Friday, September 16, 2011
Wednesday, September 14, 2011
Use Google Chrome For Better View of Contents on Blog :)
Dear Readers,
I am getting regular suggestion & useful comments from many geeks to re-arrange the question according to type, i will try to do this asap i get time as i have to edit HTML Code & also some people are saying that contents are good but unable to view them. for this problem try Google Chrome Until I change the GUI & edit Html Code.
Thanks For Giving Suggestion to Solving Problems Efficiently
Happy Coding
Shashank !!!
Monday, September 5, 2011
Find the smallest number expressible as the sum of cubes of two numbers in two different ways. Indeed, 10^3 + 9^3 =12^3+1^3=1729 , Do It Efficiently
Problem:
A mathematician G. H. Hardy was on his way to visit his collaborator S. Ramanujan an who was in the hospital. Hardy remarked to Ramanujan that he traveled in taxi cab number 1729 which seemed a dull one and he hoped it was not a bad omen. To this, Ramanujan replied that 1729 was a very interesting number-it was the smallest number expressible as the sum of cubes of two numbers in two different ways. Indeed,
103 + 93 =123 + 1 3 二1729.
Hint: This problem is very similar to another very popular problem that is asked in interviews. You are given a η ×η matrix in which both rows and columns are sorted in ascending order and you are supposed
to find a given number in the matrix.
Algorithm:
In this case, we are essentially looking for an implicit matrix A such that A(i,j) =i^3十j^3. In our case, the matrix will have n^1/3 rows and columns and this matrix of size n^1/3 * n^1/3 we try to search k=i^3+j^3 isn't it ans algorithms for searching for a number in such a matrix that are linear in the number of rows.
One approach is to start by comparing x to An,1. If x = An ,l , stop.Otherwise, there are two cases:
- x < A1,n in which case x is less than element at Column n at 1,n position. which mean we can escape whole row itself . we decrement column pointer to left.e.g. previous column.
- x > A1,n , in which case if x > a[1][n] is greater than it will be greater then all elements in Row 1.. we increment the row pointer to next row.
Efficient Solution:
Here is the Pseudo Code For The Same.
int IsNumberEqualstoSumeTwoCubes(int n)
{
int row=Ceil(Pow(n,1/3));// Pow Has Complexity of O(logn) Ceil(1729)=577
int column=row;
int i=0;int j=column;
while(i<row && j>=0)
{
int m=i*i*i+j*j*j;
if(m==n)
return 1; // print i^3 and j^3 these are two such numbers
else if ( m < n)
i++;
else
j--;
}
return 0; //such number DNE
}
Complexity Analysis:
In either case, we have a matrix with η fewer elements to search. In each iteration, we remove a row or a column, which means we inspect 2η-1 elements.
We claim that our algorithm that solves the matrix search problem will have to compare x with each of the 2n-l elements shown (i.e the diagonal elements (in case of column element matching )and the elements immediately below them (in case of row elements matching )and this is obvious just draw a diagram in which both row & column are in sorted order & try to run the above algorithm ).
Call these elemmts the # elements..
Comparing x with any other elements in matrix does not eliminate the any of the # elements.we can easily Proof this by Contradiction:Suppose X algorithm doesn't compare x with any of the above # elements.
then we could make that element x ( instead of x-1 got to prev. column or x+1 go to next row.).hence we will get wrong result.
Above Algorithm Really Requires Deep Understanding of Young_tableau :)
Time Complexity O(row+column)=O(N^1/3+N^1/3)=O(2*N^1/3)=O(N^1/3)
Space Complexity O(1)
Labels:Data
Facebook Interview
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Google Interview
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Hardy Number
Friday, September 2, 2011
Get Rid Off Suffix Tree !!!!! An Online Compilation of Helpful Tutorial
Excellent Study Material For Understanding Algorithms & implementing Suffix Tree
6.http://apps.topcoder.com/forums/?module=RevisionHistory&messageID=1171511
7.http://www.stanford.edu/class/cs97si/suffix-array.pdf
Application of Suffix Tree
1.Longest repeated substring problem:
The longest repeated substring problem is finding the longest substring of a string that occurs at least twice. This problem can be solved in linear time and space by building a suffix tree for
the string, and finding the deepest internal node in the tree. The
string spelled by the edges from the root to such a node is a longest
repeated substring. The problem of finding the longest substring with
at least k occurrences
can be found by first preprocessing the tree to count the number of
leaf descendants for each internal node, and then finding the deepest
node with at least k descendants.
