Problem Statement:
While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you'll need to defeat in this quest. Each monster i is described with two integer numbers - poweri and bonusi. To defeat this monster, you'll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately.
You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order. The quest turned out to be very hard - you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you're interested, what is the maximum possible number of monsters you can defeat?
Input:
The first line contains an integer, n, denoting the number of monsters.
The next line contains an integer, e, denoting your initial experience.
Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which
represents power of the corresponding monster.
Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.
Sample cases:
Input 2 123 78 130 10 0
Output 2
Output description
Initial experience level is 123 points.
Defeat the first monster having power of 78 and bonus of 10. Experience level is now 123+10=133.
Defeat the second monster.
What I have tried:
public static int defeat(int [] monster,int bonus[],int n,int exp){
if(n==0)
return 0;
if(n==1 && monster[0]<=exp)return 1;
if(n==1 && monster[0]>exp) return 0;
if(monster[n-1]<=exp){
return defeat(monster,bonus,n-1,bonus[n-1]+exp )+ defeat(monster,bonus,n-1,exp);
}else{
return defeat(monster,bonus,n-1,exp);
}
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int exp = s.nextInt();
int monst[] = new int[n];
int bonus[] = new int[n];
for (int i = 0; i < n; i++) {
monst[i] = s.nextInt();
}
for (int i = 0; i < n; i++) {
bonus[i] = s.nextInt();
}
System.out.println(defeat(monst,bonus,n,exp));
}
I am not getting correct answers with this solution.
I thought of this problem as 0/1 knapsack problem( correct me If I am wrong). Also can you provide me DP solution of this problem.
You can just sort the monsters from lowest to highest power required and defeat them in that order.
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int exp = s.nextInt();
int monst[] = new int[n];
int bonus[] = new int[n];
for (int i = 0; i < n; i++) {
monst[i] = s.nextInt();
}
for (int i = 0; i < n; i++) {
bonus[i] = s.nextInt();
}
class Monster {
private final int power, bonus;
public Monster(int power, int bonus){
this.power = power;
this.bonus = bonus;
}
}
Monster[] monsters = new Monster[n];
for(int i = 0; i < n; i++) monsters[i] = new Monster(monst[i], bonus[i]);
Arrays.sort(monsters, Comparator.comparingInt(m -> m.power));
int count = 0;
for(Monster m: monsters){
if(exp < m.power) break;
exp += m.bonus;
++count;
}
System.out.println(count);
}
Perhaps I am too simplistic, but I would try it as below:
First create a class Monster like this:
public class Monster
{
final int m_Power;
final int m_Bonus;
public Monster( final int power, final int bonus )
{
m_Power = power;
m_Bonus = bonus;
}
public final int getPower() { return m_Power; }
public final int getBonus() { return m_Bonus; }
}
Next, initialise a list of Monsters like this:
public static void main( String... args )
{
final var scanner = new Scanner( System.in );
final var n = scanner.nextInt();
final var experience = scanner.nextInt();
final var power [] = new int [n];
for( var i = 0; i < n; ++i )
{
power [i] = scanner.nextInt();
}
List<Monster> monsters = new ArrayList<>( n );
for( var i = 0; i < n; ++i )
{
monsters.add( new Monster( power [i], scanner.nextInt();
}
monsters.sort( (m1,m2) ->
{
final var p = m1.getPower() - m2.getPower();
return p == 0 ? m2.getBonus() - m1.getBonus(() : p;
} ); //*1
System.out.println( defeat( monsters, experience ) );
}
*1 -> this implementation of the comparator works well only for power and bonus values that are small compared to MAX_INT.
The list monsters now contains the monsters sorted by their power in ascending order; monsters with the same power are ordered by their bonus values in descending order.
