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

Modified Kaprekar Numbers

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1041 Discussions

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  • 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")

  • 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";
}

  • Kavithakumari351
     + 0 comments

    JavaCode

    public static void kaprekarNumbers(int p, int q) {
        List<Integer> list = new ArrayList<>();
        for (int i = p; i <= q; i++) { 
            long sqr = (long) i * i;
            String str = Long.toString(sqr);
            int len = str.length();
            int numLen = Integer.toString(i).length();
            
            String r = str.substring(len - numLen);    
            String l = str.substring(0, len - numLen);  
            if (l.isEmpty()) {
                l= "0";
            }
            int sum = Integer.parseInt(l) + Integer.parseInt(r);
            if (sum == i) {
                list.add(i);
            }
        }
        if (list.isEmpty()) {
            System.out.print("INVALID RANGE");
        } else {
            for (int num : list) {
                System.out.print(num + " ");
            }
        }
    }