HackerRank Modified Kaprekar Numbers problem solution

In this HackerRank Modified Kaprekar Numbers problem 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.

Problem solution in Python programming.

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))

Problem solution in Java Programming.

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);
});