[algogeeks] Sites for Interview Questions
Hi, Can someone suggest good books/websites/blogs for interview related questions. thanks-- YS -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Number problem
My code failed because it was reversing the digits and thus 0 was getting added in the front which resulted in nothing. If allowed we can use a char array else will have to find a solution which does not reverse the digits of the number. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: do this: two numbers with min diff
I hope this will work. It finds the two minimum numbers and then prints the
difference.
#include stdio.h
#include stdlib.h
void find_two_mins(int array[], int size)
{
int i,t;
int min1=array[0], min2=array[1];
for (i=2; isize; i++){
t=array[i];
if(t=min1) {
min2=min1;
min1=t;
continue;
}
}
printf(\nMin elements are 1: %d 2: %d, min1,min2);
printf(\n Absolute difference between two minimum elements are %d,
abs(min1-min2));
}
int main()
{
//int array[10]={ 9,2,3,4,1,7,5,6,8,0};
//
int array[10]={99,12,45,33,88,9098,112,33455,678,3};
find_two_mins(array,10);
return 0;
}
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[algogeeks] Re: Unbounded dictionary lookup
In case of a dictionary, can we assume that its a Sorted list of
words?
On Sep 22, 12:42 pm, ramdas kale ramda...@gmail.com wrote:
If we take high = MAX_CAPACITY
Here MAX_CAPACITY denotes the maximum no of words dictinary can index.
Actual no of words stored in dictionary could be less than MAX_CAPACITY.
On Wed, Sep 22, 2010 at 1:23 AM, Minotauraus anike...@gmail.com wrote:
high= const.(10^const)
What's const? The point of this isn't that it's a difficult prob to
solve. Point lies in working with the design to make this close to log
n.
Define what value const holds.
On Sep 21, 9:12 am, coolfrog$ dixit.coolfrog.div...@gmail.com
wrote:
its dictionary means shorted ordered arry.
let low = 1; and high= const.(10^const)
Boolean isWord(String word)
{ while(low = high)
{ mid = (low+ high)/2;
if(word = getWordAt(mid))
return true;
if( word getWordAt(mid))
{ high = mid-1
}
else
low = mid+1;
}
}
Its a simple Binary Search Algorithm ...
who's complexity is O(log n) times.
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+919983526790
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[algogeeks] Re: Inserting a box with lesser dimension into a box of bigger dimensions than that.
We can first sort the dimensions of each box.
For example if the dimensions of a box is L=10, B=12, W=6 then convert
it to L=12, B=10, W=6.
The above step is not needed if the problem states that LBH for all
boxes.
Then using a multi-key sorting we can sort all the boxes in ascending
order (or descending order).
If we sort it in ascending order:
This will give us a sorted list where in each box in the list is
smaller or equal to the next one in the list.
The boxes will fit inside the other one only if all the dimensions are
less than the other one.
Lets us have the boxes in list LIST in the sorted order (multi key)
//START
init_list (current)
init_list (discarded)
Start: // its a label.
for i = 0 to length_of_list; do //don't increment i here
T1=LIST[i]
T2=LIST[i+1]
if ( T1.L T2.L T1.B T2.B T1.W T2.W ) {
add_to_tail (current, T1 , T2)
i=i+2;
} else {
add_to_tail (discarded, T2)
i++;
}
done // for loop ends
print_list(current) // here is your one solution.
if ( elements in discarded list) {
LIST=discarded
goto Start // goto the label start
}
///END
On Sep 22, 12:11 am, Dave dave_and_da...@juno.com wrote:
Certainly having a smaller volume is necessary for a box to fit in
another box, but it is not sufficient. E.g., a box of size 1 x 1 x 1
will not fit in a box of size 2 x 2 x 1/2.
Dave
On Sep 21, 1:16 pm, rajess rajeshrules...@yahoo.com wrote:
find the volume of boxes as v=l*b*h
sort boxes in volumes in descending order and this is the way to
insert boxes one into another
On Sep 21, 7:55 pm, Rashmi Shrivastava rash...@gmail.com wrote:
If there are n number of boxes and each with different dimensions and your
job is to insert one box having lesser dimension than that to another.
Consider size of boxes as,
b1-s1(h1,l1,w1)
b2-s2(h2,l2,w2)
.
.
.
bn-sn(hn,ln,wn)- Hide quoted text -
- Show quoted text -
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[algogeeks] Re: Print 1 to n one per each line on the standard output
The question is to get an algorithm or a program. If it's a program then using execl() or fork() system call you can code this. The code will not be a recursion in a sense that we will not call functions recursively but will call the programs (executable code) recursively. On Sep 22, 8:49 pm, Divesh Dixit dixit.coolfrog.div...@gmail.com wrote: Write an algorithm that will print 1 to n, one per each line on the standard output, where n is a integer parameter to the algorithm. An algorithm should not use while, for, do-while loops, goto statement, recursion, and switch statement. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
