use stack.
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Algorithm to find the two numbers whose difference is minimum among the set
of numbers.
For example the sequence is 5, 13, 7, 0, 10, 20, 1, 15, 4, 19
The algorithm should return min diff = 20-19 = 1.
Constraint - Time Complexity O(N) Space is not a constraint [upto O(3N)]
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You
How about , using a sorting algorithm like Quick sort and return diff of
last and the before last element.
Best Regards
Abhijit
On Sun, Feb 6, 2011 at 3:29 PM, Decipher ankurseth...@gmail.com wrote:
Algorithm to find the two numbers whose difference is minimum among the set
of numbers.
I also encountered this question yesterday. This is my solution which i
tested for a few sample cases.
https://github.com/algoseeker/Interview/blob/master/Node.java
I maintained a pointer to the Node where there should be creation of a new
node. If the node created is left child, shift the
it has variable defined
On Sun, Feb 6, 2011 at 7:47 AM, Venki venkatcollect...@gmail.com wrote:
Hi Gajendra,
See the following link
http://geeksforgeeks.org/forum/topic/huwaei-interview-question-for-software-engineerdeveloper-2-5-years-about-cpuzzles#post-18725
On Feb 5, 7:28 pm, Gajendra
Hello,
I have applied for fresh grad posts in Google. I have filled up their forms
and uploaded the resume. Can anyone tell me how long do they take to process
the resume and how long should I wait?
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Write a function to check whether the Binary Tree is mirror structure.
True
(A (B (C,D), E( F,G)))
or
(A (B (C)), D(E)))
False
(A (B))
or
(A (B,C(D,E)))
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tree is represented in string form as given above
With Regards,
*Jalaj Jaiswal* (+919019947895)
Final Year Undergraduate,
IIIT ALLAHABAD
use a stack for this
{
//let preorder traversal of a tree be in a array t
for(i = t.length; i-1; i--){
if(t(i) == L){
stack.push(t[i]);
}else{
leftChild = s.pop(); // will return null if stack is empty
rightChild = s.pop(); // will return null if stack is empty
node = new Node(leftChild,
thanks..
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balaji ;-)
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test case : a = -1, b = - 1
Result : fight successfully challenged dave's problem. Method returned 1
while it shud have returned -1.
On Sun, Feb 6, 2011 at 4:12 PM, Balaji S balaji.ceg...@gmail.com wrote:
thanks..
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I gues jalaj's solun works perfect.
Thanks and regards,
Gajendra Dadheech
On Sun, Feb 6, 2011 at 4:06 PM, jalaj jalaj.jaiswa...@gmail.com wrote:
use a stack for this
{
//let preorder traversal of a tree be in a array t
for(i = t.length; i-1; i--){
if(t(i) == L){
stack.push(t[i]);
@jalal hi You Can Use This algorithm to construct the Mirror tree
(1) Call Mirror for left-subtreei.e., Mirror(left-subtree)
(2) Call Mirror for right-subtree i.e., Mirror(left-subtree)
(3) Swap left and right subtrees.
temp = left-subtree
left-subtree = right-subtree
I could not get it for recursively, but iteratively, I coded a solution. If
anyone knows recursively,
let us know please.
#includestdio.h
void main()
{
char s[18]=DGGDBCBHH;
int i=0,j=0;
int count;
while(s[i]!='\0')
{
if(s[i] == s[i+1])
{
@Decipher
Microsoft is coming in over college.. can you Give some more question which
are recently asked ...in MS test...
@everone.. do share question of MS if some body has given the recently this
Test..
Algorithm:
1. maintain min_diff so far
2. maintain max_element so far
complexity is O(n) ,
algo fails for following input set
20,18,1,7,2
Thanks and regards,
Gajendra Dadheech
On Sun, Feb 6, 2011 at 6:32 PM, coolfrog$
dixit.coolfrog.div...@gmail.comwrote:
@Decipher
Microsoft is coming in over college.. can you Give some more question which
are recently asked ...in MS test...
that should be (a+b+abs(a-b))/2
On Sun, Feb 6, 2011 at 4:15 PM, Manmeet Singh mans.aus...@gmail.com wrote:
test case : a = -1, b = - 1
Result : fight successfully challenged dave's problem. Method returned 1
while it shud have returned -1.
