All possible numbers of N digits and base B without leading zeros

Given a number of digits ‘N’ and base ‘B’, the task is to count all the ‘N’ digit numbers without leading zeros that are in base ‘B’

Examples:

Input: N = 2, B = 2
Output: 2
All possible numbers without 
leading zeros are 10 and 11.

Input: N = 5, B = 8
Output: 28672

Approach:

• If the base is ‘B’ then every digit of the number can take any value within the range [0, B-1].
• So, B‘N’ digit numbers are possible with base ‘B’ (including the numbers with leading zeros).
• And, if we fix the first digit as ‘0’ then the rest of the ‘N-1’ digits can form a total of Bnumbers.
• So, the total number of ‘N’ digit numbers with base ‘B’ possible without leading zeros are B– B.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// function to count
// all permutations
void countPermutations(int N, int B)
{
    // count of
    // all permutations
    int x = pow(B, N);

    // count of permutations
    // with leading zeros
    int y = pow(B, N - 1);

    // Return the permutations
    // without leading zeros
    cout << x - y << "\n";
}

// Driver code
int main()
{

    int N = 6;
    int B = 4;

    countPermutations(N, B);

    return 0;
}

Java

// Java implementation of the approach

class GFG
{
// function to count
// all permutations
static void countPermutations(int N, int B)
{
    // count of
    // all permutations
    int x = (int)Math.pow(B, N);

    // count of permutations
    // with leading zeros
    int y = (int)Math.pow(B, N - 1);

    // Return the permutations
    // without leading zeros
    System.out.println(x - y);
}

// Driver code
public static void main(String[] args)
{
    int N = 6;
    int B = 4;

    countPermutations(N, B);
}
}

// This code is contributed by mits

Python3

# Python3 implementation of the approach

# function to count all permutations
def countPermutations(N, B):

    # count of all permutations
    x = B ** N

    # count of permutations
    # with leading zeros
    y = B ** (N - 1)

    # Return the permutations
    # without leading zeros
    print(x - y)

# Driver code
if __name__ == "__main__":

    N, B = 6, 4
    countPermutations(N, B)

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach

using System;
class GFG
{
// function to count
// all permutations
static void countPermutations(int N, int B)
{
    // count of
    // all permutations
    int x = (int)Math.Pow(B, N);

    // count of permutations
    // with leading zeros
    int y = (int)Math.Pow(B, N - 1);

    // Return the permutations
    // without leading zeros
    Console.WriteLine(x - y);
}

// Driver code
public static void Main()
{
    int N = 6;
    int B = 4;

    countPermutations(N, B);
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP implementation of the approach

// function to count all permutations
function countPermutations($N, $B)
{
    // count of all permutations
    $x = pow($B, $N);

    // count of permutations
    // with leading zeros
    $y = pow($B, $N - 1);

    // Return the permutations
    // without leading zeros
    echo ($x - $y), "\n";
}

// Driver code
$N = 6;
$B = 4;

countPermutations($N, $B);

// This code is contributed
// by Sach_Code`
?>

Javascript

<script>

// Javascript implementation of the approach

// function to count
// all permutations
function countPermutations(N, B)
{
    // count of
    // all permutations
    var x = Math.pow(B, N);

    // count of permutations
    // with leading zeros
    var y = Math.pow(B, N - 1);

    // Return the permutations
    // without leading zeros
    document.write( x - y );
}

// Driver code
var N = 6;
var B = 4;
countPermutations(N, B);

</script>

Output:

3072

Time Complexity: O(logn), since pow function takes logn time to find the power of a number to base n.

Auxiliary Space: O(1), since no extra space has been taken.