Determine whether the binary representation of a number if a palindrome or not, code it on a white board.

This was the first question I was asked and is considered a warm up.

anon.. would this work for a number like 17 (10001)?

bool checkPalindrome(unsigned int n) { int m = n, k =0; bool ret = false; while(m!=0) { int i = 1; i = i & m; k = k > 1; } if((k^n)==0) { cout<<"Palindrome"<

I have a simple solution. Reverse the bits one by one and test equality of the two resulting numbers. (Matlab code) function [r] = isPalindrome(a) n = a; m = 0; while(n>0) m = bitshift(m, 1); m = m + mod(n, 2); n = bitshift(n, -1); end r = (m==a); end

public static boolean isPalindrome(int n) { int nb = 0, nl=n; while(nl>0) { nb=(nb>1; } return nb==n; }

@john/catalin4ever, i think it'll fail because it doesn't concat the 0s. For example: if the input is 0110, then after execution "nb" becomes to 0011. this is because of the condition: while(nl>0)

Ask interviewer if they want the MSB that is a 1 to dictate the bit range to check, have it given as a parameter, or assume sizeof(T)*8 perhaps. Little details and extras like this can make a difference to them. public static bool CheckPalindrome(uint n) { return CheckPalindrome(n, 0); } unsafe public static bool CheckPalindrome(uint n, int bits) { if (bits == 0) { // Determine bits to check as the MSB having a 1 uint temp = n; while (temp != 0) { ++bits; temp >>= 1; } } uint m1 = (uint)1 > 1; i > 0; --i) { if (((n & m1) != 0) ^ ((n & m2) != 0)) { return false; } m1 >>= 1; m2 <<= 1; } return true; } Examples: BitOps.CheckPalindrome(17, 0) is true BitOps.CheckPalindrome(17, 8) is false BitOps.CheckPalindrome(0xABD5, 0) is true BitOps.CheckPalindrome(0xABD5, 16) is true BitOps.CheckPalindrome(0x5BDA, 0) is false BitOps.CheckPalindrome(0x5BDA, 16) is true

string binary; int start=0, int end= binary.length -1; while(start < end) { if (binary[start] == binary[end]) { start ++; end- -; } else return false; } return true;

int isPalindrome( int num ) { int x = num & num; return ( x == num ) ? 1 : 0; }

/* function declaration goes here.*/ int isPalin(int orig); int main() { int x = 9; int i = isPalin(x); printf("i = %d \n", i); } int isPalin(int orig) { int copy = orig; static int reversed = 0; while(copy!=0) { reversed >= 1; } return (reversed == orig);

We can compare the leftmost and rightmost bits by left shifting and right shifting the bits and also by getting rid of those bits . If the binary number is a palindrome, the left and right bits should be equal. Here is a C# implementation of the above logic static bool IsBinaryPalindrome(byte num) { int i = 0; while (true) { i++; bool right = false, left = false; byte origNo = num; if (((num > i) != origNo) { left = true; //left most bit contains one } num >>= i; //putting the right bit back in its original position origNo = num; if (((num >>= i) << i) != origNo) { right = true; // right most bit contains one } num <<= i; //putting the left bit back in its original position if (left != right) { return false; } if (num == 0 || num == 1) break; } return true; }