Check if given number is perfect cube (original) (raw)

Last Updated : 12 Jul, 2025

Given a number **N, the task is to check whether the given number **N is a perfect cube or not.
**Examples:

**Input: N = 216
**Output: Yes
**Explanation:
As 216 = 6*6*6. Therefore the cube root of 216 is 6.

**Input: N = 100
**Output: No

Table of Content

Naive Approach:
Using Build-in Functions:
Using Prime Factors :
Using Binary Search:
Using Mathematical Fact:

Naive Approach:

The idea is to check for each number from **1 to **N if the cube of any of these numbers equals **N. If so, then that number is the cube root of **N and the N is a perfect cube.

Below is the implementation of the above approach:

C++ `

// C++ program to check whether the given // number N is the perfect cube or not #include <bits/stdc++.h> using namespace std;

// Function to check if a number // is a perfect Cube or not bool perfectCube(int n) { int cube;

// Iterate from 1-N
for (int i; i <= n; i++)
{
    // Find the cube of
    // every number
    cube = i * i * i;

    // Check if cube equals
    // N or not
    if (cube == n)
        return true;        
    else if (cube > n)
        return false;
}

}

// Driver code int main() { int n = 216; perfectCube(n)? cout << "Yes" : cout << "No "; return 0; }

Java

public class GFG {

// Function to check if a number
// is a perfect Cube or not
public static boolean perfectCube(int n) {
    int cube;

    // Iterate from 1-N
    for (int i = 1; i <= n; i++) {
        // Find the cube of
        // every number
        cube = i * i * i;

        // Check if cube equals
        // N or not
        if (cube == n)
            return true;
        else if (cube > n)
            return false;
    }
    return false;
}

// Driver code
public static void main(String[] args) {
    int n = 216;
    System.out.println(perfectCube(n) ? "Yes" : "No");
}

}

Python

Function to check if a number

is a perfect Cube or not

def perfect_cube(n): # Iterate from 1-N for i in range(1, n+1): # Find the cube of # every number cube = i ** 3

    # Check if cube equals
    # N or not
    if cube == n:
        return True
    elif cube > n:
        return False
return False

Driver code

n = 216 print("Yes" if perfect_cube(n) else "No")

C#

using System;

class GFG {

// Function to check if a number
// is a perfect Cube or not
static bool PerfectCube(int n) {
    int cube;

    // Iterate from 1-N
    for (int i = 1; i <= n; i++) {
        // Find the cube of
        // every number
        cube = i * i * i;

        // Check if cube equals
        // N or not
        if (cube == n)
            return true;
        else if (cube > n)
            return false;
    }
    return false;
}

// Driver code
static void Main() {
    int n = 216;
    Console.WriteLine(PerfectCube(n) ? "Yes" : "No");
}

}

JavaScript

// Function to check if a number // is a perfect Cube or not function perfectCube(n) { let cube;

// Iterate from 1-N
for (let i = 1; i <= n; i++) {
    // Find the cube of
    // every number
    cube = i * i * i;

    // Check if cube equals
    // N or not
    if (cube === n)
        return true;
    else if (cube > n)
        return false;
}
return false;

}

// Driver code let n = 216; console.log(perfectCube(n) ? "Yes" : "No");

**Time Complexity: O(N)
**Auxiliary Space: O(1)

Using Build-in Functions:

The idea is to use the inbuilt function (**cbrt()) to find the cube root of a number which returns floor value of the cube root of the number **N. If the cube of this number equals **N, then **N is a perfect cube otherwise **N is not a perfect cube.

Below is the implementation of the above approach:

C++ `

#include <bits/stdc++.h> using namespace std;

// Function to check if a number is // a perfect Cube using inbuilt function bool perfectCube(int n) { int x = round(cbrt(n));

// If cube of cube_root is equals to n,
// then print Yes else print No
return (x * x * x == n);

}

// Driver's code int main() { int n = 125; perfectCube(n)? cout << "Yes" : cout << "No"; return 0; }

Java

public class GFG {

// Function to check if a number is
// a perfect Cube using inbuilt function
public static boolean perfectCube(int n) {
    int x = (int) Math.round(Math.cbrt(n));

    // If cube of cube_root is equals to n,
    // then print Yes else print No
    return (x * x * x == n);
}

// Driver's code
public static void main(String[] args) {
    int n = 125;
    System.out.println(perfectCube(n) ? "Yes" : "No");
}

