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 Cubevoid 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 Codeint 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 functionvoid 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 Codeint 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 searchvoid 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 Codeint 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.

My Personal Notes arrow_drop_up