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

• Difficulty Level : Easy
• Last Updated : 19 Aug, 2020

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `#include ` ` `  `// 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 ` `#include ` ` `  `// 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(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

 `// C program for the above approach ` `#include ` `#include ` ` `  `// 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.

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