On Sun, Feb 6, 2011 at 4:12 PM, Balaji S
i use CountSort as i assuming maxn=20 :)
and then find the adjacent diff of the sorted array to find the min
On Sun, Feb 6, 2011 at 7:21 PM, Gajendra Dadheech gajju3...@gmail.com wrote:
algo fails for following input set
20,18,1,7,2
Thanks and regards,
Gajendra Dadheech
On Sun, Feb 6,
My solution in more detail (in words ):-
start from the end if you get an L
make a node with data L and push its pointer in stack,
if you get a M pop two elements from stack
make them left and right children of M and then push
back m's pointer to stack(if stack is empty then stack returns NULL
Write a function to print all unique partitions on n tht are of size m. eg:
n=10, m=4. it should print 7 1 1 1, 6 2 1 1, 5 3 1 1, 3 3 2 2,, so on
Thanks and regards,
Gajendra Dadheech
hi
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Any idea about whats the probability of getting rejected by *Hiring* *
Committee*?
-Regards
*Amit Agarwal http:///www.amitagrwal.com*
+91-779-822-8765
On Sun, Feb 6, 2011 at 8:38 PM, Badrinath Kulkarni urba...@gmail.comwrote:
what s the o/p of the followin pgm?
main()
{
int i=300;
char *ptr;
ptr=i;
*++ptr=2;
printf(%d,i);
}
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i think it will be 34...
Thanks and regards,
Gajendra Dadheech
On Sun, Feb 6, 2011 at 10:15 PM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
@ranjane
the value of i will not change as ++ptr will execute first.
but i wonder if a char pointer can be used to store integer address... will
i do'nt know why it is printing 556.
could anyone give me the logic.
Aditya Pratap
MCA II
Delhi University
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@aditya which compiler are you using
On Sun, Feb 6, 2011 at 10:21 PM, aditya pratap
contacttoadity...@gmail.comwrote:
i do'nt know why it is printing 556.
could anyone give me the logic.
Aditya Pratap
MCA II
Delhi University
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On Jan 31, 11:17 am, ritu ritugarg.c...@gmail.com wrote:
@Ralph
Build a data structure on array B[1..n] in O(n) time such that
the following problem can be solved in O(log n) time:
Given an index i and value v, find the index j of the first
element in B such that j = i and
@jalaj : gcc compiler.
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but the tree can be created from a preorder walk is more than 1 right? so
the question only ask for 1?
On Feb 6, 2011 7:31 AM, jalaj jaiswal jalaj.jaiswa...@gmail.com wrote:
My solution in more detail (in words ):-
start from the end if you get an L
make a node with data L and push its pointer
the logic is :-
int is stored in 32 bits in our systems
300 is 0001 00101100
as ptr is character pointer, it points to lower 8 bits
and when *++ptr=2 gets executed then 0001 changes to 0010(equal to
2)
so i becomes 0010 00101100 which is 556 :D
no unique tree will get created i think .. as first popped is left child and
second popped is right child in algo
On Sun, Feb 6, 2011 at 10:29 PM, yq Zhang zhangyunq...@gmail.com wrote:
but the tree can be created from a preorder walk is more than 1 right? so
the question only ask for 1?
On
Input: A unsorted array of size n.
Output: An array of size n.
Relationship:
elements of input array and output array have 1:1 correspondence.
output[i] is equal to the input[j] (ji) which is smaller than input[i]
and jth is nearest to ith ( i.e. first element which is smaller).
If no such
Both of you are right. Thanks for correcting my mistake.
Dave
On Feb 6, 8:10 am, radha krishnan radhakrishnance...@gmail.com
wrote:
that should be (a+b+abs(a-b))/2
On Sun, Feb 6, 2011 at 4:15 PM, Manmeet Singh mans.aus...@gmail.com wrote:
test case : a = -1, b = - 1
Result : fight
using this macro
size(X) ((X*)0+1)
if we give size(int) it ll return 4.
size(double) gives 8.
