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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Sherlock and Squares
  5. Discussions

Sherlock and Squares

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  • bladoncgarland
     + 0 comments

    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

    }

  • akturk_emrullah
     + 0 comments

    Here is the my Kotlin solution :

    fun squares(a: Int, b: Int): Int {

    val lowerLimit = ceil(sqrt(a.toDouble())).toInt()
val upperLimit = floor(sqrt(b.toDouble())).toInt()  

return  upperLimit-lowerLimit + 1

    }

  • illogicalsleepi1
     + 0 comments
    s=int(a**(0.5))
count=0
while b>=(s**2):
        if ((s**2)<=b and (s**2)>=a):
                count+=1
        s+=1
return count

    `

  • alban_tyrex
     + 0 comments

    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;
}

  • Mo_AATROX
     + 0 comments
    public static int squares(int a, int b) {
        int min = (int)Math.ceil(Math.sqrt(a));
        int max = (int)Math.floor(Math.sqrt(b));
        int i = min;
        int result = 0;
        while(i<=max){
            if(Math.pow(i,2)>=a || Math.pow(i,2)<=b)
                result++;
            i++;
        }
        return result;
    }