Now my implementation of defeat() would look like this:
public final int defeat( final List<Monster> m, final int initialExperience )
{
var experience = initialExperience;
var retValue = 0;
final Stack<Monster> monsters = new LinkedList<>( m );
while( !monsters.empty() )
{
var monster = monsters.pop();
if( experience > monster.getPower() )
{
experience += monster.getBonus();
++retValue;
}
else break;
}
return retValue;
}
n=int(input())
e=int(input())
P=[]
for i in range(n):
P.append(int(input()))
B=[]
for i in range(n):
B.append(int(input()))
c=0
f=True
while n>0 and f:
f=False
i=0
while i<n:
if e>=P[i]:
e+=B[i]
P.pop(i)
B.pop(i)
n-=1
c+=1
f=True
i-=1
i+=1
print(c)
import java.io.*;
import java.util.*;
class Player {
int exp;
public int getExp() {
return exp;
}
public void setExp(int exp) {
this.exp = exp;
}
}
class Monster {
int power;
int bonus;
public int getPower() {
return this.power;
}
public int getBonus() {
return this.bonus;
}
public void setBonus(int bonus) {
this.bonus = bonus;
}
public void setPower(int power) {
this.power = power;
}
}
class Calc {
public int calc() {
Scanner sc = new Scanner(System.in);
//number of monsters
System.out.println("enter the number of monsters");
int n = sc.nextInt();
int arr[] = new int[n];
System.out.println("enter the power of the player");
Player player = new Player();
player.setExp(sc.nextInt());
//declaration of array type object of monster
Monster monster[] = new Monster[n];
//value setting completed
for (int i = 0; i < n; i++) {
//allocation of object to the real
monster[i] = new Monster();
System.out.println("enter the power of monster");
monster[i].setPower(sc.nextInt());
System.out.println("enter the bonus that to be earned after killing the monster");
monster[i].setBonus(sc.nextInt());
}
//calculate win or loose
int count = 0;
int flag = 0;
//**
for (int i = 0; i < n; i++) {
if (player.getExp() >= monster[i].getPower()) {
player.setExp(player.getExp() + monster[i].getBonus());
count = count + 1;
// flag = flag + 1;
} else if (player.getExp() < monster[i].getPower()) {
for (int j = 0; j < n; j++)
arr[j] = i;
//count = count;
flag = flag + 1;
}
//**
for (int t = 0; t < flag; t++) {
if (player.getExp() >= monster[arr[t]].getPower()) {
count = count + 1;
}
}
}
return count;
}
}
public class Main {
public static void main(String[] args) {
// write your code here
System.out.println("welcome to the loosing game");
Calc c = new Calc();
System.out.println("no of monster killed" + c.calc());
}
}
#include<bits/stdc++.h>
using namespace std;
struct Monsters
{
int power;
int bonus;
};
class Solution
{
public:
//Function to find the maximum number of activities that can
//be performed by a single person.
static bool compare(Monsters a, Monsters b)
{
if(a.power < b.power) return 1;
else if(a.power > b.power) return 0;
else return 1;
}
int MonstersKill(vector<int> power, vector<int> bonus, int n, int exp)
{
Monsters arr[n];
for(int i = 0; i < n; i++)
{
arr[i].power = power[i];
arr[i].bonus = bonus[i];
}
sort(arr, arr+n, compare);
int count = 0;
for(int i = 0; i < n; i++)
{
if(arr[i].power <= exp)
{
count++;
exp += arr[i].bonus;
}
else
break;
}
return count;
}
};
int main()
{
int n, exp;
cin >> n >> exp;
vector<int> power(n), bonus(n);
//adding elements to arrays power and bonus
for(int i=0;i<n;i++)
cin>>power[i];
for(int i=0;i<n;i++)
cin>>bonus[i];
Solution ob;
cout << ob.MonstersKill(power, bonus, n, exp) << endl;
return 0;
}
Time Complexity : O(N * Log(N))
Auxilliary Space: O(N)
I need to sort my array of Money objects in ascending order, but I get 3 compiler errors.
