C Program to check whether a number is a Perfect Cube or not

Given a number N, the task is to write C program to check if the given number is 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

Method 1: Naive Approach To find the cube root of the given number iterate over all the natural numbers from 1 till N and check if cube of any number in this range is equal to the given number N then print Yes else print No Below is the implementation of the above approach: 

C

// C program for the above approach
#include <math.h>
#include <stdio.h>

// Function to check if a number is
// a perfect Cube
void perfectCube(int N)
{
    for (int i = 1; i < N; i++) {

        // If cube of i is equals to N
        // then print Yes and return
        if (i * i * i == N) {
            printf("Yes");
            return;
        }
    }

    // No number was found whose cube
    // is equal to N
    printf("No");
    return;
}

// Driver Code
int main()
{
    // Given Number
    int N = 216;

    // Function Call
    perfectCube(N);
    return 0;
}

Output

Yes

Complexity Analysis:

• Time Complexity: O(N), only one traversal of the solution is needed, so the time complexity is O(N).
• Auxiliary Space: O(1). Constant extra space is needed.

Method 2: Using inbuilt function The idea is to use the inbuilt function pow() 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

// C program for the above approach
#include <math.h>
#include <stdio.h>

// Function to check if a number is
// perfect cube using inbuilt function
void perfectCube(int N)
{
    int cube_root;
    cube_root = (int)round(pow(N, 1.0 / 3.0));

    // If cube of cube_root is equals
    // to N, then print Yes else No
    if (cube_root * cube_root * cube_root == N) {
        printf("Yes");
        return;
    }
    else {
        printf("No");
        return;
    }
}

// Driver Code
int main()
{
    // Given number N
    int N = 216;

    // Function Call
    perfectCube(N);
    return 0;
}

Output:

Yes

Complexity Analysis:

• Time Complexity: O(1), since here the pow() function takes constant time, since second argument to pow function is constant
• Auxiliary Space: O(1). Constant extra space is needed.

Method 3: Using Binary Search The idea is to use Binary Search to solve the problem. The values of i * i * i is monotonically increasing, so the problem can be solved using binary search.
Below are the steps:

1. Initialise low and high as 0 and N respectively.
2. Iterate until low ? high and do the following:
   • Find the value of mid as = (low + high)/2.
   • Check if mid*mid*mid is equals to N then print “Yes”.
   • If the cube of mid is less than N then search for a larger value in the second half of search space by updating low to mid + 1.
   • If the cube of mid is greater than N then search for a smaller value in the first half of search space by updating high to mid – 1.
3. If cube of N is not obtained in the above step then print “No”.

Below is the implementation of the above approach: 

C

// C program for the above approach
#include <math.h>
#include <stdio.h>

// Function to check if a number is
// a perfect Cube using binary search
void perfectCube(int N)
{
    int start = 1, end = N;
    while (start <= end) {

        // Calculating mid
        int mid = (start + end) / 2;

        // If N is a perfect cube
        if (mid * mid * mid == N) {
            printf("Yes");
            return;
        }

        // If mid^3 is smaller than N,
        // then move closer to cube
        // root of N
        if (mid * mid * mid < N) {
            start = mid + 1;
        }

        // If mid^3 is greater than N
        else
            end = mid - 1;
    }

    printf("No");
    return;
}

// Driver Code
int main()
{
    // Given Number N
    int N = 216;

    // Function Call
    perfectCube(N);
    return 0;
}

Output

Yes

Complexity Analysis:

• Time Complexity: O(log N). The time complexity of binary search is O(log N).
• Auxiliary Space: O(1). Constant extra space is needed.

Iterative Approach

• Initialize a variable “i” to 1.
• While “i” raised to the power of 3 is less than or equal to the given number N, increment “i”.
• If “i” raised to the power of 3 is equal to N, then the number is a perfect cube, else it is not.

C

#include <stdio.h>

int main() {
    int N = 216;
    int i = 1;
    while (i*i*i <= N) {
        if (i*i*i == N) {
            printf("Yes\n");
            return 0;
        }
        i++;
    }
    printf("No\n");
    return 0;
}

Output

Yes

Time Complexity: O(N^(1/3)).
Auxiliary Space: O(1).