HackerRank Sherlock and Squares Solution

In this post, we will solve HackerRank Sherlock and Squares Problem 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.

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.

Table of Contents

Sherlock and Squares C Solution

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

int main() {

    int t = 0, tc = -1;
    char tmp[23];
    char cstr[101][23] = {{0}};

    /* Get stdin */
    while (t != tc) {
        memset(tmp, 0, sizeof(tmp));
        fgets(tmp, sizeof(tmp), stdin);
        if (tc == -1) {
            tc = atoi(tmp);
        } else {
            if (tmp[strlen(tmp)-1] == '\n') { tmp[strlen(tmp)-1] = '\0'; }
            strcpy(cstr[t], tmp);
            t++;
        }
    }
    char *tok;
    int a = 0, b = 0, c = 0;
    for (t = 0; t < tc; t++) {
        tok = strtok(cstr[t], " ");
        a = atoi(tok);
        tok = strtok(NULL, " ");
        b = atoi(tok);
        if (sqrt(a) != (int)sqrt(a)) { a = (int)(pow(((int)sqrt(a))+1, 2)); }
        while (a <= b) {
            c++;
            a = (int)(pow(((int)sqrt(a))+1, 2));
        }
        printf("%d\n", c);
        a = 0; b = 0; c = 0;
    }
     return EXIT_SUCCESS;
}

Sherlock and Squares C++ Solution

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


int main() {
	std::vector<unsigned int> squares;
	for (int i = 1; i < 31655; ++i) {
		squares.push_back(i*i);
	}

	int n;
	cin >> n;

	for (int i = 0; i < n; ++i) {
		unsigned int a, b, c = 0;
		cin >> a >> b;

		auto lower = std::lower_bound(squares.begin(), squares.end(), a);
		auto upper = std::upper_bound(squares.begin(), squares.end(), b);

		cout << upper - lower << endl;
	}

	return 0;
}

Sherlock and Squares C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
class Solution {
    

    
    static void Main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
        int T=int.Parse(Console.ReadLine());
        for(int i =1;i<=T;i++)
            {
            String inp=Console.ReadLine();
            long []arr=Array.ConvertAll(inp.Split(null),s=>(long.Parse(s)));
            long  first,last ;
            first=(long)Math.Sqrt(arr[0]);
            last=(long)Math.Sqrt(arr[1]);
            if(first*first!=arr[0])
                {
                first=first+1;
            }
            
            Console.WriteLine(last-first+1);
        }
    }
}

Sherlock and Squares Java Solution

import java.util.*;
public class Sherlock_and_Squares {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCase = sc.nextInt();
        while(testCase-->0){
            int start =sc.nextInt();
            int end =sc.nextInt();
            int count=0;
            int square_start = (int)Math.ceil(Math.sqrt(start));
            int square_end = (int)Math.ceil(Math.sqrt(end));
            for(int i =square_start;i<=square_end;i++){
                if(i*i<=end){
                    count++;
                }
            }
            System.out.println(count);
        }
    }
}

Sherlock and Squares JavaScript Solution

function processData(input) {
    
    var data = input.split("\n");
    
    var T = data.shift();
    
    data.forEach(function(list){
        
        var range = list.split(" ");
        
        var min = Math.sqrt( parseInt(range[0], 10) );
        
        var max = Math.sqrt( parseInt(range[1], 10) );
        
        var diff = Math.floor(max) - Math.ceil(min) + 1;
        
        process.stdout.write(diff + "\n");
        
    });
    
} 

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 Python Solution

import math
n = int(input())

for _ in range(n):
    temp = input().split()
    a = int(temp[0])
    b = int(temp[1])
    i = math.ceil(math.sqrt(a))
    num = 0
    while i * i <= b:
        num = num + 1
        i = i + 1
    print(num)