}

Python

Function to check if a number is

a perfect Cube using inbuilt function

def perfect_cube(n): x = round(n ** (1/3))

# If cube of cube_root is equals to n,
# then print Yes else print No
return (x * x * x == n)

Driver's code

n = 125 print("Yes" if perfect_cube(n) else "No")

C#

using System;

class GFG {

// Function to check if a number is
// a perfect Cube using inbuilt function
static bool PerfectCube(int n) {
    int x = (int) Math.Round(Math.Cbrt(n));

    // If cube of cube_root is equals to n,
    // then print Yes else print No
    return (x * x * x == n);
}

// Driver's code
static void Main() {
    int n = 125;
    Console.WriteLine(PerfectCube(n) ? "Yes" : "No");
}

}

JavaScript

// Function to check if a number is // a perfect Cube using inbuilt function function perfectCube(n) { let x = Math.round(Math.cbrt(n));

// If cube of cube_root is equals to n,
// then print Yes else print No
return (x * x * x === n);

}

// Driver's code let n = 125; console.log(perfectCube(n) ? "Yes" : "No");

**Time Complexity: O(log N) because using inbuilt cbrt function
**Auxiliary Space: O(1)

Using Prime Factors :

Find all the Prime Factors of the given number **N using the approach in this article.

Store the frequency of all the prime factors obtained above in a Hash Map.
Traverse the Hash Map and if the frequency of every prime factors is not a multiple of **3, then the given number **N is not a perfect cube.

Below is the implementation of the above approach:

C++ `

// C++ program to check if a number // is a perfect cube using prime factors #include<bits/stdc++.h> using namespace std;

// Function to check if a // number is perfect cube bool perfectCube (int n) { // Find all prime factors and fill their // frequencies in a map unordered_map<int, int> mp; while (n % 2 == 0) { mp[2]++; n /= 2; } for(int i = 3; i <= sqrt(n); i += 2) { while (n % i == 0) { mp[i]++; n /= i; } } if (n > 2) mp[n]++;

// Check if all frequencies are multiples of 3
for(auto x : mp)
    if (x.second % 3 != 0)
        return false;
return true;

}

// Driver Code int main() { int n = 216; perfectCube(n)? cout << "True" : cout << "False"; return 0; }

Java

import java.util.*;

public class GFG {

// Function to check if a 
// number is perfect cube
public static boolean perfectCube(int n) {
    Map<Integer, Integer> mp = new HashMap<>();
    while (n % 2 == 0) {
        mp.put(2, mp.getOrDefault(2, 0) + 1);
        n /= 2;
    }
    for (int i = 3; i <= Math.sqrt(n); i += 2) {
        while (n % i == 0) {
            mp.put(i, mp.getOrDefault(i, 0) + 1);
            n /= i;
        }
    }
    if (n > 2)
        mp.put(n, mp.getOrDefault(n, 0) + 1);

    // Check if all frequencies are multiples of 3
    for (Map.Entry<Integer, Integer> entry : mp.entrySet())
        if (entry.getValue() % 3 != 0)
            return false;
    return true;
}

// Driver Code
public static void main(String[] args) {
    int n = 216;
    System.out.println(perfectCube(n) ? "True" : "False");
}

}

Python

Function to check if a

number is perfect cube

def perfect_cube(n): from collections import defaultdict import math

mp = defaultdict(int)
while n % 2 == 0:
    mp[2] += 1
    n //= 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
    while n % i == 0:
        mp[i] += 1
        n //= i
if n > 2:
    mp[n] += 1

# Check if all frequencies are multiples of 3
for freq in mp.values():
    if freq % 3 != 0:
        return False
return True

Driver Code

n = 216 print("True" if perfect_cube(n) else "False")

C#

using System; using System.Collections.Generic;

class GFG {

// Function to check if a 
// number is perfect cube
static bool PerfectCube(int n) {
    Dictionary<int, int> mp = new Dictionary<int, int>();
    while (n % 2 == 0) {
        if (!mp.ContainsKey(2))
            mp[2] = 0;
        mp[2]++;
        n /= 2;
    }
    for (int i = 3; i <= Math.Sqrt(n); i += 2) {
        while (n % i == 0) {
            if (!mp.ContainsKey(i))
                mp[i] = 0;
            mp[i]++;
            n /= i;
        }
    }
    if (n > 2) {
        if (!mp.ContainsKey(n))
            mp[n] = 0;
        mp[n]++;
    }

