Re: [algogeeks] Re: search in O(logn)
smthng like this...
*if*(arr[mid] arr[mid + 1])
return mid;
if(arr[low] arr[mid])
return findPoint(arr, low, mid-1);
else
return findPoint(arr, mid + 1, high);
On Sat, Jun 9, 2012 at 12:36 AM, Hassan Monfared hmonfa...@gmail.comwrote:
Hi pharta :
find the point where it is rotated to get 14-1 O(log(n)) how can
you find rotation point in log(n) ?
On Fri, Jun 8, 2012 at 6:57 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote:
It is easy.. find the point where it is rotated to get 14-1 O(log(n))
since 214 that means u have to find it in subarray [123].. do a binary
search here o(long(n))
final 2*O(log(n))...
On Fri, Jun 8, 2012 at 7:44 PM, Dave dave_and_da...@juno.com wrote:
@Hassan: This is not possible without additional conditions. E.g., you
cannot find the 1 in the array {0,0,0,...,0,1,0,0,...,0} in less than O(n)
time.
But with the condition that the elements of the array are pairwise
distinct, you can use a binary search to find k such that a[k-1] a[0] and
a[k] a[0]. Then if x a[k], do a binary search to find x in {a[k] ...
a[n-1]}; otherwise do a binary search to find x in {a[0] ... a[k]}.
Dave
On Friday, June 8, 2012 8:41:47 AM UTC-5, Hassan Monfared wrote:
A sorted array of integers is rotated unknown times. find an item in
O(logn) time and O(1) space complexity.
for example : 1,2,3,7,10,14 ---rotated 3 times-- 7,10,14,1,2,3
find 2 in O(logn) time in O(1) space complexity
Regards
Hassan H. Monfared
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Re: [algogeeks] first Repeating character in a string
@saurabh: why would u count all??? just see while counting if the bitmap is set.. then return the char. On Fri, Jun 8, 2012 at 4:33 PM, Saurabh Yadav saurabh...@gmail.com wrote: order of hashing and counting is important eg. abba if we do hashing by characters 'a' is stored before 'b' and count of both is 2 at the end and when we process this we give result 'a' (because 'a' comes before 'b' )which is wrong because 'b' is the first repeated character. Thanks Regards Saurabh -- 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. -- 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: search in O(logn)
It is easy.. find the point where it is rotated to get 14-1 O(log(n))
since 214 that means u have to find it in subarray [123].. do a binary
search here o(long(n))
final 2*O(log(n))...
On Fri, Jun 8, 2012 at 7:44 PM, Dave dave_and_da...@juno.com wrote:
@Hassan: This is not possible without additional conditions. E.g., you
cannot find the 1 in the array {0,0,0,...,0,1,0,0,...,0} in less than O(n)
time.
But with the condition that the elements of the array are pairwise
distinct, you can use a binary search to find k such that a[k-1] a[0] and
a[k] a[0]. Then if x a[k], do a binary search to find x in {a[k] ...
a[n-1]}; otherwise do a binary search to find x in {a[0] ... a[k]}.
Dave
On Friday, June 8, 2012 8:41:47 AM UTC-5, Hassan Monfared wrote:
A sorted array of integers is rotated unknown times. find an item in
O(logn) time and O(1) space complexity.
