Java solution :

public static int squares(int a, int b) {
    int startVal = (int)Math.sqrt(a);
    int endVal = (int)Math.sqrt(b);
    int sqrtcount = 0;

    for(int i=startVal ; i<=endVal ; i++) {
        int square = i*i;
        if(square>=a && square<=b) {
            sqrtcount++;
        }
    }

    return sqrtcount;
}

Easiest Javascript solution without any loop:

function squares(a, b) {
    let sqrtA=Math.ceil(Math.sqrt(a));
    let sqrtB=Math.floor(Math.sqrt(b));
    
    return (sqrtB-sqrtA)+1;

}

public static int squares(int start, int end) {
    int startVal = (int)Math.sqrt(start);
    int endVal = (int)Math.sqrt(end);
    int sqrtcount = 0;
    sqrtcount = (int)IntStream.rangeClosed(startVal, endVal).
    filter(a -> {
        int srt = a*a;
        return srt >= start && srt <=end;
    }).count();
    return sqrtcount;
}

Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-sherlock-and-squares-problem-solution.html

SOLUTION IN RUST:

fn squares(a: i32, b: i32) -> i32 {
    let c = (a as f64).sqrt().ceil() as i32;
    let f = (b as f64).sqrt().floor() as i32;

    (c..=f).count() as i32
}