Given a number n, we need to find the last digit of 2n
Input : n = 4
Output : 6
The last digit in 2^4 = 16 is 6
Input : n = 11
Output : 8
The last digit in 2^11 = 2048 is 8
A Naive Solution is to first compute power = pow(2, n), then find the last digit in power using power % 10. This solution is inefficient and also has an integer arithmetic issue for slightly large n.
An Efficient Solution is based on the fact that the last digits repeat in cycles of 4 if we leave 2^0 which is 1. Powers of 2 (starting from 2^1) are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, …
We can notice that the last digits are 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, …
1) We compute rem = n % 4. Note that the last rem will have a value from 0 to 3.
2) We return the last digit according to the value of the remainder.
Remainder Last Digit
1 2
2 4
3 8
0 6
Illustration : Let n = 11, rem = n % 4 = 3. Last digit in 2^3 is 8 which is same as last digit of 2^11.
C++
#include <bits/stdc++.h>
using namespace std;
int lastDigit2PowerN( int n)
{
if (n == 0)
return 1;
else if (n % 4 == 1)
return 2;
else if (n % 4 == 2)
return 4;
else if (n % 4 == 3)
return 8;
else
return 6;
}
int main()
{
for ( int n = 0; n < 20; n++)
cout << lastDigit2PowerN(n) << " " ;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG{
static int lastDigit2PowerN( int n)
{
if (n == 0 )
return 1 ;
else if (n % 4 == 1 )
return 2 ;
else if (n % 4 == 2 )
return 4 ;
else if (n % 4 == 3 )
return 8 ;
else
return 6 ;
}
public static void main(String[] args)
{
for ( int n = 0 ; n < 20 ; n++)
System.out.print(lastDigit2PowerN(n) + " " );
}
}
|
Python3
def lastDigit2PowerN(n):
if n = = 0 :
return 1
elif n % 4 = = 1 :
return 2
elif n % 4 = = 2 :
return 4
elif n % 4 = = 3 :
return 8
else :
return 6
for n in range ( 20 ):
print (lastDigit2PowerN(n), end = " " )
|
C#
using System;
class GFG{
static int lastDigit2PowerN( int n)
{
if (n == 0)
return 1;
else if (n % 4 == 1)
return 2;
else if (n % 4 == 2)
return 4;
else if (n % 4 == 3)
return 8;
else
return 6;
}
public static void Main( string [] args)
{
for ( int n = 0; n < 20; n++)
{
Console.Write(lastDigit2PowerN(n) + " " );
}
}
}
|
Javascript
<script>
function lastDigit2PowerN(n)
{
if (n == 0)
return 1;
else if (n % 4 == 1)
return 2;
else if (n % 4 == 2)
return 4;
else if (n % 4 == 3)
return 8;
else
return 6;
}
for ( var n = 0; n < 20; n++)
document.write(lastDigit2PowerN(n) + " " );
</script>
|
Output
1 2 4 8 6 2 4 8 6 2 4 8 6 2 4 8 6 2 4 8
Time Complexity: O(1)
Auxiliary Space: O(1)
Can we generalize it for any input numbers? Please refer Find Last Digit of a^b for Large Numbers
Approach: Using Bitwise Operators
Steps:
-
Firstly, we compute the remainder rem using bitwise AND operator with 3.
- Which is equivalent to compute n % 4.
-
Then, based on the rem value, return the last digit of 2^n.
Below is the code implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int last_digit_of_2n( int n) {
int rem = n & 3;
if (rem == 0) return 6;
else if (rem == 1) return 2;
else if (rem == 2) return 4;
else return 8;
}
int main() {
int n=4;
cout << last_digit_of_2n(n) << endl;
return 0;
}
|
Java
import java.io.*;
public class LastDigitOf2N {
static int lastDigitOf2N( int n) {
int rem = n & 3 ;
if (rem == 0 ) return 6 ;
else if (rem == 1 ) return 2 ;
else if (rem == 2 ) return 4 ;
else return 8 ;
}
public static void main(String[] args) {
int n = 4 ;
System.out.println(lastDigitOf2N(n));
}
}
|
Python3
def last_digit_of_2n(n):
rem = n % 4
if rem = = 0 :
return 6
elif rem = = 1 :
return 2
elif rem = = 2 :
return 4
else :
return 8
def main():
n = 4
print (last_digit_of_2n(n))
if __name__ = = "__main__" :
main()
|
C#
using System;
class Program
{
static int LastDigitOf2N( int n)
{
int rem = n & 3;
if (rem == 0) return 6;
else if (rem == 1) return 2;
else if (rem == 2) return 4;
else return 8;
}
static void Main()
{
int n = 4;
Console.WriteLine(LastDigitOf2N(n));
}
}
|
Javascript
function lastDigitOf2n(n) {
const rem = n % 4;
if (rem === 0) return 6;
else if (rem === 1) return 2;
else if (rem === 2) return 4;
else return 8;
}
const n = 4;
console.log(lastDigitOf2n(n));
|
Time Complexity: O(1)
Auxiliary Space: O(1)