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BAN USER 1of 1 vote
AnswersWrite a program to find the given new program can be scheduled or not?
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Already scheduled Programs: P1(10,5), P2(25,15)
New Programs: P3(18,7), P4(12, 10).
In P1(10, 5), where 10 is the starting time, 5 is the execution time.
As The P3(18, 7) starts at time 18 and executes for 7 mins i.e., the end time is 18+7 = 25. So this time slot is free and there is no overlap with already scheduled programs. Hence P3 can be scheduled.
As the P4 overlaps with P1, So P4 cannot be scheduled. Report Duplicate  Flag  PURGE
Google Software Engineer Data Structures  3of 3 votes
AnswersThere are N pots. Every pots have some water in it. They may be partially filled. So there is a Overflow Number 0 associated with every pot which tell how many minimum stone pieces are require for that pot to overflow. So if for a pot 0value is 5 it means minimum 5 stone pieces should be put in that pot to make it overflow. Initially a crow watched those pots and by seeing the water level he anticipated 0value correctly for every pot ( that is he knew 01 to On). But when he came back in evening he found that every pot is painted from outside and he is not able to know which pot has what 0value. Crow wants some K pots to overflow so that he can serve his child appropriately. For overflow of pots he need to search for stone in forest( assume that every stone has same size). He wants to use minimum number of stones required to overflow K pots. But only he know the 0value of pots he doesn't know now which pot has what 0value. So the task is that in what minimum number of stones he can make K pots overflow in worst case.
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Input/Output Specifications Input Specification: 1) A array 0 corresponding to 0value of N pots {01, 02, On} 2) Number of pots 3) K value ( number of pots which the crow wants to overflow}
Output Specification: Minimum number of stones required to make K pots overflow in worst case. Or 1 if input is invalid
Example: Let say there are two pots pot 1 has 0 value of 5 , 01= 5 pot 2 has 0 value of 58, 02= 58 Let say crow wants to make one of the pot to overflow. If he know which pot has what 0value he would simple search for 5 stones and put then in pot 1 to make it overflow. But in real case he doesn't know which pot has what 0value so just 5 stones may not always work. However he does know that one pot has 0value S and other has 58. So even in worst case he can make one of the pot overflow just by using 10 stones. He would put 5 stones in one pot if it doesn't overflow he would try the remaining 5 in the other pot which would definitely overflow because one of the pot has 0value of 5. So the answer for above question is minimum 10 stones even in worst case. Input : Input 1= {5,58} Input 2= 2 Input 3= 1 Output : 10 Report Duplicate  Flag  PURGE
Amazon Software Engineer  0of 0 votes
AnswersRahul is playing a very interesting game. He has some N different type of match boxes. All match boxes may have different number of matchsticks (S1, S2, S3... Sn).
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Rahul chooses two random numbers F and K. K should be less than N. The game is that Rahul wants to select any K match boxes out of N match boxes such that total number of matchsticks in these K selected match boxes should be multiple of F.
At the sametime Rahul wants that sum matchsticks of all the selected match boxes should be minimum possible.
Input Specifications:
1) Array S = {S1,S2,S3,...Sn} of size N corresponding to the number of match sticks in N matchboxes(0<=N<=1000}
2) FValue (as explained above)
3) KValue ( as explained above)
Output:
1 2 3 4 5 Here 3 is the number of matchsticks in matchbox I II III IV V minimum possible total number of matchsticks such that the conditioned explained in the problem statement is satisfied. Output 1 if it is not possible or invalid input.
For example, there are 5 match boxes i.e., N = 5
Let's say K is 3 (Rahul has to choose any 3 matchboxes)
Let's say F is 5(sum of matchsticks in 3 selected matchboxes should be multiple of 5).
Rahul can choose II, III and V matchboxes which would give the total sum of 10 which is multiple of F i.e., 5. And 10 is the minimum possible matchsticks possible in the above case.
So you have to answer the minimum possible matchstick(sum of the matchsticks in the selcted matchboxes) but the conditions given above should be satisfied. Report Duplicate  Flag  PURGE
Algorithm  1of 1 vote
AnswersWrite a class DominoChecker that has a method called addBox(int[]) that takes a box of five dominoes, described as a list of 10 integers (explained after), adds it to a collection, and returns true if a box with the same dominoes was already in the collection and false otherwise. A box of dominoes is encoded as a list of 10 integers from 0 to 9, where a pair of numbers represent a domino. For example: 0,2,9,1,3,3,7,4,5,6 represents a box containing dominoes: (0,2); (9,1); (3,3); (7,4); (5,6). http://en.wikipedia.org/wiki/Dominoes for more basic info (like pictures)
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Google Software Engineer / Developer  0of 0 votes
Answers
 veeru in IndiaFind the subsequences whose elements should not be adjacent and their sum should be maximum from the given array (contains only positive integers). Eg: int[] A = {10, 1, 3, 25} Sol: Sum: {10, 3} = 13 {1,25} = 26 {10,25} = 35 Here the Maximum subsequence is {10, 25}.
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Myntra Software Engineer / Developer Arrays  0of 0 votes
AnswersGiven an array of integers. Print a pair whose sum is closest to zero?
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Eg:
Input: arr = {2 5 8 7 2,9}
Output: => 8, 7 Report Duplicate  Flag  PURGE
Amazon Software Engineer / Developer Arrays
I have done the code almost as below. I just explained about the overriding the hashcode method but not coded. Don't know what will be the result :(
Could any one explain how to override the hashcode & compare the contents of Box object in Java?
class Domino {
int num1;
int num2;
}
class Box {
Domino[] dominos;
public Box(Domino[] dominos) {
this.dominos = dominos;
}
}
Hashtable<Box, Boolean> htable = new Hashtable<Box,Boolean>();
addBox(int[] box) {
Domino[] domino = new Domino[box.length()/2];
int domino_index = 0;
for(int i=0; i < 10; i=i+2) {
domino[domino_index] = new Domino();
domino[domino_index].num1 = box[i];
domino[domino_index].num2=box[i+1];
domino_index++;
}
Box boxobj = new Box(domino);
if(!htable.contains(boxobj) {
htable.add(boxobj)
return true;
}
else {
return false;
}
}

veeru
September 27, 2013
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Hi @Logan, It requires null check on prevProcess also bcz., if we want to add the Program like (8,1) then prevProcess will be null.
 veeru April 29, 2019