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:

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// 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;
}

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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:

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// 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;
}

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Output:

Yes

Complexity Analysis:

  • Time Complexity: O(N), since the pow() function works in O(N), so the time complexity is O(N).
  • 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

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// 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;
}

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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.

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