Given two numbers n and m, find the n-th root of m. In mathematics, the n-th root of a number m is a real number that, when raised to the power of n, gives m. If no such real number exists, return -1.
Examples:
Input: n = 2, m = 9
Output: 3
Explanation: 32 = 9
Input: n = 3, m = 9
Output: -1
Explanation: 3rd root of 9 is not integer.
Approach - Using Binary Search
The approach uses binary search to find the n-th root of a number m. The search range is between 1 and m, and in each iteration, the middle value mid is raised to the power of n.
- If mid^n equals m, mid is the root.
- If mid^n is greater than m, the search range is narrowed to smaller values, and
- if mid^n is smaller than m, the search range is adjusted to larger values.
- The process continues until the root is found or the range is exhausted. If no exact root is found, the result is -1
C++
#include <iostream>
using namespace std;
int nthRoot(int n, int m) {
if (n == 1) {
return m;
}
long long int low = 1LL, high = m, mid;
long long int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
long long int x = mid;
for (int i = 1; i < n; i++) {
x *= mid;
if (x > m * 1LL)
break;
}
if (x == m * 1LL) {
ans = mid;
break;
} else if (x > m)
high = mid - 1;
else
low = mid + 1;
}
return int(ans);
}
int main() {
int n = 1, m = 14;
int result = nthRoot(n, m);
cout << result << endl;
return 0;
}
#include <stdio.h>
int nthRoot(int n, int m) {
if (n == 1) {
return m;
}
long long int low = 1LL, high = m, mid;
long long int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
long long int x = mid;
for (int i = 1; i < n; i++) {
x *= mid;
if (x > m * 1LL)
break;
}
if (x == m * 1LL) {
ans = mid;
break;
} else if (x > m)
high = mid - 1;
else
low = mid + 1;
}
return (int)ans;
}
int main() {
int n = 1, m = 14;
int result = nthRoot(n, m);
printf("%d\n", result);
return 0;
}
public class GfG {
public static int nthRoot(int n, int m) {
if (n == 1) {
return m;
}
long low = 1, high = m, mid;
long ans = -1;
while (low <= high) {
mid = (low + high) / 2;
long x = mid;
for (int i = 1; i < n; i++) {
x *= mid;
if (x > m)
break;
}
if (x == m) {
ans = mid;
break;
} else if (x > m)
high = mid - 1;
else
low = mid + 1;
}
return (int)ans;
}
public static void main(String[] args) {
int n = 1, m = 14;
int result = nthRoot(n, m);
System.out.println(result);
}
}
def nthRoot(n, m):
if n == 1:
return m
low, high = 1, m
ans = -1
while low <= high:
mid = (low + high) // 2
x = mid
for i in range(1, n):
x *= mid
if x > m:
break
if x == m:
ans = mid
break
elif x > m:
high = mid - 1
else:
low = mid + 1
return int(ans)
n = 1
m = 14
result = nthRoot(n, m)
print(result)
using System;
class GfG{
static int NthRoot(int n, int m) {
if (n == 1) {
return m;
}
long low = 1, high = m, mid;
long ans = -1;
while (low <= high) {
mid = (low + high) / 2;
long x = mid;
for (int i = 1; i < n; i++) {
x *= mid;
if (x > m)
break;
}
if (x == m) {
ans = mid;
break;
} else if (x > m)
high = mid - 1;
else
low = mid + 1;
}
return (int)ans;
}
static void Main() {
int n = 1, m = 14;
int result = NthRoot(n, m);
Console.WriteLine(result);
}
}
function nthRoot(n, m) {
if (n === 1) {
return m;
}
let low = 1, high = m, mid;
let ans = -1;
while (low <= high) {
mid = Math.floor((low + high) / 2);
let x = mid;
for (let i = 1; i < n; i++) {
x *= mid;
if (x > m) break;
}
if (x === m) {
ans = mid;
break;
} else if (x > m)
high = mid - 1;
else
low = mid + 1;
}
return ans;
}
const n = 1, m = 14;
const result = nthRoot(n, m);
console.log(result);
Time Complexity: O(n*log m)
Auxiliary Space: O(1)
Approach - Using Newton's Method
This problem involves finding the real-valued function A^{1/N}, which can be solved using Newton's method. The method begins with an initial guess and iteratively refines the result.
By applying Newton's method, we derive a relation between consecutive iterations:
According to Newton's method:
\frac{f(x_k)}{f'(x_k)}
Here, we define the function f(x) as:
f(x) = xN- A
The derivative of f(x), denoted f'(x), is:
N\cdot x^{N - 1}
The value xk represents the approximation at the k-th iteration. By substituting f(x) and f'(x) into the Newton's method formula, we obtain the following update relation:
\frac{1}{N} \left( (N - 1) \cdot x_k + \frac{A}{x_k^{N - 1}} \right)
Using this relation, the problem can be solved by iterating over successive values of x until the difference between consecutive iterations is smaller than the desired accuracy.
