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

Modified Kaprekar Numbers

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  • zhengdx04
     + 40 comments

    I failed case 6 several times before identifying the problem in my code. My problem was due to a mix use of int and long long. I thought I only needed int for the number being check and long long for its square. Turned out it was risky without careful casting, because when you assign the multiplication of two int's to a long long variable when the multiplication exceeds the int limit, the long long variable only gets the chunked portion available in the int bits. Hope this somehow contributes to the troubleshooting options.

  • RodneyShag
     + 23 comments

    Java solution

    From my HackerRank solutions.

    It seems like the brute-force solution is the best we can do.

    Tips to avoid Integer Overflow

    If you multiply an int with another int, it may cause integer overflow. You can't simply say

    int num = 99999;
long squared = num * num; // causes integer overflow

    since that may cause integer overflow. You need a typecast to convert the int to a long before doing the multiplication.

    int num = 999999;
long squared = (long) num * num;

    Alternatively, you can just store everything as a long

    long num = 999999;
long squared = num * num;

    Also, if there are no Kaprekar numbers, you have to print "INVALID RANGE" exactly as I typed it (without the quotes).

    Working Code - passes 100% of test cases

    import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int p = scan.nextInt();
        int q = scan.nextInt();
        scan.close();
        
        boolean foundKaprekar = false;
        for (int num = p; num <= q; num++) {
            if (isKaprekar(num)) {
                foundKaprekar = true;
            }
        }
        if (!foundKaprekar) {
            System.out.println("INVALID RANGE");
        }
    }
    
    private static boolean isKaprekar(int num) {
        long squared = (long) num * num;
        String str   = String.valueOf(squared);
        String left  = str.substring(0, str.length() / 2);
        String right = str.substring(str.length() / 2);
        int numL = (left.isEmpty())  ? 0 : Integer.parseInt(left);
        int numR = (right.isEmpty()) ? 0 : Integer.parseInt(right);
        if (numL + numR == num) {
            System.out.print(num + " ");
            return true;
        } else {
            return false;
        }
    }
}

    Let me know if you have any questions.

  • NiksDesai02
     + 10 comments

    here is my code in C language. and I think

    int main() {
    long int n,m,r,f;
    int flag = 0;
    scanf("%ld\n%ld",&n,&m);
    for(long int i=n;i<=m;i++){
        long int sum;
        long int a = 0;
        long int k = i;
        while(k>0){
            k/=10 ; 
            ++a;
        }
       long int p = i * i;
        long int g = pow(10,a);
        sum = p%g + p/g;
        if(sum == i){
            printf("%ld ",sum);
            flag++;
        }
    }
    if(flag == 0)
        printf("INVALID RANGE");       
    return 0;
}

    it's very easy to understand for c programmers.

  • mohitjoshi
     + 4 comments

    Why is 10 not a Kaprekar number? or I am missing something here

  • shubhamgamer_611
     + 2 comments

    Simple and easy to undersatand

    Python code-:

    count = 0
for i in range(int(input()),int(input())+1):
    s = str(i**2)
    x = s[len(s)//2:]
    y = s[0:len(s)//2]
    if x=='':x=0
    if y=='':y=0
    if int(x)+int(y)==i:print(i,end=' ');count+=1
if count==0:print("INVALID RANGE")
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