I am trying to insert in a maxHeap in java and then bubble up the object. This is what I have done, I am not sure how I should approach the bubble up method.
I do understand the algorithm behind bubble up, which is as follows:
get parent node
see if L_childNode is less than parent Node. If Yes, then swap parent with L_child.
see if R_childNode is less than parent Node. If Yes, then swap parent with L_child.
Please point out what am I doing wrong?
private int getLeftChild(int n){
return x*2+1;
}
private int getRightChild(int n){
return x*2+2;
}
public void insert (E item) {
//Integer pos_lastEl= new Integer (heapArray.lastElement());
heapArray.add(item);
bubbleUp(item);
}
//To use to reheap up when item inserted at end of heap (complete tree)
private void bubbleUp(E x){
int place = heapArray.size()-1;
int parent=(place-1)/2;
if ((parent>=0) && (parent.compareTo(heapArray.get(getLeftChild))<0)){
swap(place,parent);
}else ((parent>=0 && (parent.compareTo(heapArray.get(getRightChild))<0))){
swap(place,parent);
}
}
//swaps two objects at index i and j
private void swap(int i, int j){
int max=heapArray.size();
if(i>=0 && i<max && j>=0 && j<max){
E temp=heapArray.get(i);
//put J item in I
heapArray.set(i,heapArray.get(j));
heapArray.set(j,temp);
}
}
Your major problem is using if instead of while to bubble up the newly added element to the proper position.
And there are also some other issues in your code, sorry I had to do some refactoring to make it clean enough:
public class MaxHeapTest<E extends Comparable<E>> {
List<E> heapArray = new ArrayList<>();
public static void main(String... args) {
int N = 13;
MaxHeapTest<Integer> maxHeap = new MaxHeapTest();
for (int i = 0; i < N; ++i) { // ascending;
maxHeap.insert(i);
}
while (!maxHeap.isEmpty()) { // descending now;
System.out.print(maxHeap.delMax() + " ");
}
}
public E delMax() {
E e = heapArray.get(0);
swap(0, heapArray.size() - 1);
heapArray.remove(heapArray.size() - 1);
sinkDown(0);
return e;
}
public void insert(E item) {
heapArray.add(item);
bubbleUp(item);
}
public boolean isEmpty() {
return heapArray.isEmpty();
}
private void bubbleUp(E x) {
int k = heapArray.indexOf(x);
int j = (k - 1) / 2;
while (j >= 0) {
if (heapArray.get(j).compareTo(heapArray.get(k)) < 0) {
swap(k, j);
k = j;
j = (j - 1) / 2;
} else break;
}
}
private void sinkDown(int k) {
int j = 2 * k + 1;
while (j < heapArray.size()) {
if (j < heapArray.size() - 1 && heapArray.get(j).compareTo(heapArray.get(j + 1)) < 0) j++;
if (heapArray.get(k).compareTo(heapArray.get(j)) < 0) {
swap(k, j);
k = j;
j = 2 * j + 1;
} else break;
}
}
private void swap(int i, int j) {
E temp = heapArray.get(i);
heapArray.set(i, heapArray.get(j));
heapArray.set(j, temp);
}
}
After the maxHeap, we can easily output the descending numbers as:
12 11 10 9 8 7 6 5 4 3 2 1 0
I have implemented Sieve of Eratosthenes for finding the list of prime number from 1 to n. My code is working fine for inputs from 1 to 10,000 but I am getting following for values >100,000:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -2146737495
at SieveOfEratosthenes.main(SieveOfEratosthenes.java:53)
I am able to find the issue, which is in the for loop when I am doing i * i as it is going out of Integer range (Integer.MAX_VALUE), but I could not find the solution. Can someone suggest me what changes can be done also I appreciate if someone suggest me any improvement for efficiency in this implementation?
public class SieveOfEratosthenes {
public static void main(String[] args) {
Integer num = Integer.parseInt(args[0]);
Node[] nodes = new Node[num + 1];
for(int i = 1; i < nodes.length; i++) {
Node n = new Node();
n.setValue(i);
n.setMarker(true);
nodes[i] = n;
}
for(int i = 1; i < nodes.length; i++) {
if(nodes[i].getMarker() && nodes[i].getValue() > 1) {
System.out.println("Prime " + nodes[i].getValue());
} else {
continue;
}
for(int j = i * i; j < nodes.length
&& nodes[i].getMarker(); j = j + i) {
nodes[j].setMarker(false);
}
}
System.out.println(l.size());
}
}
class Node {
private int value;
private boolean marker;
public void setValue(int value) {
this.value = value;
}
public int getValue() {
return this.value;
}
public void setMarker(boolean marker) {
this.marker = marker;
}
public boolean getMarker() {
return this.marker;
}
public String toString() {
return ("Value : " + marker + " value " + value);
}
}
Essentially, the for(int j = i * i;... loop is to cross out all multiples of i.