for example : 1,2,3,7,10,14 ---rotated 3 times-- 7,10,14,1,2,3
find 2 in O(logn) time in O(1) space complexity
Regards
Hassan H. Monfared
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Re: [algogeeks] Re: interview HARD problem
no if ta/om/to/am/ba/batot/amoma are words then it should be the one. no 5*2 3*3 On Tue, Jun 5, 2012 at 7:45 PM, Gene gene.ress...@gmail.com wrote: Does this sufficae? Suppose you were using a dictionary from the frapplewonk language, which has only 5 words: tab oma to am ba Then the biggest rectangle is clearly tab oma On Jun 4, 10:39 pm, Ashish Goel ashg...@gmail.com wrote: preparing a sample itself is a great problem here, that is why i called it hard all words in the rectangle horizontally as well as vertically needs to be valid dictionary words Ashish Hassan say this rectangle AH,SA,HS,IS,SA,HN should also be valid dictonary words, indeed they are not.. definitely we will need a multimap to have words of same length forming a bucket..not able to think beyond this Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 On Mon, Jun 4, 2012 at 6:38 PM, Hassan Monfared hmonfa...@gmail.com wrote: Give a sample please On Mon, Jun 4, 2012 at 5:34 PM, Ashish Goel ashg...@gmail.com wrote: Given a dictionary of millions of words, give an algorithm to find the largest possible rectangle of letter that every row forms a word(reading left to right) and every column forms a word(reading from top to bottom). Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 -- 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. -- 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. -- 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. -- 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] Google Q : all anagrams next to each other
@ashish : couldnt get u.. can u give an example?? On Tue, May 22, 2012 at 5:45 PM, Prem Krishna Chettri hprem...@gmail.comwrote: What I Could possibly think of is For each string S1 that is an anagram of some string S, use Map and Store the Key Value as (S1,S). Now there is a trick here abt how to reduce Time Complexity here... Now its easy to put all string which has correspondence S next to each other. This is Simple one. Inplace.. Hv to think abt .. I doubt, as we need some space to get the anagrams Dude.. Prem On Tue, May 22, 2012 at 5:18 PM, Ashish Goel ashg...@gmail.com wrote: Write a method to sort an array of strings so that all the anagrams are next to each other. What i could think of is preparing a multi linked list( multimap) whereby the key for each string is the sorted representation of the string(eg if string is gac, its sorted representation is acg). Walk of all lists of this multimap to give all anagrams. Is there any other better solution for this problem? Can this be done *inplace*? Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 -- 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. -- 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. -- 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] amazon
Java has something call treeMap. it stores strings lexographically.. u can
always do permutations and store them in a treeMap. and get the rank
then... just the idea.. will post the solution once i am done.. what do u
guys think.abt the idea???
On Tue, May 22, 2012 at 9:46 AM, atul anand atul.87fri...@gmail.com wrote:
actually we can think of above approach as follows :-
input : cabd
sort(input) = abcd
find first element of the input i.e 'c' in the sorted form i.e abcd
'c' is at found_index=3 ( index starting from 1 )
so permutaion stating from 'c' will come after temp_rank=(found_index -
start_index) * (total_lentgh - 1) !
now after temp_rank we know that permutation starting with 'c' will come.
so to find out the exact index sort(input-1) i.e removing 1st character
('c') from the input(cadb) we will get abd
now permute string abd and compare with input - 1 character i.e (abd).
and keep inc the counter , if match is found then you have to add this
counter to temp_rank.
so for the input = cabd
temp_rank = (3 - 1) * (4-1) !
= 2 * 3!
= 12
counter found c = 1;
rank = 12 + c = 13
but i dont think this would be good solution as be have to permute string
and then compare at last...yes we did some optimization.
i was wondering if instead of permuting at the end , we can calculate
minimum number of swaps required to convert input - 1 to sorted input -1
(removing 1st character )using edit distance ...will this work?? .
On Mon, May 21, 2012 at 11:33 PM, atul anand atul.87fri...@gmail.comwrote:
consider string input = cabd
now sort this string = abcd
now search 1st character from original string(cabd) in sorted string
abcd... index found = 3 (index starting from 1 )
now you can arrange left elements from found index i.e index-1 elements
in (index-1) ! and right elements from found index in (lastIndex - index)!
before character 'c' occurs at index 0. similarly find for other characters
and at the end subtract it from n! (n = length of the string ) to find rank
On Mon, May 21, 2012 at 11:02 PM, rahul venkat rahul911...@gmail.comwrote:
Given a string. Tell its rank among all its permutations sorted
lexicographically.
