Thanks this worked in java 8

This is giving wrong solution in c#

it doesn't checked constraints provided i.e numbers should lie between 1 & 100 (both included)

java :

int x = (a0 > b0? ascore++ : (a0 == b0 ? 0 : bscore++)) + (a1 > b1? ascore++ : (a1 == b1 ? 0 : bscore++)) + (a2 > b2? ascore++ : (a2 == b2 ? 0 : bscore++));

    System.out.println(ascore +" "+bscore);

Complete solution in Java:

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a0 = in.nextInt();
        int a1 = in.nextInt();
        int a2 = in.nextInt();
        int b0 = in.nextInt();
        int b1 = in.nextInt();
        int b2 = in.nextInt();
        int  pointsAlice = ((a0>b0)?1:0)+ ((a1>b1)?1:0)+ ((a2>b2)?1:0) ;
        int pointsBob = ((a0<b0)?1:0)+ ((a1<b1)?1:0)+ ((a2<b2)?1:0) ;
        System.out.println(pointsAlice+" " +pointsBob);
    }
}

That is actually pretty interesting can you tell me what is the name of this syntax?

I used if and if else, and its very long, whereas your code seems very short and easy. What is this sort of code called? I want to read more about these shorter codes. Thank You

how to return the values from this function?

Similar solution in Python3:

def solve(a0, a1, a2, b0, b1, b2):
    # Complete this function
    a = (1 if a0 > b0 else 0) + (1 if a1 > b1 else 0) + (1 if a2 > b2 else 0)
    b = (1 if a0 < b0 else 0) + (1 if a1 < b1 else 0) + (1 if a2 < b2 else 0)
    return (a,b)

Is there a technical name for this type of comparison? I'd like to look more into this

why its is showing this error -"missing return"??

the function returning integer pointer right? and this is received in main by a pointer result.we have to sent comparision values of two person,how is it possible to return two values to single pointer??can anyone explain plzzz

what to return back?

Dang... short and good..

can u plzz tell why you have used + operator.

what about checking limit 0

can you tell me how you solved this with explanation.

tysm for helping me

can you please tell me its working.

Better:

static int boolToInt(boolean condition){
        int retorno = (condition)?1:0;
        return retorno;
    }
    
    static int[] solve(int a0, int a1, int a2, int b0, int b1, int b2){
        int[] retorno = new int[2];
        retorno[0] = 0;     //Alice
        retorno[1] = 0;     //Bob
        
        retorno[0] = boolToInt(a0>b0)+boolToInt(a1>b1)+boolToInt(a2>b2);
        retorno[1] = boolToInt(a0<b0)+boolToInt(a1<b1)+boolToInt(a2<b2);
        
        return retorno;
    }

What a lovely solution. I never would've thought to look at it that way. What was your thought process?

this is wrong, because solve method need return int [] value

why my code wrong? but then output is right

static int[] solve(int[] a, int[] b) {

    int pointAlice = 0; 
    int pointBob = 0;
    for(int i=0; i < ((a.length<=b.length)?a.length:b.length);i++){
        pointAlice += ((a[i]>b[i])?1:0);
        pointBob += ((a[i]>b[i])?0:1);
    }
    return new int[] {pointAlice,pointBob};

}

This does not scale in case more awarding points comes into play...

her asking return statement .how will we can give it

this is not for genral use.....

This is great you are saving the time of the loop, and do applies since it's restricted to 3 games

This is not working in Java for me

That is almost exactly how I did it. haha Arrays seemed most logical to me.

static int[] solve(int a0, int a1, int a2, int b0, int b1, int b2){
    // Complete this function
    int[] alice = {a0, a1, a2};
    int[] bob = {b0, b1, b2};
    int[] result = {0,0};
    for(int i=0; i<3; i++){
        if(alice[i] > bob[i]){
            result[0]++;
        }
        else if(alice[i] < bob[i]) result[1]++;
    }
    return result;

isn't it hard coded? solve can take only two elements

Awesome!

Same problem here!

