Please send a note to me on king...@amazon.com
Thanks,
Kingston
On Fri, Jul 16, 2021 at 11:16 AM immanuel kingston <
kingston.imman...@gmail.com> wrote:
> Hi all,
>
> I am a hiring manager at Amazon. We are hiring for SDE and Applied Science
> roles in my team. Please send
Hi all,
I am a hiring manager at Amazon. We are hiring for SDE and Applied Science
roles in my team. Please send a short note about yourself, the role you
wish to apply for and your updated CV.
Thanks,
Kingston
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Our systems enable Amazon to launch
There are openings in my team for SDE-Is, SDE-IIs, SDM and QAE-1.
Please send me resumes if you or your friends are interested in any of
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@DK and Amit, thanks for correcting my understanding.
On Sun, Aug 7, 2011 at 3:51 PM, amit karmakar amit.codenam...@gmail.comwrote:
@DK
Hmm, i do understand what you said. Maybe, i should make it clear that
i just wanted to tell that implementing a non-recursive merge-sort
will not require
., take the rectangles [(0,0), (0,3),
(3,3), (3,0)] and [(2,5), (5,2), (7,4), (4,7)].
The standard test is the Separating Axis Test. You can find this with
google.
Dave
On Aug 6, 11:38 am, immanuel kingston kingston.imman...@gmail.com
wrote:
Algo goes something like this.
Rectangles
Algo goes something like this.
Rectangles are represented by two points TopLeft and bottomRight or
BottomLeft and TopRight
1. Choose Xmin and Xmax from Rectangle A. Check if any of the x-coordinates
of rectangle B fall in between Xmin and Xmax of A.
2. choose Ymin and Ymax from Rectangle A.
Yes. just remove the recursive part using 2 stacks.
Thanks,
Immanuel
On Fri, Aug 5, 2011 at 6:51 PM, Nitin Nizhawan nitin.nizha...@gmail.comwrote:
does anyone know of any in-place, iterative mergesort algorithm with nlogN
worst case complexity? It would be good if it is stable also.
TIA
Trie is better in terms of scalability and performance.
With Hashtable there is a problem of rehashing when all buckets are full and
rehashing takes O(N). Although it happens once in a blue moon. That can
impact the performance in a production environment.
With trie you dont have that problem.
I am thinking the following algo will work. Please correct me if i am wrong.
void interleaveAndPrint(char * result, char * str1, char * str2, int i ) {
if (*str1 == '\0' *str2 == '\0') {
result[i] = '\0';
printf(%s\n, result);
return;
}
First Player can always win.
For each heap
Pick heap-size - 1 coins if this is not the n-1th heap
Pick all coins from the heap if this the n-1th heap.
Please correct me if i am wrong.
Thanks,
Immanuel
On Wed, Jun 15, 2011 at 3:13 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
*There
Use a trie data structure and pre-load it with all the words of a
dictionary.
Thanks,
Immanuel
On Wed, Jun 15, 2011 at 3:06 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
*How will you design a SpellChecker for an e-mail application?*
--
*Piyush Sinha*
*IIIT, Allahabad*
The following is for LCA for 2 nodes in a n-ary tree.
A more tougher problem is to find the LCA for n nodes in the same n-ary
tree.
Node * findLCA (Node *root, Node * l, Node * r, int n) {
if (l == null || r == null) return root;
if (root == null) return null;
if (isChild(root,l) ||
.
n = 2;
heap 1 - no of coins 1
heap 2 - no of coins 2
On Wed, Jun 15, 2011 at 5:34 PM, sunny agrawal sunny816.i...@gmail.comwrote:
i think u r wrong
what if heap size -1 is 0
i think one should pick atleast one coin else game will draw
On Wed, Jun 15, 2011 at 5:17 PM, immanuel
will win i think
On Wed, Jun 15, 2011 at 6:26 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
Yes. I am wrong. As per the example, Player 2 will win if he plays
efficiently.
Let me put my solution this way,
If all the the heaps are of size 1 the Player 1 can win always.
Thanks
@Nitish,
I think it fails for this condition
4 heaps with 1,2,1,2
Player 1 starts first with picking 1 coin from heap 1
Player 2 picks 2 coins from heap 2
Player 1 picks 1 coin from heap 3
Player 2 picks 2 coins from heap 4.
Player 2 wins but XOR of the number of coins in each heap is 0(if that
yep true.
The difficult part is trying to find when Player 1 can win with heaps of
size 1.
Thanks,
Immanuel
On Wed, Jun 15, 2011 at 6:59 PM, sunny agrawal sunny816.i...@gmail.comwrote:
i think solution depends on no of heaps having single coin
if there are even number of such heaps player 1
Modified Piyush's soln to handle the above case.
int get_pivot(int a [ ],int low, int high)
{
if (low high)
{
int mid = (low+high)/2;
if(a[mid]a[mid+1])
return mid+1;
if(a[low]a[mid])
return (get_pivot(a,low,mid-1));
else
@Ankit, The input should be 2 sorted arrays
Thanks,
Immanuel
On Sun, May 29, 2011 at 10:48 AM, ankit sambyal ankitsamb...@gmail.comwrote:
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct
me if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross
Mark
On Thu, May 26, 2011 at 1:05 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*Mystery Puzzle Sherlock Holmes
*
*
*
**
*One snowy night, Sherlock Holmes was in his house sitting by a fire. All
of a sudden a snowball came crashing through his window, breaking it.
