HackerRank Sherlock and Squares problem solution

In this HackerRank Sherlock and Squares problem, you need to complete the squares function that should return an integer representing the number of square integers in the inclusive range from a to b.

Hackerrank Sherlock and Squares problem solution in Python programming.

t = int(input())

for i in range (0,t):
	count = 0
	a,b = [int(j) for j in input().strip().split()]
	square1 = a ** (.5)
	if (square1 != int(square1)):
		a1 = int(square1) + 1
	else:
		a1 = int(square1)
	square2 = b ** (.5)
	b1 = int(square2)
	count = b1 - a1 +1
	print(count)

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int square_count = 0;
        int test_cases = in.nextInt();
        int from;
        int to;
        int squareroot;
        for (int i = 0; i < test_cases; i++) {
            from = in.nextInt();
            to = in.nextInt();
            int a = (int)Math.ceil(Math.sqrt(from));
            int b = (int)Math.floor(Math.sqrt(to));
            square_count = b - a + 1;
            System.out.println(square_count);
        }
        
        in.close();
        
    }
    

}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
   int n,m;
  cin>>n;
  
  while (cin>>n>>m)
    {
        cout<<(int)(sqrt(m)+0.0000001)-(int)(sqrt(n-1)+0.0000001)<<endl;
    }
  return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() 
{
  
  //Grab the number of tests.
  int T;
  scanf( "%d", &T );
 
  //Check for SQ T times.
  for( int i = 0; i < T; i++ )
  {
    long long int A;
    scanf( "%llu", &A );
  
    long long int B;
    scanf( "%llu", &B );
    
    long long int count = 0;

     
    long long int sqA = sqrt( A );
    long long int sqB = sqrt( B );
    long double sqDoub = (long double)A;
    sqDoub = sqrt( A );
    if( sqDoub != sqA )
      count--;
    count += sqB - sqA + 1;
    printf( "%llun", count );   
  }
    return 0;
 
}

Problem solution in JavaScript programming.

function processData(input) {
    //Enter your code here
    var lines = input.split('n');
    var testCount = lines[0];
    
    for(var i=1;i<lines.length;++i){
        var test = lines[i].split(' ');
        var hit = 0;
        var min = Math.ceil(Math.sqrt(test[0]));
        var max = Math.sqrt(test[1]);
        
        for(var j = min; j <= max; ++j){
            if( (j * j)<= test[1] )
                hit += 1;
        }
        console.log(hit);
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});