TestMoney.java:44: error: bad operand types for binary operator '<'
if (list[j] < list[min]) {
^
first type: Money
second type: Money
TestMoney.java:50: error: incompatible types: Money cannot be converted to int
final int temp = list[i];
^
TestMoney.java:52: error: incompatible types: int cannot be converted to Money
list[min] = temp;
^
class TestMoney
{
public static void main(String [] args)
{
Money[] list = new Money[15];
for(int i =0; i<15; i++)
{
int dollar =(int) (Math.random() * 30 + 1);
int cent = (int) (Math.random() * 100);
list[i] = new Money(dollar, cent);
}
sortArray(list);
printArray(list);
}
public static void printArray(Money[] list)
{
for(int i =0; i <list.length; i++)
{
if(i%10 ==0)
{
System.out.println();
}
System.out.print(" " + list[i]);
}
}
public static void sortArray(Money[] list)
{
int min;
for (int i = 0; i < list.length; i++) {
// Assume first element is min
min = i;
for (int j = i + 1; j < list.length; j++) {
if (list[j] < list[min]) {
min = j;
}
}
if (min != i) {
final int temp = list[i];
list[i] = list[min];
list[min] = temp;
}
System.out.println(list[i]);// I print the in ascending order
}
}
}
class Money
{
private int dol;
private int cen;
Money()
{
dol = 0;
cen = 00;
}
Money(int dol,int cen)
{
int remainder = cen % 100;
int divisor = cen / 100;
this.dol = dol+ divisor;
this.cen = remainder;
}
public int getDollors(int dol)
{
return dol;
}
public int getCents(int cen)
{
return cen;
}
public void setDollors(int d)
{
dol = d;
}
public void setCents(int c)
{
cen = c;
}
public Money addMoney(Money m)
{
int d = this.dol + m.dol;
int c = this.cen + m.cen;
return new Money(d, c);
}
public int compareTo(Money m)
{
if(this.dol<m.dol && this.cen<m.cen)
return -1;
else if(m.dol<this.dol && m.cen<this.cen )
return 1;
return 0;
}
public Money subtract(Money m)
{
int cents1 = this.dol*100 + this.cen;
int cents2 = m.dol *100 + m.cen;
int cents = cents1 -cents2;
return new Money(cen/100,cen%100);
}
public String toString()
{
return String.format("$%d.%02d", this.dol,this.cen);
}
}
That's because you are trying to compare two Money objects with < operator which applies to Numbers only, you need to replce the following:
if (list[j] < list[min]) {
with
if (list[j].getDollors() < list[min].getDollors()
|| (list[j].getDollors() == list[min].getDollors() && list[j].getCents() < list[min].getCents())) {
Also, you don't need your getters to accept an argument, they can be zero argument methods as they just return a value.
Another way, maybe, is using Comparator interface, like this:
public class Money implements Comparator<Money>{
int n; //as an example, compare two object by them "n" variable
#Override
public int compare(Money o1, Money o2) {
int result;
if (o1.n > o2.n)
result = 1; //greater than case
else
result = o1.n < o2.n ? -1 // less than case
: 0; // equal case
return result;
}
}
In the compare method you can use the criteria of gt-ls-eq based on amount of money.
Finally do:
for (int j = i + 1; j < list.length; j++) {
if (list[j].compare(list[min]) < 0) { //you ask if is greater, less,
//or equal than 0
min = j;
}
}
I have made this program using array concept in java. I am getting Exception as ArrayIndexOutOfBound while trying to generate product.
I made the function generateFNos(int max) to generate factors of the given number. For example a number 6 will have factors 1,2,3,6. Now,i tried to combine the first and the last digit so that the product becomes equal to 6.
I have not used the logic of finding the smallest number in that array right now. I will do it later.
Question is Why i am getting Exception as ArrayIndexOutOfBound? [i couldn't figure out]
Below is my code
public class SmallestNoProduct {
public static void generateFNos(int max) {
int ar[] = new int[max];
int k = 0;
for (int i = 1; i <= max; i++) {
if (max % i == 0) {
ar[k] = i;
k++;
}
}
smallestNoProduct(ar);
}
public static void smallestNoProduct(int x[]) {
int j[] = new int[x.length];
int p = x.length;
for (int d = 0; d < p / 2;) {
String t = x[d++] + "" + x[p--];
int i = Integer.parseInt(t);
j[d] = i;
}
for (int u = 0; u < j.length; u++) {
System.out.println(j[u]);
}
}
public static void main(String s[]) {
generateFNos(6);
}
}
****OutputShown****
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6
at SmallestNoProduct.smallestNoProduct(SmallestNoProduct.java:36)
at SmallestNoProduct.generateFNos(SmallestNoProduct.java:27)
at SmallestNoProduct.main(SmallestNoProduct.java:52)
#Edit
The improved Code using array only.