Holmes got up and
http://en.wikipedia.org/wiki/Damerau%E2%80%93Levenshtein_distance
On Wed, May 25, 2011 at 11:50 AM, Akshata Sharma
akshatasharm...@gmail.comwrote:
still not getting!! :(
On Wed, May 25, 2011 at 11:38 AM, Aakash Johari aakashj@gmail.comwrote:
Sorry, it can't be because of the
Brute force Approach would be
int checkForTriangle(int a, int b, int c) {
return (a + b c) (b + c a) (a + c b);
}
int triangle (int a[], int n) {
if (a == null || n = 0) return 0;
for (int i=0 ; i n ; i++) {
for (int j=i + 1; j n; j++ ) {
for (int
#define MAX_BITS_FOR_INT=32;
int getIthBit(int n, int i) {
return (n 1i) i;
}
bool isPalindrome(int num) {
int i=0;
int j= MAX_BITS_FOR_INT - 1;
while (i j) {
int ithbit = getIthBit(num, i);
int jthbit = getIthBit(num, j);
if ( ithbit ^ jthbit)
...
For example, taking 10's binary representation as 1010...according to
question it wil be a palindrome...but according to ur algo it will
return false...
On 5/24/11, immanuel kingston kingston.imman...@gmail.com wrote:
#define MAX_BITS_FOR_INT=32;
int getIthBit(int n, int i) {
return
correct me if I am wrong.
String convertFloatToBinary(float num) {
String str = ;
int numBeforeDecimal = (int)num;
float decimalPart = num - (float)numBeforeDecimal;
int sign=1;
if (numBeforeDecimal 0 ) sign = -1;
if (sign 0) str[str.length] = '-';
@sravan,
What i meant was get the jth digit in the representation of i to the base
NUM_PER_DIGIT. ie 3rd digit of (2137)8 which is 2.. 7 is the 0th digit, 3
being 1st digit, 1 being 2nd digit and 2 being 3rd digit.Here 2137 is a base
8 representation.
Thanks,
Immanuel
On Tue, May 24, 2011 at
int fibArray[INTEGER_MAX_VALUE] = {0};
int fibonacci (int n) {
if (n = 0) {
return 0;
} else if ( n 0 fibArray[n] != 0) {
return fibArray[n];
} else {
if (n == 1) return (fibArray[n] = 1);
return (fibArray[n] = fibonacci(n - 1) + fibonacci(n-2));
http://www.java2s.com/Tutorial/Cpp/0180__Class/Initializeanarrayofobjectsbyreferencingtheconstructordirectly.htm
http://www.java2s.com/Tutorial/Cpp/0180__Class/Initializeanarrayofobjectswithoutreferencingtheconstructordirectly.htm
Thanks,
Immanuel
On Wed, May 25, 2011 at 7:34 AM,
Extending the above soln:
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {abc,def,...};
char n[] = 2156169 (number is pressed);
int k=7;
for i -- 0 to NUM_PER_DIGIT ^ k
String s=;
for j -- 0 to k
int index = getJthDigitinItotheBaseNumPerDigit(NUM_PER_DIGIT,i,j);
// ie get 1st digit
at 9:33 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
Extending the above soln:
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {abc,def,...};
char n[] = 2156169 (number is pressed);
int k=7;
for i -- 0 to NUM_PER_DIGIT ^ k
String s=;
for j -- 0 to k
int index
small correction
On Mon, May 23, 2011 at 9:46 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
A Recursive soln:
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {abc,def,...};
char n[] = 2156169 (number is pressed);
int k=7;
char * s = (char *) malloc(sizeof(char) * k);
void
, immanuel kingston
kingston.imman...@gmail.com wrote:
small correction
On Mon, May 23, 2011 at 9:46 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
A Recursive soln:
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {abc,def,...};
char n[] = 2156169 (number is pressed);
int k=7
Since the number of islands is even, either the sum of areas of islands in
the even position have a greater area or the ones in the odd position have
greater area.
Eric can start from the 1st island if the sum of areas of the islands at the
odd position is greater than that of even ones.or can
Solution:
int majorityElement(int a[], int n) {
if (a == null || a.length == 0 || n=0) return -1;
int mElement = a[0];
int count=1;
for (int i=1; i n; i++) {
if (a[i] == mElement) {
count++;
} else {
count--;
}
if (count =
A Recursive solution:
int interleaved(char *s1, char *s2, char *s3) {
if (s1 == null s2== null s3==null) return 1;
if (s3==null) return 0;
if (s1 != null *s1 == *s3) return interleaved(s1+1,s2,s3+1);
else if (s2 != null *s2 == *s3) return interleaved(s1, s2+1,s3+1);
:21 pm, immanuel kingston kingston.imman...@gmail.com
wrote:
A Recursive solution:
int interleaved(char *s1, char *s2, char *s3) {
if (s1 == null s2== null s3==null) return 1;
if (s3==null) return 0;
if (s1 != null *s1 == *s3) return interleaved(s1+1,s2,s3+1
Preprocess the Dictionary into a hash table where the key is the sorted
string of each word and the value being the A list that contains that
particular word.(O(nlogn * linked list insertion time some k) )
Now for each given word, sort it(O(nlogn)) and find get the list of words
that are
I guess a O(nk),O(k) solution exists.
Have a maxHeap of k elements in our case its 3.
Iterate through the array, compare the (difference between the position
along a number
line between ) and the top element of the maxHeap. It it happens to be
lesser than the top element, pop off the top element
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
On Thu, May 19, 2011 at 1:47 AM, Don dondod...@gmail.com wrote:
Yes. Use a sieve.
Don
On May 17, 11:36 pm, wujin chen wujinchen...@gmail.com wrote:
@Daveļ¼ thanks for your reply.
i know that, i can only check from 6*n - 1 and 6*n + 1..
I think your soln will print repetitions also.
On Mon, May 16, 2011 at 2:34 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
*int ref[] = {2,3,6,7,8};*
*void printcombination(int n,int index,int i)
{
static int a[100];
int j;
if (n == 0)
{
for(j=0;jindex;j++)
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