    // Check if all frequencies are multiples of 3
    foreach (var x in mp)
        if (x.Value % 3 != 0)
            return false;
    return true;
}

// Driver Code
static void Main() {
    int n = 216;
    Console.WriteLine(PerfectCube(n) ? "True" : "False");
}

}

JavaScript

// Function to check if a // number is perfect cube function perfectCube(n) { const mp = {}; while (n % 2 === 0) { mp[2] = (mp[2] || 0) + 1; n /= 2; } for (let i = 3; i <= Math.sqrt(n); i += 2) { while (n % i === 0) { mp[i] = (mp[i] || 0) + 1; n /= i; } } if (n > 2) mp[n] = (mp[n] || 0) + 1;

// Check if all frequencies are multiples of 3
for (const key in mp)
    if (mp[key] % 3 !== 0)
        return false;
return true;

}

// Driver Code const n = 216; console.log(perfectCube(n) ? "True" : "False");

Using Binary Search:

The idea is to use the Binary Search technique where we will be searching in a range of [1, N] iteratively by dividing the range halve every time until we get the resultant output.

Initialize two variables, left and right, to 0 and the given number, respectively.
While left is less than or equal to right follow the below steps.
Compute the mid point and cube value.
If cube is equal to N then return N, else change the left and right pointers accordingly.
If no perfect cube is found in the above steps, return false, indicating that the given number is not a perfect cube. C++ `

#include using namespace std;

bool isPerfectCube(int num) { long left = 0; long right = num; while (left <= right) { long mid = left + (right - left) / 2; long cube = mid * mid * mid; if (cube == num) { return true; } else if (cube < num) { left = mid + 1; } else { right = mid - 1; } } return false; }

int main() { if (isPerfectCube(729)) cout << "Yes" << endl; else cout << "No" << endl;

return 0;

}

Java

public class GFG {

// Function to check if a number is a perfect cube
public static boolean isPerfectCube(int num) {
    long left = 0;
    long right = num;
    while (left <= right) {
        long mid = left + (right - left) / 2;
        long cube = mid * mid * mid;
        if (cube == num) {
            return true;
        } else if (cube < num) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

public static void main(String[] args) {
    if (isPerfectCube(729))
        System.out.println("Yes");
    else
        System.out.println("No");
}

}

Python

Function to check if a number is a perfect cube

def is_perfect_cube(num): left = 0 right = num while left <= right: mid = left + (right - left) // 2 cube = mid * mid * mid if cube == num: return True elif cube < num: left = mid + 1 else: right = mid - 1 return False

if is_perfect_cube(729): print("Yes") else: print("No")

C#

using System;

class GFG {

// Function to check if a number is a perfect cube
public static bool IsPerfectCube(int num) {
    long left = 0;
    long right = num;
    while (left <= right) {
        long mid = left + (right - left) / 2;
        long cube = mid * mid * mid;
        if (cube == num) {
            return true;
        } else if (cube < num) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

public static void Main() {
    if (IsPerfectCube(729))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

}

JavaScript

// Function to check if a number is a perfect cube function isPerfectCube(num) { let left = 0; let right = num; while (left <= right) { let mid = left + Math.floor((right - left) / 2); let cube = mid * mid * mid; if (cube === num) { return true; } else if (cube < num) { left = mid + 1; } else { right = mid - 1; } } return false; }

if (isPerfectCube(729)) { console.log("Yes"); } else { console.log("No"); }

**Time Complexity : O(log N)
**Auxiliary Space : O(1)

**Using Mathematical Fact:

Every perfect cube can n (say n = x^3) can be written as sum of x consecutive odd numbers.

1^3 = 1 = 1
2^3 = 8 = 3 + 5
3^3 = 27 = 7 + 9 + 11
4^3 = 64 = 13 + 15 + 17 + 19
5^3 = 125 = 21 + 23 + 25 + 27 + 29
6^3 = 216 = 31 + 33 + 35 + 37 + 39 + 41

If you take a closer look, you can notice that we consider different set of consecutive odd numbers for every cube.