It makes sense only to cross out starting from i * i, since all lesser multiples are already crossed out by their lesser divisors.
There are at least two ways to go from here.
First, you can start crossing out from i * 2 instead of i * i.
This will get rid of the overflow.
On the bad side, the complexity of the sieve would grow from O(n log log n) to O(n log n) then.
Second, you can check whether i * i is already too much, and if it is, skip the loop completely.
Recall that it is essentially skipped anyway for i greater than the square root of nodes.length if no overflow occurs.
For example, just add an if (i * 1L * i < nodes.length) before the loop.
for(int i = 1; i < nodes.length; i++) {
if(nodes[i].getMarker() && nodes[i].getValue() > 1) {
System.out.println("Prime " + nodes[i].getValue());
} else {
continue;
}
TO
int limit = 2 << 14;
for(int i = 1; i < nodes.length; i++) {
if(nodes[i].getMarker() && nodes[i].getValue() > 1 && i <= 2 << 15) {
System.out.println("Prime " + nodes[i].getValue());
if (i > limit) {
continue;
}
} else {
continue;
}
I am trying to write a program which can solve the 8-Puzzle problem.I am using the A* algorithm to find the solution.
I have reviewed my code many times and also tried making some changes.
Even my friends tried to help me find the bug,but they couldn't. I still don't understand where i went wrong.I used javadocs to see if I did something wrong,even that din't solve my problem. I have created three classes to solve this problem.
import java.util.*;
public class Solver implements Iterable<State>
{
ArrayList<State> queue,solQueue;
public int sol[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 0 } };
int temp[][],i;
int moves;
int leastPriority,removeIndex;
State removeTemp;
public Solver(State initial)
{
queue = new ArrayList<State>();
solQueue = new ArrayList<State>();
queue.ensureCapacity(16);
solQueue.ensureCapacity(16);
temp = new int[3][3];
i=1;
leastPriority = 100;
removeTemp=initial;
queue.add(removeTemp);
Iterator<State> qu = queue.iterator();
while(removeTemp.m!=sol)
{
leastPriority = 100;
i=0;
queue.iterator();
for (State s : queue)
{
if((s.mh + s.count) <leastPriority)
{
leastPriority = (s.mh + s.count);
removeIndex = i;
}
if(qu.hasNext())
i++;
}
for(State s : removeTemp.neighbours() )
{
queue.add(s);
}
removeTemp=queue.remove(removeIndex);
solQueue.add(removeTemp);
}
this.moves();
this.solution();
}
public int moves()
{
System.out.print("Solution found out in "+ moves+" moves");
moves = removeTemp.count;
return moves;
}
public Iterable<State> solution()
{
for(State s : solQueue)
{
System.out.println(s.m);
System.out.println("");
}
return solQueue;
}
#SuppressWarnings({ "unchecked", "rawtypes" })
#Override
public Iterator iterator() {
return null;
}
}
And the JVM is throwing an exception.
Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0,Size: 0
at java.util.ArrayList.rangeCheck(Unknown Source)
at java.util.ArrayList.get(Unknown Source)
at Solver.<init>(Solver.java:41)
at Main.main(Main.java:13)
What i don't understand is that how can the size of the ArrayList be 1 when i have explicitly state it as 16.
The State Class has the heuristic function which is suppose to make the algorithm efficient.The following is the State Class.
