codewarrior
BAN USER- 2of 2 votes
AnswersGiven two strings, return true if they are one edit away from each other, else return false. An edit is insert/replace/delete a character.
- codewarrior in United States
Ex. {"abc","ab"}->true, {"abc","adc"}->true, {"abc","cab"}->false| Report Duplicate | Flag | PURGE
Amazon SDE1 - 0of 0 votes
AnswersFind the total number of connected components in a graph (there can be forests)
- codewarrior in United States| Report Duplicate | Flag | PURGE
Amazon SDE1
Good solution and clear explanation.
- codewarrior August 25, 20151. Calculate ceil(25%) from size of the list, let this be m. m customers with highest spending receive msg1, m customers with least spending receive msg2, (n-m) customers receive msg3.
2. use quick select to find i = index of m highest customers, j = index of m lowest customers - these 2m customers receive msg's 1 and 2
3. rest of the customers from i+1 to j-1 receive msg 3
Congrats on providing the worst ever description of a question.
- codewarrior August 23, 2015null
- codewarrior August 12, 2015@payal:
what do these methods do? are they similar to adding/removing facebook/linkedin relationships?
@SycophantEve:
Tried your solution with following input:
a = [{-5,-1},{1,3},{4,10}]
interv = {-2,12}
expected output: [{-5,12}]
Your solution gives output as [{-5,12}{1,3}]
Can we ban this jerk's IP?
- codewarrior July 08, 2015Cant this be improved using a trie
- codewarrior April 20, 2015can you explain a bit more clearly? What does MathedChars() do?
- codewarrior April 20, 2015@kurobi:
Not necessary. You find the root, and then you traverse based on the ordering of the children in the List. A List, as you should know, is an ordered data structure.
Doesn't this return true for input 'hhire'?
- codewarrior November 07, 2013public static boolean wordBrute(String word, int index) {
if(index>=word.length())
return true;
int[] hist = new int[4];
int i;
outerLoop:
for(i=index;i<word.length();i++) {
char histChar = word.charAt(i);
switch(histChar) {
case 'h': {
hist[0]++;
break;
}
case 'i': {
if(check(hist,1))
return false;
hist[1]++;
break;
}
case 'r': {
if(check(hist,2))
return false;
hist[2]++;
break;
}
case 'e': {
if(check(hist,3))
return false;
while(i<word.length() && word.charAt(i) == 'e' ) {
i++;
hist[3]++;
}
break outerLoop;
}
default:
return false;
}
}
int prev = hist[0];
for(int j=1;j<hist.length;j++) {
if(hist[j] != prev)
return false;
prev = hist[j];
}
return true && (wordBrute(word,i));
}
public static boolean check(int[] hist,int index) {
return hist[index-1] == 0;
}
- codewarrior November 07, 2013public static void rev(Node node,Node prev) {
if(node == null) {
list.headNode = prev;
return;
}
Node temp = node.next;
node.next = prev;
rev(temp,node);
}
Call this with
rev(list.headNode,null);
that was awesome. some explanation would have helped though!
- codewarrior February 25, 2013public static void compressString(String inp) {
StringBuilder out = new StringBuilder();
int count = 1;
for(int i=0;i<inp.length();i++)
{
if(i<inp.length()-1 && inp.charAt(i) == inp.charAt(i+1)) {
count++;
continue;
}
out.append(inp.charAt(i));
if(count>1)
out.append(count);
count=1;
}
System.out.println(out.toString());
}
This is O(n2) right?
- codewarrior February 19, 2013@Abakumoff:
Vik never said the first step is not O(N). It takes N/2 iterations for the fast pointer to reach the end of the list. Time complexity is O(N/2) which is O(N). I don't see why you should do a downvote.
Sort the array in descending order -
T(N) = O(NlgN) and S(N) = O(1)
Lookup in a set has a worst case complexity of O(N) and an average case of O(1).
- codewarrior February 15, 2013Did you even understand what was asked in the question?
- codewarrior February 15, 2013correct algorithm, but poorly written.
- codewarrior February 08, 2013
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Straightforward and simple. awesome.
- codewarrior August 30, 2015