JS

let amount = 0;
    
let number = Math.ceil(Math.sqrt(a));
    
while(number*number <= b) {
	amount += 1;
	number += 1;
}
    
return amount;

c++

int squares(int a, int b) { int result = 0;

for(int i=1; i<=b;i++){
    int square = pow(i, 2);
    if(square<= b and square>=a){
        result++;
    }else if (square > b) {
        break;
    }
}

return result;

}

c++

int squares(int a, int b) { int result = 0;

for(int i=1; i<=b;i++){
    int square = pow(i, 2);
    if(square<= b and square>=a){
        result++;
    }else if (square > b) {
        break;
    }
}

return result;

}

Here are my c++ solutions, you can watch the explanation here : https://youtu.be/LtU0ItsvbMI

Solution 1 O(√n)

int squares(int a, int b) {
    int i = ceil(sqrt(a)), result = 0;
    for(; i * i <= b; i++) result++;
    return result;
}

Solution 2 O(1)

int squares(int a, int b) {
    int i = ceil(sqrt(a)), j = floor(sqrt(b));
    return j - i + 1;
}

function squares(a: number, b: number): number { // Write your code here let squaresCount: number = 0

let lower: number = Math.ceil(Math.sqrt(a))
let upper: number = Math.floor(Math.sqrt(b))

let squaresSet: Set<number> = new Set()

for (let i = lower; i <= upper; i++) {
    squaresSet.add(i * i)
}

console.log(`lower: `${lower}, upper: $`{upper}, squareSet: ${Array.from(squaresSet)}`)

return squaresSet.size

}