import java.util.ArrayList;
import java.util.Iterator;
public class State implements Iterable<State>
{
public int sol[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 0 } };
int m[][], bi, bj, count, priority, si, sj;
int i,j,tempm[][];
int mh = 0;
boolean isInitialState, isRepeatedState;
State previousState, tempState;
ArrayList<State> neighbourStates;
public State(State s, int c, int[][] array)
{
neighbourStates = new ArrayList<State>();
neighbourStates.ensureCapacity(16);
tempState =this;
m = new int[3][3];
m=array;
if (s == null)
{
isInitialState = true;
count = 0;
previousState =null;
}
else
{
previousState = s;
count = c+1;
}
this.findZero();
this.manhattanHeuristic();
}
private void findZero()
{
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
{
if(m[i][j]==0)
{
bi=i;
bj=j;
}
}
}
private void manhattanHeuristic() {
int n = 1;
mh = 0;
for (int i = 0; i < 3; i++)
Z: for (int j = 0; j < 3; j++) {
if ((i == bi) && (j == bj)) {
continue Z;
}
else if (m[i][j] == n) {
n++;
}
else {
this.getSolutionIndex();
mh = mh + Math.abs(i - si) + Math.abs(j - sj);
}
}
}
void getSolutionIndex() {
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++) {
if (m[i][j] == 0) {
si = i;
sj = j;
}
}
}
public Iterable<State> neighbours()
{
tempm = m;
this.up();
if(!(equals(tempm)))
{
tempState = new State(this,count,tempm);
neighbourStates.add(tempState);
}
this.down();
if(!(equals(tempm)))
{
tempState = new State(this,count,tempm);
neighbourStates.add(tempState);
}
this.left();
if(!(equals(tempm)))
{
tempState = new State(this,count,tempm);
neighbourStates.add(tempState);
}
this.right();
if(!(equals(tempm)))
{
tempState = new State(this,count,tempm);
neighbourStates.add(tempState);
}
return neighbourStates;
}
public boolean equals(int s[][])
{
if((isInitialState==false)&&(previousState.m == s))
return true;
else
return false;
}
#Override
public Iterator<State> iterator() {
// TODO Auto-generated method stub
return null;
}
public void up()
{
if ((bi > 1) && (bi < 2) && (bj < 3)&& (bj > 1))
{
i = bi;
i = i + 1;
this.move(i,bj);
}
}
public void down()
{
if ((bi > 2) && (bi < 3) && (bj < 3) && (bj > 1))
{
i = bi;
i = i - 1;
this.move(i,bj);
}
}
public void left()
{
if ((bi > 1) && (bi < 3) && (bj < 2)&& (bj > 1)) {
j = bj;
j = j + 1;
this.move(bi, j);
}
}
public void right()
{
if ((bi > 1) && (bi < 3) && (bj < 3) && (bj > 2)) {
j = bj;
j = j - 1;
this.move(bi, j);
}
}
public void move(int x, int y) {
{
tempm = m;
}
if ((tempm[x + 1][y] == 0) || (tempm[x - 1][y] == 0) || (tempm[x][y + 1] == 0)|| (tempm[x][y - 1] == 0)) {
tempm[bi][bj] = tempm[x][y];
tempm[x][y] = 0;
bi = x;
bj = y;
}
}
}
And the finally the class with the main function.
import java.util.Scanner;
public class Main {
public static void main(String[] args)
{
#SuppressWarnings("resource")
Scanner sc = new Scanner(System.in);
int[][] tiles = new int[3][3];
System.out.println("Enter the elements");
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
tiles[i][j] = sc.nextInt();
State initial = new State(null,0,tiles);
Solver solver = new Solver(initial);
solver.solution();
System.out.println("Minimum number of moves = " + solver.moves());
}
}
What i don't understand is that how can the size of the ArrayList be 1 when i have explicitly state it as 16.
You did not set the size of the ArrayList to 16. You've set the capacity:
queue.ensureCapacity(16);
solQueue.ensureCapacity(16);
This does not make the ArrayList have a size of 16.
An ArrayList has an array to hold its data. When you add more elements to the ArrayList and its internal array is full, it will have to allocate a larger array and copy the content of what it currently holds plus the new element.
The capacity of the ArrayList is the minimum size that the internal array has. You can use ensureCapacity to make sure that the ArrayList doesn't have to resize too often (resizing and copying the content is an expensive operation). So, ensureCapacity is a call you make to make it work effiently.
It does not make the ArrayList have 16 elements; it only makes sure that the ArrayList has room for at least 16 elements.
If you want the ArrayList to have 16 elements, you'll have to add those elements one by one.
Size of the collection and the capacity are 2 different concepts.
capacity represents the maximum size of items a collection can hold without a reallocation.
size represents the current number of items in the collection.
IndexOutOfBoundsException is saying that you are trying to access an item with index that does not exist in the collection.
please try the below code in Solver.java
if(!queue.isEmpty())
removeTemp=queue.remove(removeIndex);
else
break;
I have been trying to solve a Java exercise on a Codility web page.