C++
#include <bits/stdc++.h>
using namespace std;
int nthRoot(int m, int n)
{
if (n == 1) {
return m;
}
double xPre = rand() % 10 + 1;
// Smaller eps denotes more accuracy
double eps = 1e-3;
// Initializing difference between two roots
double delX = INT_MAX;
// xK denotes current value of x
double xK;
// Loop until we reach the desired accuracy
while (delX > eps)
{
// Calculating the current value from the previous value by Newton's method
xK = ((n - 1.0) * xPre + (double)m / pow(xPre, n - 1)) / (double)n;
delX = abs(xK - xPre);
xPre = xK;
}
return (int)round(xK);
}
int main() {
int n = 1, m = 14;
int nthRootValue = nthRoot(m, n);
cout << nthRootValue << endl;
return 0;
}
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
int nthRoot(int m, int n) {
if (n == 1) {
return m;
}
double xPre = rand() % 10 + 1;
// Smaller eps denotes more accuracy
double eps = 1e-3;
// Initializing difference between two roots
double delX = 1e6;
// xK denotes current value of x
double xK;
// Loop until we reach the desired accuracy
while (delX > eps) {
// Calculating the current value from the previous value by Newton's method
xK = ((n - 1.0) * xPre + (double)m / pow(xPre, n - 1)) / (double)n;
delX = fabs(xK - xPre);
xPre = xK;
}
return (int)round(xK);
}
int main() {
int n = 1, m = 14;
int nthRootValue = nthRoot(m, n);
printf("%d\n", nthRootValue);
return 0;
}
import java.lang.Math;
import java.util.Random;
public class GfG {
public static int nthRoot(int m, int n) {
if (n == 1) {
return m;
}
Random rand = new Random();
// Initial guess (use m / 2 or another logic to get a reasonable starting point)
double xPre = rand.nextInt(10) + 1;
// Smaller eps denotes more accuracy
double eps = 1e-3;
// Initializing difference between two roots
double delX = Double.MAX_VALUE;
// xK denotes current value of x
double xK = xPre;
// Loop until we reach the desired accuracy
while (delX > eps) {
// Calculating the current value from the previous value by Newton's method
xK = ((n - 1.0) * xPre + (double)m / Math.pow(xPre, n - 1)) / (double)n;
delX = Math.abs(xK - xPre);
xPre = xK;
}
return (int)Math.round(xK);
}
public static void main(String[] args) {
int n = 1, m = 14;
int nthRootValue = nthRoot(m, n);
System.out.println(nthRootValue);
}
}
import random
import math
def nthRoot(m, n):
if n == 1:
return m
xPre = random.randint(1, 10)
# Smaller eps denotes more accuracy
eps = 1e-3
# Initializing difference between two roots
delX = float('inf')
# xK denotes current value of x
xK = 0.0
# Loop until we reach the desired accuracy
while delX > eps:
# Calculating the current value from the previous value by Newton's method
xK = ((n - 1.0) * xPre + m / (xPre ** (n - 1))) / n
delX = abs(xK - xPre)
xPre = xK
return round(xK)
if __name__ == '__main__':
n = 1
m = 14
nthRootValue = nthRoot(m, n)
print(nthRootValue)
using System;
class GfG {
public static int nthRoot(int m, int n) {
if (n == 1) {
return m;
}
Random rand = new Random();
double xPre = rand.Next(1, 11);
// Smaller eps denotes more accuracy
double eps = 1e-3;
// Initializing difference between two roots
double delX = double.MaxValue;
// xK denotes current value of x, initialize with xPre
double xK = xPre;
// Loop until we reach the desired accuracy
while (delX > eps) {
// Calculating the current value from the previous value by Newton's method
xK = ((n - 1.0) * xPre + (double)m / Math.Pow(xPre, n - 1)) / (double)n;
delX = Math.Abs(xK - xPre);
xPre = xK;
}
// Return the integer part of the result (rounded to nearest integer)
return (int)Math.Round(xK);
}
// Driver code to test the nthRoot method
public static void Main() {
int n = 1, m = 14;
int nthRootValue = nthRoot(m, n);
Console.WriteLine(nthRootValue);
}
}
function nthRoot(m, n) {
if (n === 1) {
return m;
}
let xPre = Math.floor(Math.random() * 10) + 1;
// Smaller eps denotes more accuracy
const eps = 1e-3;
// Initializing difference between two roots
let delX = Infinity;
// xK denotes current value of x
let xK;
// Loop until we reach the desired accuracy
while (delX > eps) {
// Calculating the current value from the previous value by Newton's method
xK = ((n - 1) * xPre + m / Math.pow(xPre, n - 1)) / n;
delX = Math.abs(xK - xPre);
xPre = xK;
}
// Return the integer part of the result
return Math.round(xK);
}
const n = 1, m = 14;
const nthRootValue = nthRoot(m, n);
console.log(nthRootValue);
Time Complexity: O(log(eps)), where eps is the desired accuracy.
Space Complexity: O(1)
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