Input number: 24
Binary representation: 00011000
Thus it's a palindrome
Input number 153:
Binary representation: 10011001
Thus it's a palindrome
Input number: 25
Binary representation: 00011001
Thus it's not a palindrome
Algorithm:
1) Take the input.
2) Create an array of length 8 to store 8 bit binary representation
of input number
3) Use bitwise operators to convert into binary from
a) Initialize i to 7 (8-1)
b) Find ith bit & store in the array
array[i]=n & 1
c) Right shift n & decrement i
n=n>>1
d) IF n=0
Break
ELSE
Repeat step b-d
4) Check the array where the binary representation is stored,
for palindrome
a) Set j= 0 & k=7 (8-1, 8 is the MAX SIZE)
b) IF (array [j]!= array [k])
It's not a palindrome
c) IF j < k
Increment j& decrement k
Repeat b, c
ELSE
It's a palindrome
Example with explanation:
Input no: 24
Converting to binary representation using bitwise operatorInitially,
N=24
Array[8]={0};
0 0 0 0 0 0 0 0 (LSB)
i=7
-----------------------------------------------------------------
first iteration,
array[7]=n&1 = 0 (refer to bitwise operators and
their working for understanding the outcome)
0 0 0 0 0 0 0 0 (current bit)
n=n>>1=12
i=6
-----------------------------------------------------------------
second iteration,
array [6]=n&1 = 0
0 0 0 0 0 0 0 (current bit) 0
n=n>>1=6
i=5
-----------------------------------------------------------------
third iteration,
array [5]=n&1 = 0
0 0 0 0 0 0 (current bit) 0 0
n=n>>1=3
i=4
-----------------------------------------------------------------
fourth iteration,
array [4]=n&1 = 1
0 0 0 0 1(current bit) 0 0 0
n=n>>1=1
i=3
-----------------------------------------------------------------
fifth iteration,
array [3]=n&1 = 1
0 0 0 1 (current bit) 1 0 0 0
n=n>>1=0
i=2
-----------------------------------------------------------------
sixth iteration,
n=0
so no more processing
thus the array finally becomes which is
the binary representation
0 0 0 1 1 0 0 0 (LSB)
Checking palindrome
Initially,
j=0 & k=7 (8-1)
-----------------------------------------------------------------
First iteration
Array[j] == array [k] (both 0)
j=j+1=1
k=k-1=6
j<k
-----------------------------------------------------------------
Second iteration
Array[j] == array [k] (both 0)
j=j+1=2
k=k-1=5
j<k
-----------------------------------------------------------------
Third iteration
Array[j] == array [k] (both 0)
j=j+1=3
k=k-1=4
j<k
-----------------------------------------------------------------
Fourth iteration
Array[j] == array [k] (both 1)
j=j+1=4
k=k-1=3
j>k
thus, stops processing & prints it's a palindrome
C Implementation
#include <stdio.h>
#define SIZE 8
int main()
{
unsigned int n;
printf("enter the no ( max range 255)\n");
scanf("%d",&n);
int c[SIZE]={0};
int i=SIZE-1;
printf("binary representation is: ");
while(n!=0){
c[i--]=n&1;
n=n>>1;
}
for(int j=0;j<SIZE;j++)
printf("%d",c[j]);
printf("\n");
for(int j=0,k=SIZE-1;j<k;j++,k--){
if(c[j]!=c[k]){
printf("Not palindrome\n");
return 0;
}
}
printf("it's palindrome\n");
return 0;
}
Output
First run:
enter the no ( max range 255)
153
binary representation is: 10011001
it's palindrome
Second run:
enter the no ( max range 255)
24
binary representation is: 00011000
it's palindrome
Third run:
enter the no ( max range 255)
-8
binary representation is: 11111000
Not palindrome
Pre-requisite: Input number n
Input Example:
Algorithm:
Example with explanation:
C Implementation
Output
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