By from Sartaj Sahni
Find the longest substring ofS
that appears at leastm > 1
times. This query can be answered inO(|S|)
time in the following way:
(a) Traverse the suffix tree labeling the branch nodes with the sum of the label lengths from the root and also with the number of information nodes in the subtrie. (b) Traverse the suffix tree visiting branch nodes with information node count>= m
. Determine the visited branch node with longest label length.
Note that step (a) needs to be done only once. Following this, we can do step (b) for as many values ofm
as is desired. Also, note that whenm = 2
we can avoid determining the number of information nodes in subtries. In a compressed trie, every subtrie rooted at a branch node has at least two information nodes in it.
Note that step (a) needs to be done only once. Following this, we can do step (b) for as many values ofm
as is desired. Also, note that whenm = 2
we can avoid determining the number of information nodes in subtries. In a compressed trie, every subtrie rooted at a branch node has at least two information nodes in it.
2. Pattern Searching/Sub-String Searching
Searching for a substring,pat[1..m]
, intxt[1..n]
, can be solved in O(m) time (after the suffix tree fortxt
has been built in O(n) time).
3. Longest Repeated Substring
4. Longest Common Subsequence
5. Longest Palindrome
http://www.cs.helsinki.fi/u/ukkonen/SuffixT1withFigs.pdf
Add a special ``end of string'' character, e.g. `$', to i.e., `issi' in the case of `mississippi'. The longest repeated substring can be found in O(n) time using a suffix tree.
The longest common substring of two strings,
Equivalently, one can build a (basic) suffix tree for the string
(Try it using the HTML FORM above.)
txt[1..n]
and build a suffix tree; the longest repeated substring oftxt[1..n]
is indicated by the deepest fork node in the suffix tree, where depth is measured by the number of characters traversed from the root, The longest common substring of two strings,
txt1
and txt2
, can be found by building a generalized suffix tree for txt1
andtxt2
: Each node is marked to indicate if it represents a suffix of txt1
or txt2
or both. The deepest node marked for both txt1
and txt2
represents the longest common substring.Equivalently, one can build a (basic) suffix tree for the string
txt1$txt2#
, where `$' is a special terminator for txt1
and `#' is a special terminator for txt2
. The longest common substring is indicated by the deepest fork node that has both `...$...' and `...#...' (no $) beneath it.(Try it using the HTML FORM above.)
Note that the `longest common substring problem' is different to the `longest common subsequence problem' which is closely related to the `edit-distance problem': An instance of a subsequence can have gaps where it appears in
txt1
and in txt2
, but an instance of a substring cannot have gaps.A palindrome is a string, P, such that P=reverse(P). e.g. `abba'=reverse(`abba'). e.g. `ississi' is the longest palindrome in `mississippi'. The longest palindrome of
txt[1..n]
can be found in O(n) time, e.g. by building the suffix tree for txt$reverse(txt)#
or by building the generalized suffix tree for txt
and reverse(txt)
.(Try it.)
Others
--
Some Awesome Algorithms/Codes For Suffix Tree Implementation
Sunday, August 28, 2011
Given an array A[] and a integer num. Find four no.s in the array whose sum is equal to given num.
Algorithm :
Naive Algorithm Will O(N^4) Use Four Loop & keep Checking for all four elements.its so Naive ,easily understood-able..
Algorithm:
Little Efficient O(N^3).
Sort the input array. O(nlogn)
Then take 4 pointers p1,p2,p3,p4 where
0<=p1<=n-3 1st loop
p1+1<=p2<=n-2 1st inner loop
in loop set p3=p1+ & p4=n-1
( e.g. Take 3 Pointer Pointing to three successive indexes from starting & 4th Pointer pointing to n-1th position. )
Now use same algo we used to check for two elements in sorted array whose sum equals to given element or not 3rd Inner Loop
Most Efficient O(N^2)
Algorithm:
Create an array of all possible PAIR sums..that would be done in O(n^2)
Sort the Above array of size n^2 that can be done in.O(n^2log(n)) .
