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Given two strings which represent different version releases of a software, print -1, 1 or 0 depending on the order of the releases.

If version1 was released after version2 print 1,
If version1 was released before version2 print -1,
otherwise print 0.

Assumptions granted : Strings may only contain digits and the ‘.'(dot) character and must contain atleast 1 character.

The ‘.’ character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 1.13 < 1.13.4

version number such as 1, 01, 1.0 are equivalent.

I would request you to change tabs or minimize your browser and give a good 30-40 minutes before seeing the solution.

This is a basic implementation based question, the only thing to take care about is the sequence of 0’s that may crept in after a dot.

WordPress is not rendering the code properly so kindly check out the code here on github.

Time Complexity   : O(n), where n is the sum of the lengths of the 2 versions.
Space Complexity  : O(n), where n is the sum of the lengths of the 2 versions.

It can also be done with the space complexity of O(1), with the help of substr function, left as an exercise. 

You can follow this blog for more such posts. Comment in your queries, report bugs, discuss! 🙂

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2. Swap bits

Question : Given a number and 2 indices return the number after swapping bits present at those indices.

Method-1 : Brute-Force
We can swap bits by following the given steps :-
1. Using bitmasks to extract the ith and jth bits.
2. Save the extracted bits to some local variables.
3. Write the jth  bit to the ith position and vice-versa.

int bitSwapBruteForce(int x, int i, int j){
	int temp1= ((x & (1<<i)) !=0);
	int temp2= ((x & (1<<j)) !=0);
	x = temp1==1 ? (x | ( temp1<<j) ) : (x & ~(1<<j));
	x = temp2==1 ? (x | ( temp2<<i) ) : (x & ~(1<<i));
	return x;
}

The Time Complexity : O(1), only a constant amount of operations are required.
The Space Complexity : O(1), only a constant amount of extra space required.

Method-2 : Oh! Its beauty.
Since a bit can take only two values we can do better, lets swap only when bits differ. We’ll use the magical XOR once again.


int bitSwapSuccinct(int num, int i, int j){
	if( ( ( num>>i ) & 1 ) != ( ( num>>j ) & 1 ) ){
		int bitMask = (1<<i) | (1<<j);
		num ^=bitMask;
	}
	return num;
}

As ith and jth bits of num differ, only 1 of them would be 1, resulting in bitMask having only one 1 in its bit representation. Now the XOR would swap the bits with help of bitMask.

For num=(1110)2, i=0, j=2;
bitMask=(1)2 | (100)2 = (101)2
num = (1110)2 XOR (0101)2 = (1011)2

The Time Complexity : O(1), only a constant amount of operations are required.
The Space Complexity : O(1), only a constant amount of extra space required.

The full working code is available at Github, you can star the repository to get updates. You can also follow this blog for more such posts. Comment in your queries, report bugs, discuss!:)

1. Parity of a Word

Question : Compute the parity of a 64-bit word.
Parity of a number refers to whether it contains an odd or an even number of 1’s in its binary representation. The parity of a number is 1 if it contains odd number of 1’s, else 0.

Method-1 : Brute Force
We can iteratively test the value of each bit while tracking the number of 1’s seen so far.

int Parity (long x){
	int result=0;
	while(x){
		result+=(x&1);
		x>>1;
	}
	return result%2;
}

 

The Time Complexity : O(n), where n is the number of bits in the word.
The Space Complexity : O(1), only a constant amount of extra space required.

Method-2 : A better Approach
We can apply the method that is used to compute the number of 1’s in the bit representation of a number.

int Parity (int x){
	int result=0;
	while(x){
		x=x&(x-1); // removes the lowest set bit.
		result^=1;
	}
	return result;
}

x-1 contains last set bit of x as unset, hence there AND unsets that bit.
The Time Complexity : O(k), where k is the number of SET bits.
The Space Complexity : O(1), only a constant amount of extra space is required.

Method-3: CACHE!
The original question asks us the compute the parity of a 64 bit number, previous methods are sufficiently fast to compute parity for such numbers, though if the computation has to take place for many numbers it is a better idea to cache our solutions and use them when required. Caching 2^64(9223372036854775808) numbers? Ah! does not look like a good idea. Better we can cache answers to first 2^16 numbers and then compute solution for larger numbers by combining the solutions of smaller segments. For a 64 bit number we’ll break the number into 4 segments of 16 bits. We’ll use the table to compute parities of all the 4 segments and then XOR the values to compute the final answer.

void precompute (void){
	for (int i=0;i<=int(pow(2,(sizeOfInt/4))-1);i++){
		Table[i]=parity(i);
		//cout<<i<<' '<<Table[i]<<endl; Uncomment this to check for yourself
	}
	return ;
}

 

int parityAns( int ParityOf, int BitMask, int wordSize){
	return ( Table [ParityOf>>( 3 * wordSize)]
	^ Table[( ParityOf >>  ( 2 * wordSize ) ) & BitMask]
	^ Table[( ParityOf >> ( wordSize ) ) & BitMask]
	^ Table[ ParityOf & BitMask] );
}


The Time Complexity : O(n/L + 2^L)
, where n is the number of bits in the word and L is the width of the word for which we cache the results. 2^L time will be taken to prepare the table, if L is known beforehand we can remove it from the notation, as we only talk about the growth of function in Big-Oh Notation.
The Space Complexity : O(2^L), only a constant amount of extra space required.

Method-4 The Final Blow
We can use the property of XOR to compute the parity.
say X consists of 4 bits(a,b,c,d), we ran the following steps on it.

X =        a     b     c     d
X=X^X>>1 = a    a^b   b^c   c^d
X=X^X>>2 = a    a^b  a^b^c a^b^c^d

The last bit contains the parity which we can extract by ANDing X with 1.

int parity(int ParityOf){
	int factorOfShift = sizeOfInt/2;
	while(factorOfShift){
		ParityOf^=(ParityOf>>factorOfShift);
		factorOfShift=factorOfShift/2;
	}
	return ParityOf & 0x1;
}

The Time Complexity : O(log n), where n is the number of bits in the word.
The Space Complexity : O(1), only a constant amount of extra space required.

Also we can combine the last 2 methods, the full working code to which is available at Github, you can star the repository to get updates. You can also follow this blog for more such posts. Comment in your queries, report bugs, discuss! 🙂

A series on Programming Interview Questions

I have decided to write complete solutions and analysis of the questions that are frequently asked in programming interviews at major tech. companies for the position of Software Engineer/Software Engineering Intern.

This series will contain exhaustive list of solutions for all the problems posed in The Elements Of Programming Interviews, I will also maintain a Github repository for all the codes. They can be accessed directly using this EPI link.