Same here :(

same here. I kept losing my cool, thinking I did something wrong. I tried multiple solutions, arrived at the correct output 3 -way times, but getting a runtime error with result.join as you say. Same for python. LAME!

Yea if you chose the JavaScript (nodejs) template for the response, the solve method expects you to return an array.

[scoreAlice scoreBob]

But tbh you should be able to figure that out by reading the error ;) Good luck.

should use function

Even better c++ code:

#include <vector>
#include <iterator>
#include <algorithm>
#include <numeric>
#include <iostream>
using namespace std;

int main() {
    vector<int> a(3), b(3);
    copy_n(istream_iterator<int>(cin), 3, begin(a));
    copy_n(istream_iterator<int>(cin), 3, begin(b));
    
    cout << inner_product(begin(a), end(a), begin(b), 0, plus<int>(), greater<int>()) << ' ' 
         << inner_product(begin(b), end(b), begin(a), 0, plus<int>(), greater<int>())
         << endl;
    return 0;
}

Over here why have you multiplied the brackets by 1.I mean does this signify anything?

super

how can we return value in this code?

" int alice = (a0 > b0)*1+(a1>b1)*1+(a2>b2)*1; int bob = (b0>a0)*1+(b1>a1)*1+(b2>a2)*1;" the above line will show up a syntax error.

simplest code in c here:

(https://github.com/akkshaychauhan/hacckerank-code/blob/master/Algorithms:Compare%20the%20Triplets)

Now is working.

**using namespace std;

int a0, a1, a2, b0, b1, b2; ** int main () {

cin >> a0 >> a1 >> a2;

cin >> b0 >> b1 >> b2;


int alice = (a0 > b0)*1+(a1>b1)*1+(a2>b2)*1;

int bob = (b0>a0)*1+(b1>a1)*1+(b2>a2)*1;

cout << alice << " " << bob << endl;

}

A = list(map(int, raw_input().strip().split()))
B = list(map(int, raw_input().strip().split()))

Ascore = len([1 for a,b in zip(A,B) if a>b])
Bscore = len([1 for a,b in zip(A,B) if b>a])
print Ascore, Bscore

Above code will work for any length of vector. You can use sum instead of len as well.

Another python 2 solution. Not better, because less readable but works independantly of the range:

a = map(int,raw_input().split())
b = map(int,raw_input().split())
print ' '.join(map(str,map(sum,zip(*[[x>y,y>x] for x,y in zip(a,b)]))))

A little modification for any length:

print(sum(1 for x in range(len(a)) if a[x] > b[x]), sum(1 for x in range(len(b)) if b[x] > a[x]))

New to Python 3. Just learned about list comprehensions so I'm using them everywhere. Feedback please.

a = [int(x) for x in input().strip().split(' ')]
b = [int(x) for x in input().strip().split(' ')]

# 1 is a point for Alice, -1 a point for Bob
score = [1 if x > y else 0 if x == y else -1 for x, y in zip(a, b)]

print (score.count(1), score.count(-1))

edit: replaced sum() with .count

this is what I aimed at for

The zip function is vital.

Good work!

this is perfect. Anything shorter becomes less readable. Using just 1 special function (zip) which is quite familiar in Python. Also, scalable to more items, and uses clear variable names.

One liners are not for professional coders, but for script kiddies showing off their code-fu.

Great solution, I'd remove the ifs though:

for a, b in zip(alice, bob):
    score[0] += a > b
    score[1] += a < b

static int[] solve(int a0, int a1, int a2, int b0, int b1, int b2) {

    int[] aScore = {a0,a1,a2};
    int[] bScore = {b0,b1,b2};
    int aliceScore=0, bobScore = 0;
    for(int i=0; i < 3; i++){

        if(aScore[i] > bScore[i]){
            aliceScore += 1;
        }else if (bScore[i] > aScore[i]){
            bobScore += 1;
        }
    }
    int[] result = {aliceScore,bobScore};
    return result;
}

can you please explain how vector v works ?