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Reverse of an array Hackerrank


By : maphey
Date : July 30 2020, 01:00 PM
To fix the issue you can do You are creating an array with n elements before you read n, i.e n is having garbage value. Here is correct working code.
code :

#include <iostream>

using namespace std; 
int main(){
    int n,i;
   cout<<"Enter number of elements"<<endl; // Comment it if you don't want debug console outut
    cin>>n;
     int arr[n];
     cout<<"Enter "<<n<<" elements"<<endl; // Comment it if you don't want debug  console outut
    for(int i=0;i<n;i++)
    {
        cin>>arr[i];

    }
        for(int i=n-1;i>=0;i--)
        {
            cout<<arr[i];
        }
}


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hackerrank.com find-hackerrank solution on Obj-C


By : rvagin
Date : March 29 2020, 07:55 AM
Any of those help The difference between the run in Xcode and on HackerRank is that when you run locally, the call to
code :
[input availableData];
#import <Foundation/Foundation.h>

int main()
{
    NSFileHandle *input;

    NSString *match = @"hackerrank";
    int amount;

    input = [NSFileHandle fileHandleWithStandardInput];
    NSArray *inputData = [[[NSString alloc] initWithData:[input availableData] encoding:NSUTF8StringEncoding] componentsSeparatedByString: @"\n"];
    amount = [inputData[0] intValue];

    for (int j = 1; j <= amount; j++)
    {
        NSString *str = inputData[j];

        NSArray *redexArr = @[match,
                              [NSString stringWithFormat:@"^%@.+", match],
                              [NSString stringWithFormat:@".+%@$", match]
                              ];

        for (int i = 2; i>=-1; i--)
        {
            if (i <= -1)
            {
                printf("-1\n");
            } else
            {
                NSPredicate *pred = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", redexArr[i]];
                if ([pred evaluateWithObject:str])
                {
                    printf("%d\n", i);
                    break;
                }
            }
        }
    }
    return 0;
}

Pass an array to a function, reverse it without using the Array.Reverse() method and return the reversed array


By : user1967888
Date : March 29 2020, 07:55 AM
To fix the issue you can do
I am actually trying to do it without using the temp variable in this case.
code :
static void Main(string[] args)
{
    int[] array = { 1, 2, 3, 4, 5 };
    int[] reverse = new int[5];
    ReverseArray(array, reverse);

    foreach (int item in reverse)
    {
        Console.WriteLine(item);
    }
}

public static void ReverseArray(int[] arr, int[] rev)
{
    int x = arr.Length - 1;
    for (int i = 0; i < arr.Length; i++, x--)
    {
        rev[i] = arr[x];
    }
}
static void Main(string[] args)
{
    int[] array = { 1, 2, 3, 4, 5 };
    int[] reverse = ReverseArray(array);

    foreach (int item in reverse)
    {
        Console.WriteLine(item);
    }
}

public static int[] ReverseArray(int[] arr)
{
    int[] rev = new int[arr.Length];
    int x = arr.Length - 1;
    for (int i = 0; i < arr.Length; i++, x--)
    {
        rev[i] = arr[x];
    }

    return rev;
}
static void Main(string[] args)
{
    int[] array = { 1, 2, 3, 4, 5 };
    IEnumerable<int> reverse = ReverseArray(array);

    foreach (int item in reverse)
    {
        Console.WriteLine(item);
    }
}

public static IEnumerable<int> ReverseArray(int[] arr)
{
    for (int i = arr.Length - 1; i >= 0; i--)
    {
        yield return arr[i];
    }
}
static void Main(string[] args)
{
    int[] array = { 1, 2, 3, 4, 5 };
    ReverseArray(array);

    foreach (int item in array)
    {
        Console.WriteLine(item);
    }
}

public static void ReverseArray(int[] arr)
{
    int x = arr.Length - 1;
    for (int i = 0; i < x; i++, x--)
    {
        int temp = arr[x];
        arr[x] = arr[i];
        arr[i] = temp;
    }
}
static void Main(string[] args)
{
    int[] array = { 1, 2, 3, 4, 5 };
    ReverseArray(array);

    foreach (int item in array)
    {
        Console.WriteLine(item);
    }
}

public static void ReverseArray(int[] arr)
{
    int x = arr.Length - 1;
    for (int i = 0; i < x; i++, x--)
    {
        arr[x] ^= arr[i];
        arr[i] ^= arr[x];
        arr[x] ^= arr[i];
    }
}

Hackerrank Algorithms Challenge Problem - showing type error in hackerrank environment (node) but works fine on the brow


By : user2285954
Date : March 29 2020, 07:55 AM
wish help you to fix your issue The function accepts only one parameter, not two. Look at how it's invoked:
https://www.hackerrank.com/challenges/birthday-cake-candles/problem
code :
let result = birthdayCakeCandles(ar);
function birthdayCakeCandles(ar) {

hackerrank problem runs fine in jupyter but fails in hackerrank


By : Flavio
Date : March 29 2020, 07:55 AM
it fixes the issue This code returns but doesn't print the output to stdout that the HR code runner is looking for. Try print(checkMagazine(magazine, note)).
In general, HR is a bit fussy about IO. Data will be read through stdin and will be printed to stdout, often in bizarre formats like "Yes" or "Impossible!" for a function that would normally return a boolean.

Segmentation fault when assigning to Array? HackerRank 2D Array - DS


By : hiteshgupta9193
Date : March 29 2020, 07:55 AM
To fix this issue This error occurs because of an out-of-scope array index that is causing a buffer overflow, an incorrectly initialised pointer, etc.
A signal is generated when a program either tries to read or write outside the memory that is allocated for it or to write memory that can only be read.
code :
#include <bits/stdc++.h>

using namespace std;

// Complete the hourglassSum function below.
int hourglassSum(vector<vector<int>> arr) {
    vector <int> sum(16);

    for(int i=0, s=0, countI=0;countI<4&&s<16;i++){  //Sums hourglass values one column  
        for(int j=0,countJ=0;countJ<4;j++){          //at a time at each row into vector sum
            if(i+2<6 && j+2<6){
                sum[s]+=arr[0+i][0+j];
                sum[s]+=arr[0+i][1+j];
                sum[s]+=arr[0+i][2+j];
                sum[s]+=arr[1+i][1+j];
                sum[s]+=arr[2+i][0+j];
                sum[s]+=arr[2+i][1+j];
                sum[s]+=arr[2+i][2+j];
                countJ++;
                s++;
            }
        }
        countI++;
    }
    sort (sum.begin(), sum.end());

    return sum.back();
}
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