Largest N digit number in Base B

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Given a positive integer N and base B, the task is to find the largest N-digit numbers of Base B in decimal form.

Examples:

Input: N = 2, B = 10
Output: 99

Input: N = 2, B = 5
Output: 24

Approach:
Since there are B digits in base B, so with these digits we can create B^N strings of length N. They represent the integers in range 0 to B^N – 1
Therefore, the largest N-digits number of base B in decimal form is given by B^N – 1.

Below is the implementation of the above approach:

C++

// C++ program for the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the largest
// N-digit numbers of base b
void findNumbers(int n, int b)
{
 
    // Find the largest N digit
    // number in base b using the
    // formula B^N - 1
    int largest = pow(b, n) - 1;
 
    // Print the largest number
    cout << largest << endl;
}
 
// Driver Code
int main()
{
    // Given Number and Base
    int N = 2, B = 5;
 
    // Function Call
    findNumbers(N, B);
 
    return 0;
}

Java

// Java program for the approach
import java.util.*;
class GFG{
     
// Function to print the largest
// N-digit numbers of base b
static void findNumbers(int n, int b)
{
 
    // Find the largest N digit
    // number in base b using the
    // formula B^N - 1
    double largest = Math.pow(b, n) - 1;
 
    // Print the largest number
    System.out.println(largest);
}
 
// Driver Code
public static void main(String []args)
{
    // Given Number and Base
    int N = 2, B = 5;
 
    // Function Call
    findNumbers(N, B);
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 program for the above approach
 
# Function to print the largest
# N-digit numbers of base b
def findNumbers(n, b):
     
    # Find the largest N digit
    # number in base b using the
    # formula B^N - 1
    largest = pow(b, n) - 1
 
    # Print the largest number
    print(largest)
 
# Driver Code
 
# Given number and base
N, B = 2, 5
 
# Function Call
findNumbers(N, B)
 
# This code is contributed by jrishabh99

C#

// C# program for the approach
using System;
class GFG{
     
// Function to print the largest
// N-digit numbers of base b
static void findNumbers(int n, int b)
{
 
    // Find the largest N digit
    // number in base b using the
    // formula B^N - 1
    double largest = Math.Pow(b, n) - 1;
 
    // Print the largest number
    Console.Write(largest);
}
 
// Driver Code
public static void Main(String []args)
{
    // Given Number and Base
    int N = 2, B = 5;
 
    // Function Call
    findNumbers(N, B);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

<script>
// javascript program for the approach
 
// Function to print the largest
// N-digit numbers of base b
function findNumbers( n,  b)
{
 
    // Find the largest N digit
    // number in base b using the
    // formula B^N - 1
    let largest = Math.pow(b, n) - 1;
 
    // Print the largest number
     document.write(largest);
}
 
// Driver Code
 
    // Given Number and Base
    let N = 2, B = 5;
 
    // Function Call
    findNumbers(N, B);
     
    // This code is contributed by aashish1995 
 
</script>

Output

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

METHOD 2:Using Math formula

APPROACH:

We can find the largest N-digit number in base B by simply calculating the sum of (B-1) multiplied by B to the power of i (where i ranges from 0 to N-1).

ALGORITHM:

1.Take the input values of N and B from the user.
2.Initialize a variable ‘result’ to 0.
3.Run a loop from 0 to N-1 and for each iteration, calculate the sum of (B-1) multiplied by B to the power of i and add it to ‘result’.
4.Display the value of ‘result’ as the largest N-digit number in base B.

C++

// C++ program for the approach
#include <iostream>
#include <cmath> // Include the cmath library to use pow() function.
 
int main() {
    int N = 2; // Number of digits in the number.
    int B = 10; // Base of the number system.
 
    int result = 0; // Initialize the variable to store the result.
 
    // Loop to calculate the largest N-digit number in base B.
    for (int i = 0; i < N; i++) {
        // Calculate (B^i) * (B - 1) and add it to the result.
        result += pow(B, i) * (B - 1);
    }
 
    // Print the result.
    std::cout << "The largest " << N << "-digit number in base " << B << " is " << result << std::endl;
 
    N = 2; // Reset N to 2.
    B = 5; // Change the base to 5.
 
    result = 0; // Reset the result variable.
 
