Determine whether the kth largest element of the heap is greater than x or not.

Consider a binary heap containing n numbers (the root stores the greatest number). You are given a positive integer k < n and a number x . You have to determine whether the kth largest element of the heap is greater than x or not. Your algorithm must take O(k) time. You may use O(k) extra storage.

Searching for a friend

You are standing at a crossing from where there emerge four roads extending to infinity. Your friend is somewhere on one of the four roads. You do not know on which road he is and how far he is from you. You have to walk to your friend and the total distance traveled by you must be at most a constant times the actual distance of your friend from you. In terminology of algorithms, you should traverse O(d) distance, where d is the distance of your friend from you.

Searching For Celebrity

Celebrity is a person whom everybody knows but he knows nobody. You have gone to a party. There are total n persons in the party. Your job is to find the celebrity in the party. You can ask questions of the form Does Mr. X know Mr. Y ?. You will get a binary answer for each such question asked. Find the celebrity by asking only O(n) questions.

You are given n real numbers in an array. A number in the array is called a decimal dominant if it occurs more than n/10 times in the array. Give an O(n) time algorithm to determine if the given array has a decimal dominant.

Generating lexial predecessor of a k-subset of set of n elements

Given a file containing 4,300,000,000 integers, how can you find one that appears at least twice

A Theoritical Approach From Programming Pearls

Binary search find an element that occurs at least twice by recursively searching the subinterval that contains more than half of the integers. My original solution did not guarantee that the number of integers is halved in each iteration, so the worst case run time of its log2 n passes was proportional to n·log n. Jim Saxe reduced that to linear time by observing that the search can avoid carrying too many duplicates.

When his search knows that a duplicate must be in a current range of m integers, it will only store m+1 integers on its current work tape;

If more integers would have gone on the tape, his program discards them. Although his method frequently ignores input variables, its strategy is conservative enough to ensure that it finds at least one duplicate.

The algorithm that Bentley is talking about works by repeatedly halving
the candidate range in which the duplicate element must lie. Initially
this range is 0..2^32-1. On each pass it throws away the half of the
range that contains fewer data values and it also throws away the data
lying in that half. Eventually the range decreases to a single value,
which must be a duplicate because the remaining data has at least two
elements. The problem that Bentley notes is that the data may still
have 4 billion elements at this stage! The final improvement he
mentions is that you can throw away data _inside_ the candidate range
as long as you keep enough data around to ensure that at least one
duplicate occurs in it. This is equal to the size of the current
candidate range plus 1!


Another Similer Approach I Found

Create a bit array of length 2^32 bits (initialize to zero), that would be about 512MB and will fit into RAM on any modern machine.

Start reading the file, int by int, check bit with the same index as the value of the int, if the bit is set you have found a duplicate, if it is zero, set to one and proceed with the next int from the file.

The trick is to find a suitable data structure and algorithm. In this case everything fits into RAM with a suitable data structure and a simple and efficient algorithm can be used.
If the numbers are int64 you need to find a suitable sorting strategy or make multiple passes, depending on how much additional storage you have available.

The Pigeonhole Principle -- If you have N pigeons in M pigeonholes, and N>M, there are at least 2 pigeons in a hole. The set of 32-bit integers are our 2^32 pigeonholes, the 4.3 billion numbers in our file are the pigeons. Since 4.3x10^9 > 2^32, we know there are duplicates.

You can apply this principle to test if a duplicate we're looking for is in a subset of the numbers at the cost of reading the whole file, without loading more than a little at a time into RAM-- just count the number of times you see a number in your test range, and compare to the total number of integers in that range. For example, to check for a duplicate between 1,000,000 and 2,000,000 inclusive:

int pigeons = 0;
int pigeonholes = 2000000 - 1000000 + 1; // include both fenceposts
for (each number N in file) {
if ( N >= 1000000 && N <= 2000000 ) { pigeons++ } } if (pigeons > pigeonholes) {
// one of the duplicates is between 1,000,000 and 2,000,000
// try again with a narrower range
}

Picking how big of range(s) to check vs. how many times you want to read 16GB of data is up to you :)

As far as a general algorithm category goes, this is a combinatorics (math about counting) problem.

Given an array[] of integers (+ive , -ive, 0) find the subarray which gives the largest product.

Data Structure Used: Array

Algorithm & Expliantion

Lets us take an array = { 2,-25,4,5,-3,-5}

We take 3 variables P , N , Val

P=1 , N=0 , Val=0

first value is 2 , A[0] which is +ve . So we multiply both P & N with A[0] .
P=2 , N=0

now V[1] = -25 -ve .

We multiply P with -V[1] & N with -V[1] .