Below is the link to the mentioned exercise and my solution.
https://codility.com/demo/results/demoH5GMV3-PV8
Can anyone tell what can I correct in my code in order to improve the score?
Just in case here is the task description:
A small frog wants to get to the other side of a river. The frog is currently located at position 0, and wants to get to position X. Leaves fall from a tree onto the surface of the river.
You are given a non-empty zero-indexed array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in minutes.
The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X.
For example, you are given integer X = 5 and array A such that:
A[0] = 1
A[1] = 3
A[2] = 1
A[3] = 4
A[4] = 2
A[5] = 3
A[6] = 5
A[7] = 4
In minute 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.
Write a function:
class Solution { public int solution(int X, int[] A); }
that, given a non-empty zero-indexed array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.
If the frog is never able to jump to the other side of the river, the function should return −1.
For example, given X = 5 and array A such that:
A[0] = 1
A[1] = 3
A[2] = 1
A[3] = 4
A[4] = 2
A[5] = 3
A[6] = 5
A[7] = 4
the function should return 6, as explained above. Assume that:
N and X are integers within the range [1..100,000];
each element of array A is an integer within the range [1..X].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(X), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
And here is my solution:
import java.util.ArrayList;
import java.util.List;
class Solution {
public int solution(int X, int[] A) {
int list[] = A;
int sum = 0;
int searchedValue = X;
List<Integer> arrayList = new ArrayList<Integer>();
for (int iii = 0; iii < list.length; iii++) {
if (list[iii] <= searchedValue && !arrayList.contains(list[iii])) {
sum += list[iii];
arrayList.add(list[iii]);
}
if (list[iii] == searchedValue) {
if (sum == searchedValue * (searchedValue + 1) / 2) {
return iii;
}
}
}
return -1;
}
}
You are using arrayList.contains inside a loop, which will traverse the whole list unnecessarily.
Here is my solution (I wrote it some time ago, but I believe it scores 100/100):
public int frog(int X, int[] A) {
int steps = X;
boolean[] bitmap = new boolean[steps+1];
for(int i = 0; i < A.length; i++){
if(!bitmap[A[i]]){
bitmap[A[i]] = true;
steps--;
if(steps == 0) return i;
}
}
return -1;
}
Here is my solution. It got me 100/100:
public int solution(int X, int[] A)
{
int[] B = A.Distinct().ToArray();
return (B.Length != X) ? -1 : Array.IndexOf<int>(A, B[B.Length - 1]);
}
100/100
public static int solution (int X, int[] A){
int[]counter = new int[X+1];
int ans = -1;
int x = 0;
for (int i=0; i<A.length; i++){
if (counter[A[i]] == 0){
counter[A[i]] = A[i];
x += 1;
if (x == X){
return i;
}
}
}
return ans;
}
A Java solution using Sets (Collections Framework) Got a 100%
import java.util.Set;
import java.util.TreeSet;
public class Froggy {
public static int solution(int X, int[] A){
int steps=-1;
Set<Integer> values = new TreeSet<Integer>();
for(int i=0; i<A.length;i++){
if(A[i]<=X){
values.add(A[i]);
}
if(values.size()==X){
steps=i;
break;
}
}
return steps;
}
Better approach would be to use Set, because it only adds unique values to the list. Just add values to the Set and decrement X every time a new value is added, (Set#add() returns true if value is added, false otherwise);
have a look,
public static int solution(int X, int[] A) {
Set<Integer> values = new HashSet<Integer>();
for (int i = 0; i < A.length; i++) {
if (values.add(A[i])) X--;
if (X == 0) return i;
}
return -1;
}
do not forget to import,
import java.util.HashSet;
import java.util.Set;
Here's my solution, scored 100/100:
import java.util.HashSet;
class Solution {
public int solution(int X, int[] A) {
HashSet<Integer> hset = new HashSet<Integer>();
for (int i = 0 ; i < A.length; i++) {
if (A[i] <= X)
hset.add(A[i]);
if (hset.size() == X)
return i;
}
return -1;
}
}
Simple solution 100%
public int solution(final int X, final int[] A) {
Set<Integer> emptyPosition = new HashSet<Integer>();
for (int i = 1; i <= X; i++) {
emptyPosition.add(i);
}
// Once all the numbers are covered for position, that would be the
// moment when the frog will jump
for (int i = 0; i < A.length; i++) {
emptyPosition.remove(A[i]);
if (emptyPosition.size() == 0) {
return i;
}
}
return -1;
}
Here's my solution.
