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# Fifth root of a number

• Difficulty Level : Medium
• Last Updated : 19 Oct, 2021

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.

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## C++14

 `// A C++ program to find floor of 5th root``#include``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

 ``

## Javascript

 ``

Output:

`Floor of 5'th root is 3`

Time complexity of above solution is O(n1/5). 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``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

 ``

## Javascript

 ``

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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