Then Search this array for two pairs..that sum to the required value..this can be done by maintaining two pointer One at the starting index..and other at the highest index..and moving them accordingly..(if sum of pair exceeds given value..move up highest value pointer..else move down..lowest value pointer) .it Will take O(n)
Total Time Complexity Will be O(n^2logn) +O(nlogn) +O(n)=O(n^2logn).
Space Complexity O(N^2).
Naive Algorithm Will O(N^4) Use Four Loop & keep Checking for all four elements.its so Naive ,easily understood-able..
Algorithm:
Little Efficient O(N^3).
Sort the input array. O(nlogn)
Then take 4 pointers p1,p2,p3,p4 where
0<=p1<=n-3 1st loop
p1+1<=p2<=n-2 1st inner loop
in loop set p3=p1+ & p4=n-1
( e.g. Take 3 Pointer Pointing to three successive indexes from starting & 4th Pointer pointing to n-1th position. )
Now use same algo we used to check for two elements in sorted array whose sum equals to given element or not 3rd Inner Loop
Most Efficient O(N^2)
Algorithm:
Create an array of all possible PAIR sums..that would be done in O(n^2)
Sort the Above array of size n^2 that can be done in.O(n^2log(n)) .
Then Search this array for two pairs..that sum to the required value..this can be done by maintaining two pointer One at the starting index..and other at the highest index..and moving them accordingly..(if sum of pair exceeds given value..move up highest value pointer..else move down..lowest value pointer) .it Will take O(n)
Total Time Complexity Will be O(n^2logn) +O(nlogn) +O(n)=O(n^2logn).
Space Complexity O(N^2).
Labels:Data
Amazon Interview
find the minimum number of platforms so that all the buses can be placed as per their schedule.
At a bus-station, you have time-table for buses arrival and departure. You need to find the minimum number of platforms so that all the buses can be placed as per their schedule.
Example
Now modify the array as put 1 where you see A and -1 where you see D.
So new array will be like this:
1 1 -1 1 1 -1 -1 -1
And finally make a cumulative array out of this:
1 2 1 2 3 2 1 0
Your solution will be the maximum value in this array. Here it is 3.
I think that code for this will not be complex so I am skipping that part.
The complexity of this solution depends on the complexity of sorting.
Also we don not need to create a cumulative array or an array with 1 and -1;
you just need a counter (cnt) initialized at '0'. Whenever, you find an 'A' in
arrival-departure array, increment cnt by 1. Compare it with maximum value (max);
if it is greater than max, make max equal to cnt. If you get a 'D' in arrival-departure
array, decrement cnt by 1. At the end, return 'max'.
Example
Bus Arrival Departure
BusA 0900 hrs 0930 hrs
BusB 0915 hrs 1300 hrs
BusC 1030 hrs 1100 hrs
BusD 1045 hrs 1145 hrs
Algorithm
Its simple dynamic programming question that calculate the
number of buses at station at any time(when a bus comes or
leaves). Maximum number in that pool will be nothing but
the maximum number of buses at the bus-station at any time
,which is same as max number of platforms required.
So first sort
all the arrival(A) and departure(D) time in an int array.
Please save the corresponding arrival or departure in the
array also.Either you can use a particular bit for this
purpose or make a structure. After sorting our array will
look like this:
0900 0915 1930 1030 1045 1100 1145 1300
A A D A A D D D
Now modify the array as put 1 where you see A and -1 where you see D.
So new array will be like this:
1 1 -1 1 1 -1 -1 -1
And finally make a cumulative array out of this:
1 2 1 2 3 2 1 0
Your solution will be the maximum value in this array. Here it is 3.
I think that code for this will not be complex so I am skipping that part.
The complexity of this solution depends on the complexity of sorting.
Also we don not need to create a cumulative array or an array with 1 and -1;
you just need a counter (cnt) initialized at '0'. Whenever, you find an 'A' in
arrival-departure array, increment cnt by 1. Compare it with maximum value (max);
if it is greater than max, make max equal to cnt. If you get a 'D' in arrival-departure
array, decrement cnt by 1. At the end, return 'max'.
Time Compelxity O(nlogn)
Space Complexity O(1)
Labels:Data
Adobe Question
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Amazon Interview
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