It isn't perfect, but it's good enough to score 100/100.
(I think that it shouldn't have passed a test with a big A and small X)
Anyway, it fills a new counter array with each leaf that falls
counter has the size of X because I don't care for leafs that fall farther than X, therefore the try-catch block.
AFTER X leafs fell (because it's the minimum amount of leafs) I begin checking whether I have a complete way - I'm checking that every int in count is greater than 0.
If so, I return i, else I break and try again.
public static int solution(int X, int[] A){
int[] count = new int[X];
for (int i = 0; i < A.length; i++){
try{
count[A[i]-1]++;
} catch (ArrayIndexOutOfBoundsException e){ }
if (i >= X - 1){
for (int j = 0; j< count.length; j++){
if (count[j] == 0){
break;
}
if (j == count.length - 1){
return i;
}
}
}
}
return -1;
}
Here's my solution with 100 / 100.
public int solution(int X, int[] A) {
int len = A.length;
if (X > len) {
return -1;
}
int[] isFilled = new int[X];
int jumped = 0;
Arrays.fill(isFilled, 0);
for (int i = 0; i < len; i++) {
int x = A[i];
if (x <= X) {
if (isFilled[x - 1] == 0) {
isFilled[x - 1] = 1;
jumped += 1;
if (jumped == X) {
return i;
}
}
}
}
return -1;
}
Here's what I have in C#. It can probably still be refactored.
We throw away numbers greater than X, which is where we want to stop, and then we add numbers to an array if they haven't already been added.
When the count of the list has reached the expected number, X, then return the result. 100%
var tempArray = new int[X+1];
var totalNumbers = 0;
for (int i = 0; i < A.Length; i++)
{
if (A[i] > X || tempArray.ElementAt(A[i]) != 0)
continue;
tempArray[A[i]] = A[i];
totalNumbers++;
if (totalNumbers == X)
return i;
}
return -1;
below is my solution. I basically created a set which allows uniques only and then go through the array and add every element to set and keep a counter to get the sum of the set and then using the sum formula of consecutive numbers then I got 100% . Note : if you add up the set using java 8 stream api the solution is becoming quadratic and you get %56 .
public static int solution2(int X, int[] A) {
long sum = X * (X + 1) / 2;
Set<Integer> set = new HashSet<Integer>();
int setSum = 0;
for (int i = 0; i < A.length; i++) {
if (set.add(A[i]))
setSum += A[i];
if (setSum == sum) {
return i;
}
}
return -1;
}
My JavaScript solution that got 100 across the board. Since the numbers are assumed to be in the range of the river width, simply storing booleans in a temporary array that can be checked against duplicates will do. Then, once you have amassed as many numbers as the quantity X, you know you have all the leaves necessary to cross.
function solution(X, A) {
covered = 0;
tempArray = [];
for (let i = 0; i < A.length; i++) {
if (!tempArray[A[i]]) {
tempArray[A[i]] = true;
covered++
if(covered === X) return i;
}
}
return -1;
}
Here is my answer in Python:
def solution(X, A):
# write your code in Python 3.6
values = set()
for i in range (len(A)):
if A[i]<=X :
values.add(A[i])
if len(values)==X:
return i
return -1
Just tried this problem as well and here is my solution. Basically, I just declared an array whose size is equal to position X. Then, I declared a counter to monitor if the necessary leaves have fallen at the particular spots. The loop exits when these leaves have been met and if not, returns -1 as instructed.
class Solution {
public int solution(int X, int[] A) {
int size = A.length;
int[] check = new int[X];
int cmp = 0;
int time = -1;
for (int x = 0; x < size; x++) {
int temp = A[x];
if (temp <= X) {
if (check[temp-1] > 0) {
continue;
}
check[temp - 1]++;
cmp++;
}
if ( cmp == X) {
time = x;
break;
}
}
return time;
}
}
It got a 100/100 on the evaluation but I'm not too sure of its performance. I am still a beginner when it comes to programming so if anybody can critique the code, I would be grateful.