public class SmallestNoProduct {
public static void generateFNos(int max) {
int s = 0;
int ar[] = new int[max];
int k = 0;
for (int i = 1; i <= max; i++) {
if (max % i == 0) {
ar[k] = i;
k++;
s++;
}
}
for (int g = 0; g < s; g++) {
System.out.println(ar[g]);
}
smallestNoProduct(ar, s);
}
public static void smallestNoProduct(int x[], int s) {
int j[] = new int[x.length];
int p = s - 1;
for (int d = 0; d < p;) {
String t = x[d++] + "" + x[p--];
System.out.println(t);
int i = Integer.parseInt(t);
j[d] = i;
}
/*for (int u = 0; u < j.length; u++) {
System.out.println(j[u]);
}*/
}
public static void main(String s[]) {
generateFNos(6);
}
}
Maybe it better:
public class SmallestNoProduct {
public static int smallest(int n) {
int small = n*n;
for(int i = 1; i < Math.sqrt(n); i++) {
if(n%i == 0) {
int temp = Integer.parseInt(""+i+""+n/i);
int temp2 = Integer.parseInt(""+n/i+""+i);
temp = temp2 < temp? temp2: temp;
if(temp < small) {
small = temp;
}
}
}
return small;
}
public static void main(String[] args) {
System.out.println(smallest(6)); //6
System.out.println(smallest(10)); //25
System.out.println(smallest(100)); //205
}
}
Problem lies in this line
String t=x[d++]+""+x[p--];
x[p--] will try to fetch 7th position value, as p is length of array x i.e. 6 which results in ArrayIndexOutOfBound exception. Array index starts from 0, so max position is 5 and not 6.
You can refer this question regarding postfix expression.
Note: I haven't checked your logic, this answer is only to point out the cause of exception.
We are unnecessarily using array here...
below method should work....
public int getSmallerMultiplier(int n)
{
if(n >0 && n <10) // if n is 6
return (1*10+n); // it will be always (1*10+6) - we cannot find smallest number than this
else
{
int number =10;
while(true)
{
//loop throuogh the digits of n and check for their multiplication
number++;
}
}
}
int num = n;
for(i=9;i>1;i--)
{
while(n%d==0)
{
n=n/d;
arr[i++] = d;
}
}
if(num<=9)
arr[i++] = 1;
//printing array in reverse order;
for(j=i-1;j>=0;j--)
system.out.println(arr[j]);
In the codility test NumberOfDiscIntersections I am getting perf 100% and correctness 87% with the one test failing being
overflow
arithmetic overflow tests
got -1 expected 2
I can't see what is causing that given that I am using long which is 64-bit. And even if I can get it to 100% perf 100% correctness I am wondering if there is a better way to do this that is not as verbose in Java.
edit: figured out a much better way to do with with two arrays rather than a pair class
// you can also use imports, for example:
import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
int j = 0;
Pair[] arr = new Pair[A.length * 2];
for (int i = 0; i < A.length; i++) {
Pair s = new Pair(i - A[i], true);
arr[j] = s;
j++;
Pair e = new Pair(i + A[i], false);
arr[j] = e;
j++;
}
Arrays.sort(arr, new Pair(0, true));
long numIntersect = 0;
long currentCount = 0;
for (Pair p: arr) {
if (p.start) {
numIntersect += currentCount;
if (numIntersect > 10000000) {
return -1;
}
currentCount++;
} else {
currentCount--;
}
}
return (int) numIntersect;
}
static private class Pair implements Comparator<Pair> {
private long x;
private boolean start;
public Pair(long x, boolean start) {
this.x = x;
this.start = start;
}
public int compare(Pair p1, Pair p2) {
if (p1.x < p2.x) {
return -1;
} else if (p1.x > p2.x) {
return 1;
} else {
if (p1.start && p2.start == false) {
return -1;
} else if (p1.start == false && p2.start) {
return 1;
} else {
return 0;
}
}
}
}
}
Look at this line:
Pair s = new Pair(i + A[i], true);
This is equivalent with Pair s = new Pair((long)(i + A[i]) , true);
As i is integer, and A[i] is also integer, so this can cause overflow, as value in A[i] can go up to Integer.MAX_VALUE, and the cast to long happened after add operation is completed.