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Re: [algogeeks] symmetric binary tree
recurse on left and right will do it and keep comparing the values On Tue, May 22, 2012 at 2:50 PM, atul anand atul.87fri...@gmail.com wrote: no need of creating another mirror tree you just need to call the function func(root-left,root-right); now left sub tree and right sub tree will be considered as if you are checking 2 different trees same code to check if 2 tree are structurally similar will work. On Tue, May 22, 2012 at 10:50 AM, algogeek dayanidhi.haris...@gmail.comwrote: How to check if a given binary tree is structurally symmetric ie. the left sub tree should be mirror image of right sub tree and vice-versa. -- 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. -- 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. -- 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] amazon
sorry it was treeset Here is the code..
public class asd1 {
public static TreeSetString t = new TreeSetString();
public static int j = 0;
public static void main(String args[]) {
String s= edcba;
permute(, s);
System.out.println(t.toString());
int length=t.size();
String[] arr=(String[]) t.toArray(new String[length]);
for(int i =0;ilength;i++)
{
System.out.println(arr[i]);
if(arr[i].equals(s)){
System.out.println(i+1);
break;
}
}
}
public static void permute(String st, String chars) {
if (chars.length() = 1)
t.add(st+chars);
else
for (int i = 0; i chars.length(); i++) {
try {
String newString = chars.substring(0, i)
+ chars.substring(i + 1);
permute(st + chars.charAt(i), newString);
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
On Tue, May 22, 2012 at 2:19 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote:
Java has something call treeMap. it stores strings lexographically.. u can
always do permutations and store them in a treeMap. and get the rank
then... just the idea.. will post the solution once i am done.. what do u
guys think.abt the idea???
On Tue, May 22, 2012 at 9:46 AM, atul anand atul.87fri...@gmail.comwrote:
actually we can think of above approach as follows :-
input : cabd
sort(input) = abcd
find first element of the input i.e 'c' in the sorted form i.e abcd
'c' is at found_index=3 ( index starting from 1 )
so permutaion stating from 'c' will come after temp_rank=(found_index -
start_index) * (total_lentgh - 1) !
now after temp_rank we know that permutation starting with 'c' will come.
so to find out the exact index sort(input-1) i.e removing 1st character
('c') from the input(cadb) we will get abd
now permute string abd and compare with input - 1 character i.e (abd).
and keep inc the counter , if match is found then you have to add this
counter to temp_rank.
so for the input = cabd
temp_rank = (3 - 1) * (4-1) !
= 2 * 3!
= 12
counter found c = 1;
rank = 12 + c = 13
but i dont think this would be good solution as be have to permute string
and then compare at last...yes we did some optimization.
i was wondering if instead of permuting at the end , we can calculate
minimum number of swaps required to convert input - 1 to sorted input -1
(removing 1st character )using edit distance ...will this work?? .
On Mon, May 21, 2012 at 11:33 PM, atul anand atul.87fri...@gmail.comwrote:
consider string input = cabd
now sort this string = abcd
now search 1st character from original string(cabd) in sorted string
abcd... index found = 3 (index starting from 1 )
now you can arrange left elements from found index i.e index-1 elements
in (index-1) ! and right elements from found index in (lastIndex - index)!
before character 'c' occurs at index 0. similarly find for other characters
and at the end subtract it from n! (n = length of the string ) to find rank
On Mon, May 21, 2012 at 11:02 PM, rahul venkat rahul911...@gmail.comwrote:
Given a string. Tell its rank among all its permutations sorted
lexicographically.
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Re: [algogeeks] Re: Microsoft interview question
@ashish.. it wont be constant space then.. surely it will be o(n) though
On Mon, May 21, 2012 at 7:23 PM, Ashish Goel ashg...@gmail.com wrote:
Dave,
Cant we have a hash table with the item as key and its count as value
(walk over array A and build HT).