Maybe it is not perfect but its straightforward. Just made a counter Array to track the needed "leaves" and verified on each iteration if the path was complete. Got me 100/100 and O(N).
public static int frogRiver(int X, int[] A)
{
int leaves = A.Length;
int[] counter = new int[X + 1];
int stepsAvailForTravel = 0;
for(int i = 0; i < leaves; i++)
{
//we won't get to that leaf anyway so we shouldnt count it,
if (A[i] > X)
{
continue;
}
else
{
//first hit!, keep a count of the available leaves to jump
if (counter[A[i]] == 0)
stepsAvailForTravel++;
counter[A[i]]++;
}
//We did it!!
if (stepsAvailForTravel == X)
{
return i;
}
}
return -1;
}
This is my solution. I think it's very simple. It gets 100/100 on codibility.
set.contains() let me eliminate duplicate position from table.
The result of first loop get us expected sum. In the second loop we get sum of input values.
class Solution {
public int solution(int X, int[] A) {
Set<Integer> set = new HashSet<Integer>();
int sum1 = 0, sum2 = 0;
for (int i = 0; i <= X; i++){
sum1 += i;
}
for (int i = 0; i < A.length; i++){
if (set.contains(A[i])) continue;
set.add(A[i]);
sum2 += A[i];
if (sum1 == sum2) return i;
}
return -1;
}
}
Your algorithm is perfect except below code
Your code returns value only if list[iii] matches with searchedValue.
The algorithm must be corrected in such a way that, it returns the value if sum == n * ( n + 1) / 2.
import java.util.ArrayList;
import java.util.List;
class Solution {
public int solution(int X, int[] A) {
int list[] = A;
int sum = 0;
int searchedValue = X;
int sumV = searchedValue * (searchedValue + 1) / 2;
List<Integer> arrayList = new ArrayList<Integer>();
for (int iii = 0; iii < list.length; iii++) {
if (list[iii] <= searchedValue && !arrayList.contains(list[iii])) {
sum += list[iii];
if (sum == sumV) {
return iii;
}
arrayList.add(list[iii]);
}
}
return -1;
}
}
I think you need to check the performance as well. I just ensured the output only
This solution I've posted today gave 100% on codility, but respectivly #rafalio 's answer it requires K times less memory
public class Solution {
private static final int ARRAY_SIZE_LOWER = 1;
private static final int ARRAY_SIZE_UPPER = 100000;
private static final int NUMBER_LOWER = ARRAY_SIZE_LOWER;
private static final int NUMBER_UPPER = ARRAY_SIZE_UPPER;
public static class Set {
final long[] buckets;
public Set(int size) {
this.buckets = new long[(size % 64 == 0 ? (size/64) : (size/64) + 1)];
}
/**
* number should be greater than zero
* #param number
*/
public void put(int number) {
buckets[getBucketindex(number)] |= getFlag(number);
}
public boolean contains(int number) {
long flag = getFlag(number);
// check if flag is stored
return (buckets[getBucketindex(number)] & flag) == flag;
}
private int getBucketindex(int number) {
if (number <= 64) {
return 0;
} else if (number <= 128) {
return 1;
} else if (number <= 192) {
return 2;
} else if (number <= 256) {
return 3;
} else if (number <= 320) {
return 4;
} else if (number <= 384) {
return 5;
} else
return (number % 64 == 0 ? (number/64) : (number/64) + 1) - 1;
}
private long getFlag(int number) {
if (number <= 64) {
return 1L << number;
} else
return 1L << (number % 64);
}
}
public static final int solution(final int X, final int[] A) {
if (A.length < ARRAY_SIZE_LOWER || A.length > ARRAY_SIZE_UPPER) {
throw new RuntimeException("Array size out of bounds");
}
Set set = new Set(X);
int ai;
int counter = X;
final int NUMBER_REAL_UPPER = min(NUMBER_UPPER, X);
for (int i = 0 ; i < A.length; i++) {
if ((ai = A[i]) < NUMBER_LOWER || ai > NUMBER_REAL_UPPER) {
throw new RuntimeException("Number out of bounds");
} else if (ai <= X && !set.contains(ai)) {
counter--;
if (counter == 0) {
return i;
}
set.put(ai);
}
}
return -1;
}
private static int min(int x, int y) {
return (x < y ? x : y);
}
}
This is my solution it got me 100/100 and O(N).
