Smallest number whose square has N digits

Given a number N, the task is to find the smallest number whose square has N digits.
Examples: 
 

Input: N = 2 
Output: 4 
Explanation: 
3² = 9, which has 1 digit. 
4² = 16, which has 2 digits. 
Hence, 4 is the smallest number whose square has N digits.
Input: N = 3 
Output: 10 
Explanation: 
10² = 100, which has 3 digits. 
 

Naive Approach: The simplest approach to solve the problem is to calculate the square of each number starting from and count the number of digits in its square. Print the first number whose square is obtained to be of N digits. 
Time Complexity: O(?(10^N))
Efficient Approach: To solve the problem, we need to make the following observations: 
 

The smallest number whose square has 1 digit = 1 
The smallest number whose square has 2 digits = 4 
The smallest number whose square has 3 digits = 10 
The smallest number whose square has 4 digits = 32 
The smallest number whose square has 5 digits = 100
Hence, these numbers form a series 1, 4, 10, 32, 100, 317, ....... 
 

Now, we need to find a formula for the N^th term of the series. 
The terms of the series can be expressed in the following form: 
 

If N = 1, Smallest number possible is 1. 
T(1) = \lceil 10^{\frac{1-1}{2}}\rceil = 1   
If N = 2, Smallest number possible is 41. 
T(2) = \lceil 10^{\frac{2-1}{2}}\rceil = 4   
If N = 3, Smallest number possible is 10. 
T(3) = \lceil 10^{\frac{3 -1}{2}}\rceil = 10   
 

Hence, we can conclude that the N^th of the series can be expressed as 
 

T(N) = \lceil 10^{\frac{N-1}{2}}\rceil   
 

Hence, in order to solve the problem, we just need to calculate ceil(10^{(N - 1)/ 2}) for the given integer N.
Below is the implementation of the above approach: 
 

// C++ Program to find the smallest
// number whose square has N digits

#include <bits/stdc++.h>
using namespace std;

// Function to return smallest number
// whose square has N digits
int smallestNum(int N)
{

    // Calculate N-th term of the series
    float x = pow(10.0, (N - 1) / 2.0);
    return ceil(x);
}

// Driver Code
int main()
{
    int N = 4;
    cout << smallestNum(N);

    return 0;
}

// Java program for above approach 
class GFG{ 

// Function to return smallest number
// whose square has N digits
static int smallestNum(int N)
{
 
    // Calculate N-th term of the series
    float x = (float)(Math.pow(10, (N - 1) / 2.0));
    return (int)(Math.ceil(x));
}

// Driver code 
public static void main(String[] args) 
{ 
    int N = 4;
    System.out.print(smallestNum(N)); 
} 
} 

// This code is contributed by spp 

# Python3 Program to find the smallest
# number whose square has N digits
import math;

# Function to return smallest number
# whose square has N digits
def smallestNum(N):

    # Calculate N-th term of the series
    x = pow(10.0, (N - 1) / 2.0);
    return math.ceil(x);

# Driver Code
N = 4;
print(smallestNum(N));

# This code is contributed by Code_mech

// C# program for above approach 
using System;
class GFG{ 

// Function to return smallest number
// whose square has N digits
static int smallestNum(int N)
{

    // Calculate N-th term of the series
    float x = (float)(Math.Pow(10, (N - 1) / 2.0));
    return (int)(Math.Ceiling(x));
}

// Driver code 
public static void Main() 
{ 
    int N = 4;
    Console.Write(smallestNum(N)); 
} 
} 

// This code is contributed by Code_Mech

<script>
// java script  Program to find the smallest
// number whose square has N digits
// Function to return smallest number
// whose square has N digits
function smallestNum(N)
{

    // Calculate N-th term of the series
    x = Math.pow(10.0, (N - 1) / 2.0);
    return Math.ceil(x);
}

// Driver Code
let N = 4;
document.write(smallestNum(N));
    
// This code is contributed by Gottumukkala Bobby
</script>

32

Time Complexity: O(log(N)) 
Auxiliary Space: O(1)