HackerRank Modified Kaprekar Numbers problem solution

In this HackerRank Modified Kaprekar Numbers problem solution Consider a positive whole number n with d digits. We square n to arrive at a number that is either 2 x d digits long or (2 x d) – 1 digit long. Split the string representation of the square into two parts, l, and r. The right-hand part, r must be d digits long. The left is the remaining substring. Convert those two substrings back to integers, add them and see if you get n.

Given two positive integers p and q where p is lower than q, write a program to print the modified Kaprekar numbers in the range between p and q, inclusive. If no modified Kaprekar numbers exist in the given range, print INVALID RANGE.

Function Description

Complete the kaprekarNumbers function in the editor below.

kaprekarNumbers has the following parameter(s):

• int p: the lower limit
• int q: the upper limit

Prints

It should print the list of modified Kaprekar numbers, space-separated on one line and in ascending order. If no modified Kaprekar numbers exist in the given range, print INVALID RANGE. No return value is required.

Hackerrank Modified Kaprekar Numbers problem solution in Python.

def kaprekar(n):
    d = len(str(n))
    n_sqr = str(n*n)
    right = int(n_sqr[len(n_sqr)-d:])
    left = n_sqr[:len(n_sqr)-d]
    if left == '':
        left = 0
    left = int(left)
    #print("left:", left, "right:", right)
    return left+right == n
    
p = int(input())
q = int(input())
a = []
for i in range(p, q+1):
    if kaprekar(i):
        a.append(i)
if len(a) == 0:
    print("INVALID RANGE")
else:
    print(' '.join(str(c) for c in a))

Modified Kaprekar Numbers problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {   
    
    
    public static void main(String[] args) throws Exception {
              
        Scanner sc = new Scanner(System.in); 
        int p = sc.nextInt();
        int q = sc.nextInt();
        boolean isFirst = true;
        for(int i = p; i <= q; i++)
            if(isKaprekar(i)) {
                System.out.printf((isFirst) ? "%d" : " %d", i);
                isFirst = false;
            }
        System.out.println((isFirst) ? "INVALID RANGE" : "");
    }
    
    public static boolean isKaprekar(long a) {
        int d = String.valueOf(a).length();
        String sqr = String.valueOf(a * a);
        if(sqr.length() == 1) return a == 1;
        int idx = sqr.length() - d;
        long x = Long.valueOf(sqr.substring(0, idx));        
        long y = Long.valueOf(sqr.substring(idx));
        return x + y == a;
    }
}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    long long p;
    long long q;
    cin>>p;
    cin>>q;
    vector<long long> res;
    for(long long i=p; i<=q; ++i){
        long long sq = i*i;
        string s = to_string(sq);
        int d = s.size()/2;
        if(d == 0){
            if(i == sq){
                res.push_back(i);
            }
            continue;
        }
        if(stoll(s.substr(0,d))+stoll(s.substr(d,s.size()-d)) == i){
            res.push_back(i);
        }
           }
    if(res.size()>0){
           for(int i=0; i<res.size(); ++i){
               cout<<res[i]<<" ";
           }
        cout<<endl;}
    else{
        cout<<"INVALID RANGE"<<endl;
    }
           return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <math.h>

#define SIZE 100000

int K[SIZE];

void kaprekar() {
    long i, j, n, sq, dNo, mid, left, right, d[12];
   
    for(n = 1; n < SIZE; n++) {
        sq = n * n;
        
        dNo = 0;
        while(sq) {
            d[++dNo] = sq % 10;
            sq /= 10;
        }
        
        mid = dNo / 2;
        for(j = dNo, i = 1; i <= dNo/2; i++, j--) d[i] ^= d[j] ^= d[i] ^= d[j];
        
        for(left = j = 0, i = mid; i >= 1; i--, j++) left += d[i] * pow(10, j);
        for(right = j = 0, i = dNo; i >= mid + 1; i--, j++) right += d[i] * pow(10, j);
        
        if(left + right == n) K[n] = 1;
        else                  K[n] = 0;
    }
}

int main() {
   
    kaprekar();
    
    int i, p, q, flag;
    scanf(" %d %d", &p, &q);
    
    for(flag = 1, i = p; i <= q; i++) {
        if(K[i]) {
            printf("%d ", i);
            flag = 0;
        }
    }
    if(flag) printf("INVALID RANGE");
    
    return 0;
}

Problem solution in JavaScript programming.

function splitNum(num){
    var num = (num * num ).toString()
    var center;
    var res;
    
    if(num.length > 1){
        center = Math.floor(num.length/2)
        res = [parseInt(num.substr(0,center)),parseInt(num.substr(center,num.length))]
    }else{
        center = 0
        res = [parseInt(num),0]
    }
    return res
}


function processData(input) {
    var inputs = input.split("n")
    var p = parseInt(inputs[0])
    var q = parseInt(inputs[1])
    var kaprekar = []
    
    for(var i = p; i <= q; i++){
        var sq = splitNum(i)
        if (sq[0]+sq[1] == i){
            kaprekar.push(i)
        }
    }
    
    if(kaprekar.length == 0){
        console.log( "INVALID RANGE")
    }else{
        console.log( kaprekar.join(" "))
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});