public int solution(int X, int[] A) {
Map<Integer, Integer> leaves = new HashMap<>();
for (int i = A.length - 1; i >= 0 ; i--)
{
leaves.put(A[i] - 1, i);
}
return leaves.size() != X ? -1 : Collections.max(leaves.values());
}
This is my solution
public func FrogRiverOne(_ X : Int, _ A : inout [Int]) -> Int {
var B = [Int](repeating: 0, count: X+1)
for i in 0..<A.count {
if B[A[i]] == 0 {
B[A[i]] = i+1
}
}
var time = 0
for i in 1...X {
if( B[i] == 0 ) {
return -1
} else {
time = max(time, B[i])
}
}
return time-1
}
A = [1,2,1,4,2,3,5,4]
print("FrogRiverOne: ", FrogRiverOne(5, &A))
Actually I re-wrote this exercise without seeing my last answer and came up with another solution 100/100 and O(N).
public int solution(int X, int[] A) {
Set<Integer> leaves = new HashSet<>();
for(int i=0; i < A.length; i++) {
leaves.add(A[i]);
if (leaves.contains(X) && leaves.size() == X) return i;
}
return -1;
}
I like this one better because it is even simpler.
This one works good on codality 100% out of 100%. It's very similar to the marker array above but uses a map:
public int solution(int X, int[] A) {
int index = -1;
Map<Integer, Integer> map = new HashMap();
for (int i = 0; i < A.length; i++) {
if (!map.containsKey(A[i])) {
map.put(A[i], A[i]);
X--;
if (X == 0) {index = i;break;}
}
}
return index;
}
%100 with js
function solution(X, A) {
let leafSet = new Set();
for (let i = 0; i < A.length; i += 1) {
if(A[i] <= 0)
continue;
if (A[i] <= X )
leafSet.add(A[i]);
if (leafSet.size == X)
return i;
}
return -1;
}
With JavaScript following solution got 100/100.
Detected time complexity: O(N)
function solution(X, A) {
let leaves = new Set();
for (let i = 0; i < A.length; i++) {
if (A[i] <= X) {
leaves.add(A[i])
if (leaves.size == X) {
return i;
}
}
}
return -1;
}
100% Solution using Javascript.
function solution(X, A) {
if (A.length === 0) return -1
if (A.length < X) return -1
let steps = X
const leaves = {}
for (let i = 0; i < A.length; i++) {
if (!leaves[A[i]]) {
leaves[A[i]] = true
steps--
}
if (steps === 0) {
return i
}
}
return -1
}
C# Solution with 100% score:
using System;
using System.Collections.Generic;
class Solution {
public int solution(int X, int[] A) {
// go through the array
// fill a hashset, until the size of hashset is X
var set = new HashSet<int>();
int i = 0;
foreach (var a in A)
{
if (a <= X)
{
set.Add(a);
}
if (set.Count == X)
{
return i;
}
i++;
}
return -1;
}
}
https://app.codility.com/demo/results/trainingXE7QFJ-TZ7/
I have a very simple solution (100% / 100%) using HashSet. Lots of people check unnecessarily whether the Value is less than or equal to X. This task cannot be otherwise.
public static int solution(int X, int[] A) {
Set<Integer> availableFields = new HashSet<>();
for (int i = 0; i < A.length; i++) {
availableFields.add(A[i]);
if (availableFields.size() == X){
return i;
}
}
return -1;
}
public static int solutions(int X, int[] A) {
Set<Integer> values = new HashSet<Integer>();
for (int i = 0; i < A.length; i++) {
if (values.add(A[i])) {
X--;
}
if (X == 0) {
return i;
}
}
return -1;
}
This is my solution. It uses 3 loops but is constant time and gets 100/100 on codibility.
class FrogLeap
{
internal int solution(int X, int[] A)
{
int result = -1;
long max = -1;
var B = new int[X + 1];
//initialize all entries in B array with -1
for (int i = 0; i <= X; i++)
{
B[i] = -1;
}
//Go through A and update B with the location where that value appeared
for (int i = 0; i < A.Length; i++)
{
if( B[A[i]] ==-1)//only update if still -1
B[A[i]] = i;
}
//start from 1 because 0 is not valid
for (int i = 1; i <= X; i++)
{
if (B[i] == -1)
return -1;
//The maxValue here is the earliest time we can jump over
if (max < B[i])
max = B[i];
}
result = (int)max;
return result;
}
}
Short and sweet C++ code. Gets perfect 100%... Drum roll ...
#include <set>
int solution(int X, vector<int> &A) {
set<int> final;
for(unsigned int i =0; i< A.size(); i++){
final.insert(A[i]);
if(final.size() == X) return i;
}
return -1;
}