# HackerRank Sherlock and Squares Solution

#### ByBrokenprogrammers

Dec 12, 2022

Hello Programmers, In this post, you will know how to solve the HackerRank Sherlock and Squares Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## HackerRank Sherlock and Squares Solution

Watson likes to challenge Sherlock’s math ability. He will provide a starting and ending value that describe a range of integers, inclusive of the endpoints. Sherlock must determine the number of square integers within that range.

Note: A square integer is an integer which is the square of an integer, e.g. 1, 4, 9, 16, 25.

Example
a = 24
b = 49

There are three square integers in the range: 25, 36 and 49. Return 3.

Function Description

Complete the squares function in the editor below. It should return an integer representing the number of square integers in the inclusive range from a to b.

squares has the following parameter(s):

• int a: the lower range boundary
• int b: the upper range boundary

Returns

• int: the number of square integers in the range

Input Format

The first line contains q, the number of test cases.
Each of the next q lines contains two space-separated integers, a and b, the starting and ending integers in the ranges.

Constraints

• 1 <= q <= 100
• 1 <= a <= b <= 109

Sample Input

2
3 9
17 24

Sample Output

2
0

Explanation

Test Case #00: In range [3, 9]4 and 9 are the two square integers.
Test Case #01: In range [17, 24], there are no square integers.

## HackerRank Sherlock and Squares Solution

### Sherlock and Squares Solution in C

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t;
int a,b,sa,sb;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&a,&b);
sa = sqrt(a), sb = sqrt(b);
if(sa*sa == a)sa--;
printf("%d\n",sb - sa);
}
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}```

### Sherlock and Squares Solution in Cpp

```#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;
}```

### Sherlock and Squares Solution in Java

```//package contest;
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) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
try{
int num = Integer.parseInt(line);
for(int i = 0; i < num; i++){
int a = Integer.parseInt(inp[0]);
int b = Integer.parseInt(inp[1]);
int count = 0;
for(int j = 1; j*j <= b; j++){
if(j*j >= a){
count++;
}
}
System.out.println(count);
}

} catch(Exception e) {

}

}
}```

### Sherlock and Squares Solution in Python

```from math import sqrt
T = int(input())
for i in range(T):
count = 0
inp = [int(a) for a in input().split()]
c = inp[0]
while int(sqrt(c)) != sqrt(c) and c<inp[1]:
c+=1
while c <= inp[1] and int(sqrt(c)) == sqrt(c):
count += 1
c += 2*sqrt(c)+1
print(count)```

### Sherlock and Squares Solution using JavaScript

```function processData(input) {
lines = input.trim().split("\n");

for(var i=1;i<lines.length && lines[i].length > 0;i++){
start = parseInt(lines[i].trim().split(" ")[0]);
end = parseInt(lines[i].trim().split(" ")[1]);
squares =0;

min = Math.sqrt(start);

if(min != Math.floor(min)){
min = Math.floor(min) + 1;
}

max = Math.sqrt(end);
if(max != Math.floor(max)){
max = Math.floor(max);
}

if(max < min){
console.log("0");
} else {

console.log((max-min+1));
}

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

### Sherlock and Squares Solution in Scala

```object Solution {
def main(args: Array[String]) {
for(i <- 1 to t) println(f(readLine.split(" ")));
}

def f(n:Array[String]):Long = {
return Math.sqrt(n(1).toInt).toLong - Math.sqrt(n(0).toInt-1).toLong;
}
}```

### Sherlock and Squares Solution in Pascal

```program PSherlockandSquares;
Uses math;
var T,A,B,i:Longint;
begin
for i:=1 to T do
begin
WriteLn(floor(sqrt(B))-ceil(sqrt(A))+1);
end;
end.```

Disclaimer: This problem (Sherlock and Squares) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.