Merge pull request codeharborhub#3682 from Ishitamukherjee2004/new-branch

Nth Root of M of gfg problem is added

Author: ajay-dhangar

dsa-solutions/gfg-solutions/Easy problems/Nth-Number-Of-Root.md

@@ -0,0 +1,136 @@
+---
+id: nth-root-of-m
+title: Nth root of M
+sidebar_label: Nth-root-of-M
+tags:
+  - Mathematical
+  - Algorithm
+description: "This tutorial covers the solution to the Nth root of M problem from the GeeksforGeeks."
+---
+## Problem Description
+You are given 2 numbers `(n, m)`; the task is to find `n√m` (`nth` root of `m`).
+
+
+## Examples
+
+**Example 1:**
+
+```
+Input: n = 2, m = 9
+Output: 3
+Explanation: 3^2 = 9
+```
+
+**Example 2:**
+
+```
+Input: n = 3, m = 9
+Output: -1
+Explanation: 3rd root of 9 is not
+integer.
+```
+
+## Your Task
+
+You don't need to read or print anyhting. Your task is to complete the function NthRoot() which takes n and m as input parameter and returns the nth root of m. If the root is not integer then returns -1.
+
+
+## Constraints
+
+* `1 <= n <= 30`
+* `1 <= m <= 10^9`
+
+## Problem Explanation
+You are given 2 numbers (n , m); the task is to find n√m (nth root of m).
+
+## Code Implementation
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@arunimad6yuq"/>
+
+  ```py
+ import math
+
+n = 3
+m = 27
+result = m ** (1/n)
+print(result)
+
+  ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@YourUsername"/>
+
+  ```cpp
+ #include <cmath>
+#include <iostream>
+
+int main() {
+    int n = 3;
+    int m = 27;
+    double result = pow(m, 1.0 / n);
+    std::cout << result << std::endl;
+    return 0;
+}
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Javascript" label="Javascript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```javascript
+
+function nthRoot(n, m) {
+    return Math.pow(m, 1/n);
+}
+
+
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Typescript" label="Typescript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```typescript
+function nthRoot(n: number, m: number): number {
+    return Math.pow(m, 1/n);
+}
+
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```java
+public class Main {
+    public static void main(String[] args) {
+        int n = 3;
+        int m = 27;
+        double result = Math.pow(m, 1.0 / n);
+        System.out.println(result);
+    }
+}
+
+
+  ```
+
+  </TabItem>
+</Tabs>
+
+
+## Time Complexity
+
+* The iterative approach has a time complexity of $O(1)$.
+
+## Space Complexity
+
+* The space complexity is $O(1)$ since we are using only a fixed amount of extra space.