DEV Community: Shashi Vardhan

Bit Counting Codewars

Shashi Vardhan — Thu, 03 Feb 2022 14:29:56 +0000

Write a function that takes an integer as input, and returns the number of bits that are equal to one in the binary representation of that number. You can guarantee that input is non-negative.

Example: The binary representation of 1234 is 10011010010, so the function should return 5 in this case

Code(Java)

public static int countBits(int n)

     {

         int count=0;

         String BinaryValues=Integer.toBinaryString(n);

         //System.out.println(BinaryValues);

         for(int i=0;i<BinaryValues.length();i++)

         {

              if(BinaryValues.charAt(i)=='1')

              {

                  count++;

              }

         }

         return count;

     }

Solution 2:

public static int countBits(int n){



    return Integer.bitCount(n);

  }

More at : https://onlylang.blogspot.com/

Popular Contest

Shashi Vardhan — Thu, 03 Feb 2022 09:06:53 +0000

Codeground is hosting the World's largest coding competition. There are N participants in the contest sitting in a single row.

The energy of i’th participant from left is A[i]. To raise the popularity of the contest, some rules are added.

A participant numbered i will challenge other participant numbered j if and only if j is ahead of him (j>i) and the distance between i and j (j-i) is prime.

The contest popularity value of participant i challenging participant j is A[j]-A[i]. The total contest popularity value of the competition is sum of popularity value of all such challenges.

Given the energy of all participants, you have to find the total popularity value.

INPUT

The first line contains N, the number of participants. The next line contains N space separated integers representing the energy of all the participants.

OUTPUT

Print a single line containing the total popularity value.

*CONSTRAINTS
*
1<=N<=1000

1 <=A[i]<= 1000000000

EXPLANATION OF SAMPLE

1 Sample input has N as 7 and participants energy as 24 6 8 10 12 14. 2. The contest popularity based on rules described is as below:

j-i val diff

5-0=5 12-2=10

3-0=3 8-2 = 6

2-0=2 6-2 = 4

6-1=5 14-4 = 10

4-1=3 10-4 = 6

3-1=2 8-4 = 4

5-2=3 12-6 = 6

4-2=2 10-6 = 4

6-3=3 14-8= 6

5-3=2 12-8 = 4

6-4=2 14-10=4

Total 64

Sample Input

7 2 4 6 8 10 12 14

Sample Output

Code(python 3)

def solution(n,A):

    sum=0

    for i in range(n):

        for j in range(n):

            if(j>i):

                num=j-i

                if(prime(num)):

                    sum=sum+(A[j]-A[i])

    return sum

def prime(n):

    if n==1:

        return False

    for i in range(2,int(n/2)+1):

        if n%i==0:

            return False

    return True

n=int(input())

A=list(map(int,input().split()))[:n]

print(solution(n,A))

More at : https://onlylang.blogspot.com/

Two Sum

Shashi Vardhan — Thu, 03 Feb 2022 09:03:32 +0000

Write a function that takes an array of numbers (integers for the tests) and a target number. It should find two different items in the array that, when added together, give the target value. The indices of these items should then be returned in a tuple like so: (index1, index2).

For the purposes of this kata, some tests may have multiple answers; any valid solutions will be accepted.

The input will always be valid (numbers will be an array of length 2 or greater, and all of the items will be numbers; target will always be the sum of two different items from that array).

Based on: http://oj.leetcode.com/problems/two-sum/

twoSum [1, 2, 3] 4 === (0, 2)

Code(Java Solution)

static int[] twoSum(int[] numbers, int target) {

         // TODO Auto-generated method stub

         int[] x = new int[2];

         for(int i=0;i<numbers.length;i++)

         {

              for(int j=0;j<numbers.length;j++)

              {

                  if(i!=j)

                  {

                       if(numbers[i]+numbers[j]==target)

                       {

                            x[0]=i;

                            x[1]=j;

                       }

                  }



              }

         }



        return x;

     }

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