Beautiful Triplets , Equalize the Array, Sock Merchant Hacker Rank Solution in C/c++ ,Weapon value code chef solution in c

Beautiful Triplets

here a[j]-a[i] must be equal to d and 

a[k]-[j] = d

so solving a[j]-a[i] =d we get a[i] = a[j] -d

and solving a[k]-a[j] =d we get a[k] = a[j] + d

so according to problem statement a[j]-d and a[j] +d must be in the array...

but if we write the if condition as h[a[i]-d] >=1 and h[a[i]+d >=1 then this logic will fail wen there are repetitions

Example:

• 10 3

• 1 6 7 7 8 10 12 13 14 19

here 10-3 = 7 and 7 exists in array

10+3 is 13 and 13 also exists in array but the if condition will only count once

but since there is another 7, another beautiful triplet (7,10,13) also exists.

No we must change the if condition

a[j]-d, a[j] and a[j]+d must be in array

it is as good as

a[i], a[i]+d, a[i]+2d must exist in array (since a[i] = a[j]-d)

so our for loop would be

h[a[i]+d] >=1 and h[a[i]+2*d] >=1

Solution:

#include <stdio.h>

int main() 

{

    int n,d; 

    scanf("%d%d", &n,&d);

    long long int a[n],h[2000001]={0},c=0,i;

    for (i = 0; i < n; i++)

    {

       scanf("%lld",&a[i]);

        h[a[i]]++;

    }

    for( i=0;i<n;i++)

        if( h[a[i]+d] >=1 && h[a[i]+2*d]>=1)

            c++;

    printf("%lld\n", c);

}

Equalize the Array

https://www.hackerrank.com/challenges/equality-in-a-array/problem

#include <stdio.h>

int main()

{

    int n,h[101]={0};

    scanf("%d", &n);

    int a[n],max=0,c=0;

    for(int i=0;i<n;i++)

    {

        scanf("%d", &a[i]);

        h[a[i]]++;

        if(h[a[i]]>max)

            max=h[a[i]];

    }

    printf("%d",n-max);

}

Sock Merchant

https://www.hackerrank.com/challenges/sock-merchant/copy-from/62004605

In C :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main() {
    int n,i,s=0;
    scanf("%i", &n);
    int *ar = malloc(sizeof(int) * n);
    int b[101]={0};
    for(int ar_i = 0; ar_i < n; ar_i++){
       scanf("%i",&ar[ar_i]);
        b[ar[ar_i]]++;
    }
    for(i=0;i<101;i++)
        s=s+b[i]/2;
    printf("%d\n", s);
    return 0;
}

In C++

int main() {

    map<int,int> socks;

    int i, total=0;

    cin.ignore();

    while(cin>>i){

       socks[i]++;

       if(!(socks[i]%2))

           total++;

    }

    cout<<total;

    return 0;

}

In Java

static int sockMerchant(int n , int ar[])
{
         Map<Integer, Integer> map = new HashMap();
         int ans=0;
         for(int val: ar)
         {
               Integer c = map.get(val);
               if(c == null)
                      map.put(val,1);
               else
                      map.put(val,c+1);
          }
          for(int z: map.values())
                    ans=ans+z/2;
          return ans;
}

========================================================================
https://www.hackerrank.com/challenges/beautiful-triplets/problem

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int h[20001]={0};
int beautifulTriplets(int d, int arr_size, int* arr) {
    // Complete this function
    int c=0;
    for(int i=1;i<arr_size;i++)
        if(h[arr[i]-d] ==1 && h[arr[i]+d]==1)
            c++;
    return c;
}

int main() {
    int n;
    int d;
    scanf("%i %i", &n, &d);
    int *arr = malloc(sizeof(int) * n);
    for (int arr_i = 0; arr_i < n; arr_i++) {
       scanf("%i",&arr[arr_i]);
        h[arr[arr_i]]++;
    }
    int result = beautifulTriplets(d, n, arr);
    printf("%d\n", result);
    return 0;
}
========================================================================

Weapon Value CodeChef solution 

#include<stdio.h>

int main() {
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,j,c=0;
        scanf("%d",&n);
        char st[n][11];
        int h[26]={0};
        for(i=0;i<n;i++)
            scanf("%s",&st[i]);
        for(i=0;i<n;i++)
        {
            for(j=0;st[i][j]!='\0';j++)
            {
                if(st[i][j] == '1')
                    h[j]++;
            }
        }
        for(i=0;i<11;i++)
            if(h[i] !=0 && h[i]%2 == 1)
                c++;
        printf("%d\n",c);
    }
}

Other Wonderful problems on hashing

https://www.hackerearth.com/practice/basic-programming/implementation/basics-of-implementation/practice-problems/algorithm/car-names-4/

https://enthusiaststudent.blogspot.com/2018/02/anagram-hacker-rank-solution-in-c.html
https://enthusiaststudent.blogspot.com/2019/07/sherlock-and-valid-string-hacker-rank.html

https://www.hackerrank.com/contests/w38/challenges/minute-to-win-it/problem