C++ `

#include using namespace std;

bool isPerfectCube(int n) { int sum = 0; int odd = 1; for (int count = 1; sum < n; count++) { sum = 0;

    // Check if current value of count is 
    // cube root of n
    for (int i = 0; i < count; i++) {
        sum += odd;
        odd += 2;
    }
    if (sum == n) return true;
}
return false;

}

int main() { for (int n = 1; n <= 100; n++) { if (isPerfectCube(n)) { cout << n << " "; } }

return 0;

}

Java

public class Main { public static boolean isPerfectCube(int n) { int sum = 0; int odd = 1; for (int count = 1; sum < n; count++) { sum = 0;

        // Check if current value of count is 
        // cube root of n
        for (int i = 0; i < count; i++) {
            sum += odd;
            odd += 2;
        }
        if (sum == n) return true;
    }
    return false;
}

public static void main(String[] args) {
    for (int n = 1; n <= 100; n++) {
        if (isPerfectCube(n)) {
            System.out.print(n + " ");
        }
    }
}

}

Python

def isPerfectCube(n): sum = 0 odd = 1 count = 1 while sum < n: sum = 0

    # Check if current value of count is 
    # cube root of n
    for i in range(count):
        sum += odd
        odd += 2
    
    if sum == n:
        return True
    count += 1
return False

def main(): for n in range(1, 101): if isPerfectCube(n): print(n, end=" ")

main()

C#

using System;

class Program { public static bool IsPerfectCube(int n) { int sum = 0; int odd = 1; for (int count = 1; sum < n; count++) { sum = 0;

        // Check if current value of count is 
        // cube root of n
        for (int i = 0; i < count; i++) {
            sum += odd;
            odd += 2;
        }
        if (sum == n) return true;
    }
    return false;
}

static void Main() {
    for (int n = 1; n <= 100; n++) {
        if (IsPerfectCube(n)) {
            Console.Write(n + " ");
        }
    }
}

}

JavaScript

function isPerfectCube(n) { let sum = 0; let odd = 1; for (let count = 1; sum < n; count++) { sum = 0;

    // Check if current value of count is 
    // cube root of n
    for (let i = 0; i < count; i++) {
        sum += odd;
        odd += 2;
    }
    if (sum === n) return true;
}
return false;

}

function main() { for (let n = 1; n <= 100; n++) { if (isPerfectCube(n)) { console.log(n + " "); } } }

main();

**How does this work?

Let n be perfect square and its square root be x.

We can write x^3 = n

We can also say the following
(x^2) * x = n^3
(x^2 + x^2 + x^2 .......... x-times) = n^3

**If x is an odd number:
Examples:
x = 3, we can write the above as ((x^2 - 1) + x^2 + (x^2 + 1)) = n [Sum of 3 Consecutive Odd numbers]
For example say x = 5, we can write it as ((x^2 - 2) + (x^2 - 1) + x^2 + (x^2 + 1) + (x^2 + 2)) = n [Sum of 5 Consecutive Odd Numbers]
In General, ((x^2 - i + 1) + (x^2 - i + 2) ........ + x^2 + .......... (x^2 + i - 1) + (x^2 + i - 2)) where i is floor(x/2)

**If x is an even number:
Examples:
x = 2, we can write the above as ((x^2 - 1) + (x^2 + 1)) = n [Sum of 2 Consecutive Odd numbers]
For example say x = 4, we can write it as ((x^2 - 2) + (x^2 - 1) + (x^2 + 1) + (x^2 + 2)) = n [Sum of 4 Consecutive Odd Numbers]
In General, ((x^2 - i) + (x^2 - i + 1) ........ (x^2 + i ) + (x^2 + i)) where i is x/2

**Time Complexity : If we take a closer look, we mainly generate odd numbers 1 to a number. The number of times the innermost statement is executed is equal to the number of odd numbers generated. For example of n = 125, we generate all numbers from 1 to 29. If n = x^3, then the maximum odd numbers generated is close to x^2 + x (as explained above). We can upper bound the total iterations by O(Log n). We have explained how is the time complexity of checking if a number is perfect square is O(Log n) using the mathematical approach in the last approach discussed here.

Check if given number is perfect cube (original) (raw)

Naive Approach:

Function to check if a number

is a perfect Cube or not

Driver code

Using Build-in Functions:

Function to check if a number is

a perfect Cube using inbuilt function

Driver's code

**Using **Prime Factors :

Function to check if a

number is perfect cube

Driver Code

Using Binary Search:

Function to check if a number is a perfect cube

**Using Mathematical Fact:

Using Prime Factors :