    // Loop to calculate the largest N-digit number in base B.
    for (int i = 0; i < N; i++) {
        // Calculate (B^i) * (B - 1) and add it to the result.
        result += pow(B, i) * (B - 1);
    }
 
    // Print the result.
    std::cout << "The largest " << N << "-digit number in base " << B << " is " << result << std::endl;
 
    return 0;
}

Java

import java.lang.Math;
 
public class Main {
    public static void main(String[] args) {
        int N = 2; // Number of digits in the number.
        int B = 10; // Base of the number system.
 
        int result = 0; // Initialize the variable to store the result.
 
        // Loop to calculate the largest N-digit number in base B.
        for (int i = 0; i < N; i++) {
            // Calculate (B^i) * (B - 1) and add it to the result.
            result += Math.pow(B, i) * (B - 1);
        }
 
        // Print the result.
        System.out.println(
          "The largest " + N + "-digit number in base " + B + " is " + result);
 
        N = 2; // Reset N to 2.
        B = 5; // Change the base to 5.
 
        result = 0; // Reset the result variable.
 
        // Loop to calculate the largest N-digit number in base B.
        for (int i = 0; i < N; i++) {
            // Calculate (B^i) * (B - 1) and add it to the result.
            result += Math.pow(B, i) * (B - 1);
        }
 
        // Print the result.
        System.out.println(
          "The largest " + N + "-digit number in base " + B + " is " + result);
    }
}

Python3

N = int(2)
B = int(10)
 
result = 0
for i in range(N):
    result += (B ** i) * (B - 1)
 
print(f"The largest {N}-digit number in base {B} is {result}")
  
N = int(2)
B = int(5)
 
result = 0
for i in range(N):
    result += (B ** i) * (B - 1)
 
print(f"The largest {N}-digit number in base {B} is {result}")

C#

using System;
 
class GFG
{
    static void Main()
    {
        int N = 2; // Number of digits in the number.
        int B = 10; // Base of the number system.
 
        int result = 0; // Initialize the variable to store the result.
 
        // Loop to calculate the largest N-digit number in base B.
        for (int i = 0; i < N; i++)
        {
            // Calculate (B^i) * (B - 1) and add it to the result.
            result += (int)Math.Pow(B, i) * (B - 1);
        }
 
        // Print the result.
        Console.WriteLine("The largest " + N + "-digit number in base " + B + " is " + result);
 
        N = 2; // Reset N to 2.
        B = 5; // Change the base to 5.
 
        result = 0; // Reset the result variable.
 
        // Loop to calculate the largest N-digit number in base B.
        for (int i = 0; i < N; i++)
        {
            // Calculate (B^i) * (B - 1) and add it to the result.
            result += (int)Math.Pow(B, i) * (B - 1);
        }
 
        // Print the result.
        Console.WriteLine("The largest " + N + "-digit number in base " + B + " is " + result);
    }
}

Javascript

// JavaScript program to calculate the largest N-digit number in base B
 
// Function to calculate the largest N-digit number in base B
function calculateLargestNDigitNumber(N, B) {
    let result = 0;
 
    // Loop through each digit position from 0 to N-1
    for (let i = 0; i < N; i++) {
        // Calculate the value of the digit position by multiplying B to the power of i
        // and then subtracting 1 (since the largest digit in base B is B-1)
        result += Math.pow(B, i) * (B - 1);
    }
 
    // Return the result, which is the largest N-digit number in base B
    return result;
}
 
// Test case 1: N = 2, B = 10 (decimal)
let N = 2;
let B = 10;
let result = calculateLargestNDigitNumber(N, B);
console.log(`The largest ${N}-digit number in base ${B} is ${result}`);
 
// Test case 2: N = 2, B = 5 (quinary)
N = 2;
B = 5;
result = calculateLargestNDigitNumber(N, B);
console.log(`The largest ${N}-digit number in base ${B} is ${result}`);

Output

The largest 2-digit number in base 10 is 99
The largest 2-digit number in base 5 is 24

Time complexity: O(N)
Auxiliary Space: O(1)

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