floorRoot5(n):

    # Base cases
    if n == 0 or n == 1:
        return n

    # Do Binary Search for floor of
    # 5th square root
    low, high, ans = 1, n, 0
    while low <= high:

        # Find the middle point and its power 5
        mid = (low + high) // 2
        mid5 = mid * mid * mid * mid * mid

        # If mid is the required root
        if mid5 == n:
            return mid

        # Since we need floor, we update answer
        # when mid5 is smaller than n, and move
        # closer to 5'th root
        if mid5 < n:

            low = mid + 1
            ans = mid

        else: # If mid^5 is greater than n
            high = mid - 1

    return ans

# Driver Code
if __name__ == "__main__":

    n = 250
    print("Floor of 5'th root is", floorRoot5(n))

# This code is contributed by Rituraj Jain

C#

// A C# program to find
// floor of 5'th root
using System;

class GFG {

    // Returns floor of 5'th
    // root of n
    static int floorRoot5(int n)
    {

        // Base cases
        if (n == 0 || n == 1)
        return n;

        // Do Binary Search for
        // floor of 5th square root
        int low = 1, high = n, ans = 0;
        while (low <= high)
        {

            // Find the middle point
            // and its power 5
            int mid = (low + high) / 2;
            long mid5 = mid * mid * mid *
                            mid * mid;

            // If mid is the required root
            if (mid5 == n)
                return mid;

            // Since we need floor,
            // we update answer when
            // mid5 is smaller than n,
            // and move closer to
            // 5'th root
            if (mid5 < n)
            {
                low = mid + 1;
                ans = mid;
            }

            // If mid^5 is greater
            // than n
            else
                high = mid - 1;
        }
        return ans;
    }

    // Driver Code
    public static void Main(String []args)
    {
        int n = 250;
        Console.WriteLine("Floor of 5'th root is " +
                                     floorRoot5(n));
    }
}

// This code is contributed by Anshul Aggarwal.

PHP

<?php
// A PHP program to find floor of 5'th root

// Returns floor of 5'th root of n
function floorRoot5($n)
{
    // Base cases
    if ($n == 0 || $n == 1)
    return $n;

    // Do Binary Search for floor of 5th square root
    $low = 1;
    $high = $n;
    $ans = 0;
    while ($low <= $high)
    {
        // Find the middle point and its power 5
        $mid = (int)(($low + $high) / 2);
        $mid5 = $mid*$mid*$mid*$mid*$mid;

        // If mid is the required root
        if ($mid5 == $n)
            return $mid;

        // Since we need floor, we update answer when
        // mid5 is smaller than n, and move closer to
        // 5'th root
        if ($mid5 < $n)
        {
            $low = $mid + 1;
            $ans = $mid;
        }
        else // If mid^5 is greater than n
            $high = $mid - 1;
    }
    return $ans;
}

    // Driver code
    $n = 250;
    echo "Floor of 5'th root is ".floorRoot5($n);

// This code is contributed by mits
?>

Javascript

<script>

// A javascript program to find
// floor of 5'th root
// Returns floor of 5'th
// root of n
function floorRoot5(n)
{

    // Base cases
    if (n == 0 || n == 1)
    return n;

    // Do Binary Search for
    // floor of 5th square root
    var low = 1, high = n, ans = 0;
    while (low <= high)
    {

        // Find the middle point
        // and its power 5
        var mid = parseInt((low + high) / 2);
        var mid5 = mid * mid * mid *
                        mid * mid;

        // If mid is the required root
        if (mid5 == n)
            return mid;

        // Since we need floor,
        // we update answer when
        // mid5 is smaller than n,
        // and move closer to
        // 5'th root
        if (mid5 < n)
        {
            low = mid + 1;
            ans = mid;
        }

        // If mid^5 is greater
        // than n
        else
            high = mid - 1;
    }
    return ans;
}

// Driver Code
var n = 250;
document.write("Floor of 5'th root is " +
                             floorRoot5(n));

// This code contributed by Princi Singh

</script>

Output:

Floor of 5'th root is 3

Time Complexity: O(logN)
Auxiliary Space: O(1) 
We can also use Newton Raphson Method to find exact root. See this for implementation.
Source : http://qa.geeksforgeeks.org/7487/program-calculate-fifth-without-using-mathematical-operators
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