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got a O(n) soln guys
Starts from an empty stack and scan the input array (IN) from the bottom (i
= n-1):
for(int i = n-1; i = 0; i--) {
while (!stack.isEmpty() IN[i] = stack.top()) {
stack.pop();
}
if (stack.isEmpty()) {
OUT[i] = 0;
stack.add(IN[i]);
continue;
}
how do we work for a rectangle here..
the only way i could think of is once we have found a square, backtrack from
right bottom towards left and up to see if the value is not changing
10101
11100
1
11000
it will be
10101
11100
12211
12000
now here if (3,1) is considered, there there is a
you are given a bst where each node has a int value , parent pointer , and
left and right pointers , write a function to find a path with a given sum
value. Path can go from left subtree tree , include root and go to right
tree as well . we need to find these paths also .
5
/ \
110
/ \/ \
Someone please share The C Programming Language, 2nd edition, Kernighan
and Ritchie Solutions
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public static ArrayListArrayListInteger Partition(int val, int start,
int size) {
ArrayListArrayListInteger r = new ArrayListArrayListInteger();
if (start * size val) {
return null;
}
if(size == 1)
{
r.add(new ArrayListInteger());
r.get(0).add(val);
return r;
}
for
If duplicate values are allowed ::
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class PartionNumber {
public static void main(String[] args) {
System.out.println(Enter the number);
InputStreamReader ifn = new InputStreamReader(System.in);
There is a language with only two elements ie 'a' and 'bc'. How many words
can be
made out of it if whose Length is n.
I think it is n+1.
if the len = 3
abc
bca
if the len is 6
bc
aaabca
aabcaa
abcaaa
ba
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There is a firm that provides Mobile Operator 3 functions.
1. char[] get_number: to get a new number(10 digit)
2. bool is_allocated(char[]) : if the number is already allocated.
3. bool allocate(char[]) : allocate the number.
What data structure to use?
We can use hash table to store
If [a1,a2,a3...,an,b1,b2...bn] is given input change this to
[a1,b1,a2,b2.an,bn]
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For len = 6, isn't bcbcbc also valid?
On Sun, Feb 6, 2011 at 9:08 PM, Anand anandut2...@gmail.com wrote:
There is a language with only two elements ie 'a' and 'bc'. How many words
can be
made out of it if whose Length is n.
I think it is n+1.
if the len = 3
abc
bca
if the len is 6
certainly it is not n+1.
for n=1, ans is 'a'.No other word can be formed.
On Sun, Feb 6, 2011 at 7:10 PM, nphard nphard nphard.nph...@gmail.comwrote:
For len = 6, isn't bcbcbc also valid?
On Sun, Feb 6, 2011 at 9:08 PM, Anand anandut2...@gmail.com wrote:
There is a language with only two
@Anand: Let N(k) be the number of strings of length k. Since you can
make a string of length k by appending 'a' onto a string of length k-1
or by appending 'bc' onto a string of length k-2, it follows that
N(k) = N(k-1) + N(k-2)
Furthermore, there are no strings of length -1 and only one string
Can someone explain the logic behind this
Thanks and regards,
Gajendra Dadheech
On Mon, Feb 7, 2011 at 7:10 AM, Sreeprasad Govindankutty
sreeprasad...@gmail.com wrote:
If duplicate values are allowed ::
import java.io.BufferedReader;
import java.io.InputStreamReader;
public
ya i also want the explaination of gcc compiler output...thanx in
advance
On Sun, Feb 6, 2011 at 10:13 AM, tech rascal techrascal...@gmail.comwrote:
but can nybody explain why the output is coming like this on gcc compiler??
On Sun, Feb 6, 2011 at 10:04 AM, Manish Verma
int a=10,b;
b=a++ + ++a; (Till here I think it is clear the value of a is 12 and b = 22)
printf(%d,%d,%d,%d,b,a++,a,++a); (Evaluate from right to left. So ++a
makes the value of a as 13, then a and then a++ which is post increment
still makes a as 13)
Now it try to display the value from left
@anand:while printing post fix gave 13 but rest two why 14
On Mon, Feb 7, 2011 at 10:28 AM, Anand anandut2...@gmail.com wrote:
int a=10,b;
b=a++ + ++a; (Till here I think it is clear the value of a is 12 and b =
22)
printf(%d,%d,%d,%d,b,a++,a,++a); (Evaluate from right to left. So ++a
a++ when it gets printed. it is 13 and now it gets increment to 14. So now
if you print a it will 14.