For permutation check, walk over second array and start reducing the count
and remove when count becomes zero for that particular key. If some char
not there in HT, return false, else return true.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Mon, May 21, 2012 at 6:14 PM, Dave dave_and_da...@juno.com wrote:
@Piyush: Did you even try this on any examples? If not, try a = {0,1,2,3}
and b = {0,2,2,2}.
Dave
On Sunday, May 20, 2012 1:39:25 AM UTC-5, Kalyan wrote:
Piyush. I think we can use your logic. But You should check the product
also.
Have 4 variables, sum_a,sum_b , prod_a, prod_b
Calculate Sum and product of array 'a' and store it in sum_a,prod_a
Calculate Sum and product of array 'b' and store it in sum_b,prod_b
if sum_a=sum_b prod_a==prod_b then these 2 arrays are permutations
of each other.
Space = O(1)
Time=O(n)
I think this should work. Please correct me if you find mistakes.
On 5/20/12, anuj agarwal coolbuddy...@gmail.com wrote:
U are checking if the sum is same or not.. which can be same even if
the
elements are different.
On Sun, May 20, 2012 at 11:54 AM, Piyush Khandelwal
piyushkhandelwal...@gmail.com wrote:
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [ 3,5,4 ]
diff=0;
for (i=0; in;i++)
{ diff= diff+a[i]-b[i];
}
if diff == 0
print: permutation
else
print: not permutation
On 20 May 2012 07:2 0, Dave dave_and_da...@juno.com wrote:
@Harshit: These are a few unanswered questions that came to mind
when I
read your solution attempt: What do you do with negative elements?
What
is
the -12th prime number? How do you deal with overflow in the cases
where
you have a lot of large prime numbers and the product exceeds your
native
data types?
Dave
On Saturday, May 19, 2012 2:29:52 PM UTC-5, harshit pahuja wrote:
given 2 unsorted integer arrays a and b of equal size. Determine if
b is
a permutation of a. Can this be done in O(n) time and O(1) space ?
please help me with my solution
suppose a -- 3 5 4
b -- 4 3 5
now we replace a[i] with a[i]..th prime number and b with b[i] ..
th
prime number
now array a becomes 5 11 7
array b becomes 7 5 11
now we take product of elements of array a and do the same with
array b
elements
if product is equal then b is a permutation of a
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Re: [algogeeks] Re: Microsoft interview question
a[] = [-1,-3,4,0,7,0,36,2,-3] b[] = [0,0,6,2,-1,9,28,1,6] b1[] = [0,7,0,36,4,-6,3,0,0] b2[] =[-1,-3,11,0,0,0,35,0,0] suma = 42 proda = -84*72*3 sumb = 51 prodb = -84*72*3 sumb1 = 44 prodb1 = -84*72*3 sumb2 = 42 prodb2 = 33*35 do the sum and prod operation w/o 0s and compare the values.. if both are equal they are pormutations if i am missing any corner cases related to 0 or -e numbers u can keep a track of them while traversalO(N) and constant space On Mon, May 21, 2012 at 6:40 PM, karthikeya s karthikeya.a...@gmail.comwrote: No way u can do it in O(1) space and O(n) time.sols above are not gonna work..yeah, it is possible in O(n) space and O(n) time. On May 20, 12:29 am, HARSHIT PAHUJA hpahuja.mn...@gmail.com wrote: given 2 unsorted integer arrays a and b of equal size. Determine if b is a permutation of a. Can this be done in O(n) time and O(1) space ? please help me with my solution suppose a -- 3 5 4 b -- 4 3 5 now we replace a[i] with a[i]..th prime number and b with b[i] .. th prime number now array a becomes 5 11 7 array b becomes 7 5 11 now we take product of elements of array a and do the same with array b elements if product is equal then b is a permutation of a -- 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. -- 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.
