Given a number A in decimal base, the task is to find the Nth digit from last in base B
Examples:
Input: A = 100, N = 3, B = 4
Output: 2
Explanation:
(100)4 = 1210
3rd digit from last is 2
Input: A = 50, N = 3, B = 5
Output: 2
Explanation:
(50)5 = 200
3rd digit from last is 2
Naive Approach: The basic idea is to first convert the decimal number A into base B and then extract the Nth digit from the right of the resulting number.
Implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int nthDigit(int a, int n, int b)
{
int num = 0;
int power = 0;
while (a > 0) {
int digit = a % b;
num += digit * pow(10, power);
a /= b;
power++;
}
int digit;
while (n > 0) {
digit = num % 10;
num /= 10;
n--;
}
return digit;
}
int main()
{
int a = 100;
int n = 3;
int b = 4;
cout << nthDigit(a, n, b) << endl;
return 0;
}
|
Java
import java.lang.Math;
public class Main {
public static int nthDigit(int a, int n, int b)
{
int num = 0;
int power = 0;
while (a > 0) {
int digit = a % b;
num += digit * (int)Math.pow(10, power);
a /= b;
power++;
}
int digit = 0;
while (n > 0) {
digit = num % 10;
num /= 10;
n--;
}
return digit;
}
public static void main(String[] args)
{
int a = 100;
int n = 3;
int b = 4;
System.out.println(nthDigit(a, n, b));
}
}
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Python3
def nthDigit(a, n, b):
num = 0
power = 0
while a > 0:
digit = a % b
num += digit * (10 ** power)
a //= b
power += 1
digit = 0
while n > 0:
digit = num % 10
num //= 10
n -= 1
return digit
a = 100
n = 3
b = 4
print(nthDigit(a, n, b))
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C#
using System;
public class GFG {
public static int nthDigit(int a, int n, int b)
{
int num = 0;
int power = 0;
while (a > 0) {
int digit_ = a % b;
num += digit_ * (int)Math.Pow(10, power);
a /= b;
power++;
}
int digit = 0;
while (n > 0) {
digit = num % 10;
num /= 10;
n--;
}
return digit;
}
public static void Main()
{
int a = 100;
int n = 3;
int b = 4;
Console.WriteLine(nthDigit(a, n, b));
}
}
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Javascript
function nthDigit(a, n, b) {
let num = 0;
let power = 0;
while (a > 0) {
let digit = a % b;
num += digit * Math.pow(10, power);
a = Math.floor(a / b);
power++;
}
let digit;
while (n > 0) {
digit = num % 10;
num = Math.floor(num / 10);
n--;
}
return digit;
}
let a = 100;
let n = 3;
let b = 4;
console.log(nthDigit(a, n, b));
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Time Complexity: O(log b(a) + a)
Auxiliary Space: O(1)
Efficient Approach: The idea is to skip (N-1) digits of the given number in base B by dividing the number with B, (N – 1) times and then return the modulo of the current number by the B to get the Nth digit from the right.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int nthDigit(int a, int n, int b)
{
for (int i = 1; i < n; i++)
a = a / b;
return a % b;
}
int main()
{
int a = 100;
int n = 3;
int b = 4;
cout << nthDigit(a, n, b);
return 0;
}
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Java
import java.util.*;
class GFG
{
static int nthDigit(int a, int n, int b)
{
for (int i = 1; i < n; i++)
a = a / b;
return a % b;
}
public static void main(String[] args)
{
int a = 100;
int n = 3;
int b = 4;
System.out.print(nthDigit(a, n, b));
}
}
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Python3
def nthDigit(a, n, b):
for i in range(1, n):
a = a // b
return a % b
a = 100
n = 3
b = 4
print(nthDigit(a, n, b))
|
C#
using System;
class GFG
{
static int nthDigit(int a, int n, int b)
{
for (int i = 1; i < n; i++)
a = a / b;
return a % b;
}
public static void Main()
{
int a = 100;
int n = 3;
int b = 4;
Console.Write(nthDigit(a, n, b));
}
}
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Javascript
<script>
function nthDigit(a, n, b)
{
for (var i = 1; i < n; i++)
a = parseInt(a / b);
return a % b;
}
var a = 100;
var n = 3;
var b = 4;
document.write(nthDigit(a, n, b));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)