On Sun, Feb 6, 2011 at 9:04 PM, jagannath prasad das jpdasi...@gmail.comwrote:
@anand:while printing post fix gave 13 but rest two why 14
On Mon, Feb 7, 2011 at 10:28 AM, Anand
Hi
Try following code
Suppose we need to find size of variable *obj*
int size = (char*)(obj +1 ) - (char*)(obj);
*
*Thanks Regards,
Ricky
On Mon, Feb 7, 2011 at 12:19 AM, albert theboss alberttheb...@gmail.comwrote:
using this macro
size(X) ((X*)0+1)
if we give size(int) it ll
@juver:in f(a)+f(b) i guess f(a) is evaluated first because fucntion calls
are evaluated from left to right
On Mon, Feb 7, 2011 at 10:37 AM, Anand anandut2...@gmail.com wrote:
a++ when it gets printed. it is 13 and now it gets increment to 14. So now
if you print a it will 14.
On Sun, Feb
@rajiv:i guess obj is a pointer here
On Mon, Feb 7, 2011 at 10:51 AM, Rajiv Podar rajeevpo...@gmail.com wrote:
Hi
Try following code
Suppose we need to find size of variable *obj*
int size = (char*)(obj +1 ) - (char*)(obj);
*
*Thanks Regards,
Ricky
On Mon, Feb 7, 2011 at
A very common interview question
let the array be from 0 to 2n-1
now,
every element of array has its initial position and final position.. start
from beginning
if the elemnt you r processing is the first half of array then f=i+i;
else f=2*i-(2n-1);
replace the elemnt at final position with the
@jagan: that's true obj is a pointer to the data type for which size
need to be calculated.
On Feb 7, 10:38 am, jagannath prasad das jpdasi...@gmail.com wrote:
@rajiv:i guess obj is a pointer here
On Mon, Feb 7, 2011 at 10:51 AM, Rajiv Podar rajeevpo...@gmail.com wrote:
Hi
write the program to add two numbers without using arithmetic and bit
operation..
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@Ricky: if increment and decrement operators are not considered
arithmetic, try
int sum(int m, int n)
{
while( m 0 )
{
m--;
n++;
}
while( m 0 )
{
m++;
n--;
}
return n;
}
On Feb 6, 11:49 pm, Ricky rajeevpo...@gmail.com wrote:
write
try this
let nos be m n
char * p;
p=m;
int sum = (int)p[n] ;
sum is m+n :)
On Mon, Feb 7, 2011 at 11:41 AM, Dave dave_and_da...@juno.com wrote:
@Ricky: if increment and decrement operators are not considered
arithmetic, try
int sum(int m, int n)
{
while( m 0 )
{
m--;
@Dave: ++ and -- are arithmetic operations.
@Jalaj: I agree with the above solution
Thanks Regards,
Rajiv Podar
On Mon, Feb 7, 2011 at 11:47 AM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
try this
let nos be m n
char * p;
p=m;
int sum = (int)p[n] ;
sum is m+n :)
On Mon,
bit operation means what?
On Mon, Feb 7, 2011 at 12:00 PM, Rajiv Podar rajeevpo...@gmail.com wrote:
@Dave: ++ and -- are arithmetic operations.
@Jalaj: I agree with the above solution
Thanks Regards,
Rajiv Podar
On Mon, Feb 7, 2011 at 11:47 AM, jalaj jaiswal
for e.g. u can use operator.
int a = 1;
a = a 1;
a become 2;
Or u cannot use any , |, ^ operations.
Thanks Regards,
Rajiv Podar
On Mon, Feb 7, 2011 at 12:33 PM, jagannath prasad das
jpdasi...@gmail.comwrote:
bit operation means what?
On Mon, Feb 7, 2011 at 12:00 PM, Rajiv Podar
thank u jalaj
On Feb 6, 10:03 pm, jalaj jaiswal jalaj.jaiswa...@gmail.com wrote:
the logic is :-
int is stored in 32 bits in our systems
300 is 0001 00101100
as ptr is character pointer, it points to lower 8 bits
and when *++ptr=2 gets executed then 0001 changes to
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