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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Modified Kaprekar Numbers
  5. Discussions

Modified Kaprekar Numbers

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1042 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can what the explanation here : https://youtu.be/MttrQCHGu3w

    int getSum(long a){
    long sq = a * a;
    int digit = (int)log10(a) + 1;
    int result = sq % (int)pow(10, digit);
    int rest = (int)log10(sq) + 1 - digit;
    if(rest > 0) result += stoi(to_string(sq).substr(0, rest));
    return result;
}

void kaprekarNumbers(int p, int q) {
    bool valid_range = false;
    for(int i = p; i <= q; i++){
        int s = getSum(i);
        if(s == i){
           cout << i << " ";
           valid_range = true; 
        }  
    }
    if(!valid_range) cout << "INVALID RANGE";
}

  • robertbielicki
     + 0 comments

    RUST:

    fn kaprekarNumbers(p: i32, q: i32) {
    let mut results: Vec<String> = Vec::new();
    for n in p..=q {
        let square = (n as u128).pow(2).to_string();
        let middle = square.len() - n.to_string().len();

        let (left, right) = square.split_at(middle);
        let left_int = left.parse::<i32>().unwrap_or(0);
        let right_int = right.parse::<i32>().unwrap_or(0);
        if left_int + right_int == n {
            results.push(n.to_string());
        }
    }
    if results.len() > 0 {
        println!("{}", results.join(" "))
    } else {
        println!("INVALID RANGE")
    };
}

  • patrickgao2020
     + 0 comments

    Here is my rather long Python solution! The first function checks if a number is a modified kaprekar number, and the second function puts all numbers in the specified into the first function, and returns the results.

    def isk(num):
    d = len(str(num))
    num = str(num ** 2)
    if len(num) == 2 * d:
        if int(num[-d:]) + int(num[:d]) == math.sqrt(int(num)):
            return True
    elif len(num) == 2 * d - 1:
        if d != 1:
            if int(num[-d:]) + int(num[:d - 1]) == math.sqrt(int(num)):
                return True
        elif num == "1":
            return True
    return False

def kaprekarNumbers(p, q):
    knums = []
    for num in range(p, q + 1):
        if isk(num):
            knums.append(str(num))
    if len(knums) == 0:
        print("INVALID RANGE")
        return
    print(" ".join(knums))

  • qaisabusamak1
     + 0 comments

    in c++ code hackerrank.com

    in engligh

    void kaprekarNumbers(int p, int q) { for(int i;i

  • akshay_vijay123
     + 0 comments

    Here is my python code (simple brute force):

    def kaprekarNumbers(p, q):
    arr = []
    for i in range(p, q + 1):
        curr = str(i ** 2)
        d1 = curr[:len(curr) // 2] if len(curr) > 1 else "0"
        d2 = curr[len(curr) // 2:] if len(curr) > 1 else curr
        if int(d1) + int(d2) == i:
            arr.append(str(i))
    print(" ".join(arr) if arr != [] else "INVALID RANGE")