P=50
N=0

As V[1] is -ve we swap P & N .
P=0 , N=50

if V[i] == 0
We initialise P=1 , N=0

if( P < 1 ) /* analogous to kadane's algo's if( sumSoFar < 0 ) */ { P=1; N=0; } at every step val = max( val , P ) We proceed in the same fashion till the end . What we are trying to do is to maintain max Positive product so far & max -ve product so far . Hence when we encounter a -ve value we multiply both with - V[i] & then swap as -ve * -ve = +ve -ve * +ve = -ve Working Code #include

int swap(int *a,int *b)
{
int t;
t=*a;
*a=*b;
*b=t;
}

int main()
{
int a[]={6,-1,2,-33,4,15,-7,28,-9,-10};
int maxp=1,maxn=0;
int i=0;
for(i=0;i<10;i++) { if(a[i]>0)
{
maxp*=a[i];
maxn*=a[i];
}
else
{
maxp*=-a[i];
maxn*=-a[i];
swap(&maxp,&maxn);
}


}

printf("%d",maxp);
}

Time Complexity O(N)
Space Conmplexity O(1)
Run Here https://ideone.com/VaKgk

Design An Algorithm to Exhibit Did You Mean Feature of Google Search Engine

When people search at Google, they often type two words without space. For example, instead of typing "binary tree", someone can type "binarytree". Given an input, what is a good way of finding if it is a combination of two valid words e.g. In Directly You have Add Space between two such or if are using some data structure , whene words would have been inserted , you would have mainted a boolen varible to defined the end of word (eod)


An Exanmple Will be Let's say you have a phrase without any spaces - eg. "thisisawesome". Given a dictionary, how would you add spaces in this string?

When people search at Google, they often type two words without space. For example, instead of typing "binary tree", someone can type "binarytree". Given an input, what is a good way of finding if it is a combination of two valid words e.g. In Directly You have Add Space between two such or if are using some data structure , where words would have been inserted , you would have maintained a boolean variable to defined the end of word (eod)
An Example Will be Let's say you have a phrase without any spaces - eg. "thisisawesome". Given a dictionary, how would you add spaces in this string?

        Trie trie = new Trie();
        trie.insert("i");
        trie.insert("this");
        trie.insert("saw");
        trie.insert("is");
        trie.insert("a");
        trie.insert("some");
        trie.insert("awesome");

        String inputString = "thisisawesome";
        String outputString = "";
      
        int left = 0;
        int right = 1;

        while ( right <= inputString.length()) 

        {
            {
                String str = inputString.substring(left,right);
                boolean found = trie.containsKey(str);
                if ( found) 

                {
                    outputString += str;
                    outputString += " ";
                    left += str.length();
                    right = left + 1;
                }

                 else

                 {
                    right++;
                }
            }
        }
      
   

Another Good Working Code  You Can Refer Avinash Solution


Time Complexity O(K) k is length of input string

Given a set of letters and a length N, produce all possible output.(Not permutation).

For example, give the letter (p,o) and length of 3, produce the following output(in any order you want, not just my example order)

ppp ppo poo pop opp opo oop ooo

another example would be given (a,b) and length 2

answer: ab aa bb ba

Division Operation without using division operator!!! FAQ in Core Dev Compny Telephonic

 We can use bit logic to reduce the time complexity to O(logn) where n is Quotient

Algorithm will be as follow as we know

1. Left shifting an unsigned number by 1 multiplies that number by 2.
2. Right shifting an unsigned number by 1 divides that number by 2.

Therefore the procedure for the division algorithm, given a dividend and a divisor .
core logic will be to left shift (multiply by 2) untill its greater then dividend , then continue this routine with the the difference between the dividend and divisor and divisor till the point where dividend is less than divisor or their difference is zero.

Lets see one example: dividend=23 divisor=3

then left shift for 3 i.e 3 will converted to3, 6,12,24,… since 12 <23 hence now dividend is 11 and quotient in 4(two time shift operation) now again 3,6,12.. 6<11 hence dividend is 11-6=5 and quotient =4+2=6 now only 3<5 hence remainder =2 quotient =6+1=7 so answer.

Time Complexity O(logn) Number of bits in Quotient

#include
using namespace std;

int dividend,divisor,remainder;
int division(int p,int q){
int quotient=1;
/*if divisor and diviend are equal then quotient=1*/
if(p==q){
remainder=0;
return 1;
}
/*if dividend is smaller than divisor then remainder=dividend*/
if(p dividend*/
while(p>=q){
q<<=1; quotient<<=1; } /*shift right for one time so that divisor become smaller than dividend*/ q>>=1;
quotient>>=1;
/*again call division recurcively*/
quotient+=division(p-q,divisor);
return quotient;
}
int main(){
cout<<"\nEnter dividend:"; cin>>dividend;
cout<<"\nEnter divisor:"; cin>>divisor;
cout<<"\nQuotient:"<<<"\nRemainder:"

}

Time Compelxity O(log(Quotient))
Space Complexity O(1)

(left shift ) one time i.e divide by 2 and if we do <<(right shift) one time i.e multiplied by 2 if we do n times the 2power n time it will do same operation) simultanuosly we are doing left shift for quotient also. Thus we get quotient for that division. After doing this one time again we call the function recursively after subtracting the with remaining numbers.

Lets see one example: dividend=23 divisor=3

then left shift for 3 i.e 3 will converted to3, 6,12,24,… since 12 <23 hence now dividend is 11 and quotient in 4(two time shift operation) now again 3,6,12.. 6<11 hence dividend is 11-6=5 and quotient =4+2=6 now only 3<5 hence remainder =2 quotient =6+1=7 so answer got

Similar  Iterative program for doing the same is


int div(int a, int b)
{
if(!b) return -1;
if(a<b) {
printf("%d / %d = %d , remainder %d\n", a, b, 0, a);
return 0;
}
int q = 1, t=d, d=1;
while (t<a)
{
d=t;
t=t<<2;
q=q<<1;
}
while (d+b<a)
{
d=d+b;
++q;
}
printf("%d / %d = %d , remainder %d\n", a, b, q, a-d);
return 0;
}


A Good Discussion You Can Also Found here Algogeek