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Find the total count of numbers up to N digits in a given base B

  • Last Updated : 05 Apr, 2021

Given two integers N and B, the task is to find the count of natural numbers of Base B up to N digits.

Examples:  

Input: N = 2, B = 10 
Output: 99 
Explanation: 
1, 2, 3, 4, 5, 6, 7, 8, 9 are 1 digit Natural numbers of Base 10. 
10, 11, 12………99 are 2 digit Natural numbers of Base 10 
So, total = 9 + 90 = 99

Input: N = 2, B = 16 
Output: 255 
Explanation: 
There are a total of 240 two digit hexadecimal numbers and 15 one digit hexadecimal numbers. 
Therefore, 240 + 15 = 255. 
 

Approach: On observing carefully the count of numbers with N digits in base B is a geometric progression formed with the first term being (B – 1) and a common ratio of B.
Therefore,  



Nth term = Number of natutal numbers of N digits in Base B = (B – 1) * BN – 1 
 

Finally, count of all natural numbers in Base B up to N digits can be found out by iterating a loop from 1 to N and calculating the sum of ith term using the above formula. 

Below is the implementation of the above approach: 

C++




// C++ implementation to find the count
// of natural numbers upto N digits
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to return the count of
// natural numbers upto N digits
int count(int N, int B)
{
    int sum = 0;
 
    // Loop to iterate from 1 to N
    // and calculating number of
    // natural numbers for every 'i'th digit.
    for (int i = 1; i <= N; i++) {
        sum += (B - 1) * pow(B, i - 1);
    }
    return sum;
}
 
// Driver Code
int main()
{
    int N = 2, B = 10;
    cout << count(N, B);
 
    return 0;
}

Java




// Java implementation to find the count
// of natural numbers upto N digits
 
class GFG{
 
// Function to return the count of
// natural numbers upto N digits
static int count(int N, int B)
{
    int sum = 0;
 
    // Loop to iterate from 1 to N
    // and calculating number of
    // natural numbers for every 'i'th digit.
    for (int i = 1; i <= N; i++){
        sum += (B - 1) * Math.pow(B, i - 1);
    }
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, B = 10;
    System.out.print(count(N, B));
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 implementation to find the count
# of natural numbers up to N digits
 
from math import pow
 
# Function to return the count of
# natural numbers upto N digits
def count(N, B):
    sum = 0
 
    # Loop to iterate from 1 to N
    # and calculating number of
    # natural numbers for every 'i'th digit.
    for i in range(1, N+1):
        sum += (B - 1) * pow(B, i - 1)
    return sum
 
# Driver Code
if __name__ == '__main__':
    N = 2
    B = 10
    print(int(count(N, B)))
 
# This code is contributed by Bhupendra_Singh

C#




// C# implementation to find the count
// of natural numbers upto N digits
using System;
using System.Collections.Generic;
class GFG{
 
// Function to return the count of
// natural numbers upto N digits
static int count(int N, int B)
{
    int sum = 0;
 
    // Loop to iterate from 1 to N
    // and calculating number of
    // natural numbers for every
    // 'i'th digit.
    for(int i = 1; i <= N; i++)
    {
       sum += (int)((B - 1) * Math.Pow(B, i - 1));
    }
    return sum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 2, B = 10;
     
    Console.Write(count(N, B));
}
}
 
// This code is contributed by amal kumar choubey

Javascript




<script>
 
// Javascript implementation to find the count
// of natural numbers upto N digits
 
// Function to return the count of
// natural numbers upto N digits
function count(N, B)
{
    var sum = 0;
 
    // Loop to iterate from 1 to N and
    // calculating number of natural
    // numbers for every 'i'th digit.
    for(var i = 1; i <= N; i++)
    {
        sum += (B - 1) * Math.pow(B, i - 1);
    }
    return sum;
}
 
// Driver code
var N = 2, B = 10;
 
document.write(count(N, B));
 
// This code is contributed by Ankita saini
    
</script>
Output: 
99

 

Time Complexity: O(N)
 

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