Count Squares

Problem

Consider a sample space S consisting of all perfect squares starting from 1, 4, 9 and so on. You are given a number N, you have to output the number of integers less than N in the sample space S.

Examples:

Example 1:

Input :
N = 9
Output:
2
Explanation:
1 and 4 are the only Perfect Squares less than 9. So, the Output is 2.

Example 2:

Input :
N = 3
Output:
1
Explanation:
1 is the only Perfect Square less than 3. So, the Output is 1.

Your task:

You don't need to read input or print anything. Your task is to complete the function countSquares() which takes an Integer N as input and returns the answer.

• Expected Time Complexity: $O(sqrt(N))$
• Expected Auxiliary Space: $O(1)$

Constraints:

• $1<=N<=10^8$

Solution

Python

def countSquares(self, N):
    return (math.floor(math.sqrt(N-1)) - math.ceil(math.sqrt(1)) + 1)

Java

static int countSquares(int N) {
    return (int)(Math.floor(Math.sqrt(N - 1)) - Math.ceil(Math.sqrt(1)) + 1);
}

C++

int countSquares(int N) {
    return (floor(sqrt(N - 1)) - ceil(sqrt(1)) + 1);
}

C

int countSquares(int N) {
    return (int)(floor(sqrt(N - 1)) - ceil(sqrt(1)) + 1);
}

• Time Complexity: $O(sqrt(N))$
• Auxiliary Space: $O(1)$