To fix:
Pair s = new Pair((long)i + (long)A[i], true);
Note: I have submitted with my fixed and got 100%
https://codility.com/demo/results/demoRRBY3Q-UXH/
My todays solution. O(N) time complexity. Simple assumption that number of availble pairs in next point of the table is difference between total open circle to that moment (circle) and circles that have been processed before. Maybe it's to simple :)
public int solution04(int[] A) {
final int N = A.length;
final int M = N + 2;
int[] left = new int[M]; // values of nb of "left" edges of the circles in that point
int[] sleft = new int[M]; // prefix sum of left[]
int il, ir; // index of the "left" and of the "right" edge of the circle
for (int i = 0; i < N; i++) { // counting left edges
il = tl(i, A);
left[il]++;
}
sleft[0] = left[0];
for (int i = 1; i < M; i++) {// counting prefix sums for future use
sleft[i]=sleft[i-1]+left[i];
}
int o, pairs, total_p = 0, total_used=0;
for (int i = 0; i < N; i++) { // counting pairs
ir = tr(i, A, M);
o = sleft[ir]; // nb of open till right edge
pairs = o -1 - total_used;
total_used++;
total_p += pairs;
}
if(total_p > 10000000){
total_p = -1;
}
return total_p;
}
int tl(int i, int[] A){
int tl = i - A[i]; // index of "begin" of the circle
if (tl < 0) {
tl = 0;
} else {
tl = i - A[i] + 1;
}
return tl;
}
int tr(int i, int[] A, int M){
int tr; // index of "end" of the circle
if (Integer.MAX_VALUE - i < A[i] || i + A[i] >= M - 1) {
tr = M - 1;
} else {
tr = i + A[i] + 1;
}
return tr;
}
My take on this, O(n):
public int solution(int[] A) {
int[] startPoints = new int[A.length];
int[] endPoints = new int[A.length];
int tempPoint;
int currOpenCircles = 0;
long pairs = 0;
//sum of starting and end points - how many circles open and close at each index?
for(int i = 0; i < A.length; i++){
tempPoint = i - A[i];
startPoints[tempPoint < 0 ? 0 : tempPoint]++;
tempPoint = i + A[i];
if(A[i] < A.length && tempPoint < A.length) //first prevents int overflow, second chooses correct point
endPoints[tempPoint]++;
}
//find all pairs of new circles (combinations), then make pairs with exiting circles (multiplication)
for(int i = 0; i < A.length; i++){
if(startPoints[i] >= 2)
pairs += (startPoints[i] * (startPoints[i] - 1)) / 2;
pairs += currOpenCircles * startPoints[i];
currOpenCircles += startPoints[i];
currOpenCircles -= endPoints[i];
if(pairs > 10000000)
return -1;
}
return (int) pairs;
}
The explanation to Helsing's solution part:
if(startPoints[i] >= 2) pairs += (startPoints[i] * (startPoints[i] - 1)) / 2;
is based on mathematical combinations formula:
Cn,m = n! / ((n-m)!.m!
for pairs, m=2 then:
Cn,2 = n! / ((n-2)!.2
Equal to:
Cn,2 = n.(n-1).(n-2)! / ((n-2)!.2
By simplification:
Cn,2 = n.(n-1) / 2
Not a very good performance, but using streams.
List<Long> list = IntStream.range(0, A.length).mapToObj(i -> Arrays.asList((long) i - A[i], (long) i + A[i]))
.sorted((o1, o2) -> {
int f = o1.get(0).compareTo(o2.get(0));
return f == 0 ? o1.get(1).compareTo(o2.get(1)) : f;
})
.collect(ArrayList<Long>::new,
(acc, val) -> {
if (acc.isEmpty()) {
acc.add(0l);
acc.add(val.get(1));
} else {
Iterator it = acc.iterator();
it.next();
while (it.hasNext()) {
long el = (long) it.next();
if (val.get(0) <= el) {
long count = acc.get(0);
acc.set(0, ++count);
} else {
it.remove();
}
}
acc.add(val.get(1));
}
},
ArrayList::addAll);
return (int) (list.isEmpty() ? 0 : list.get(0) > 10000000 ? -1 : list.get(0));
This one in Python passed all "Correctness tests" and failed all "Performance tests" due to O(n²), so I got 50% score. But it is very simple to understand. I just used the right radius (maximum) and checked if it was bigger or equal to the left radius (minimum) of the next circles. I also avoided to use sort and did not check twice the same circle. Later I will try to improve performance, but the problem here for me was the algorithm. I tried to find a very easy solution to help explain the concept. Maybe this will help someone.
def solution(A):
n = len(A)
cnt = 0
for j in range(1,n):
for i in range(n-j):
if(i+A[i]>=i+j-A[i+j]):
cnt+=1
if(cnt>1e7):
return -1
return cnt
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;
}