Smallest root of the equation x^2 + s(x)*x – n = 0, where s(x) is the sum of digits of root x.
Last Updated :
08 Jun, 2022
You are given an integer n, find the smallest positive integer root of equation x, or else print -1 if no roots are found.
Equation: x^2 + s(x)*x – n = 0
where x, n are positive integers, s(x) is the function, equal to the sum of digits of number x in the decimal number system.
1 <= N <= 10^18
Examples:
Input: N = 110
Output: 10
Explanation: x = 10 is the minimum root.
As s(10) = 1 + 0 = 1 and
102 + 1*10 - 110=0.
Input: N = 4
Output: -1
Explanation: there are no roots of the
equation possible
A naive approach will be to iterate through all the possible values of X and find out if any such root exists but this won’t be possible as the value of n is very large.
An efficient approach will be as follows
Firstly let’s find the interval of possible values of s(x). Hence x^2 <= N and N <= 10^18, x <= 109. In other words, for every considerable solution x the decimal length of x does not extend 10 digits. So Smax = s(9999999999) = 10*9 = 90.
Let’s brute force the value of s(x) (0 <= s(x) <= 90). Now we have an ordinary square equation. The deal is to solve the equation and to check that the current brute forced value of s(x) is equal to sum of digits of the solution. If the solution exists and the equality holds, we should get the answer and store the minimum of the roots possible.
Below is the implementation of the above approach
C++
#include <bits/stdc++.h>
using namespace std;
bool check(long long a, long long b)
{
long long int c = 0;
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
long long root(long long n)
{
bool found = 0;
long long mx = 1e18;
for (long long i = 0; i <= 90; i++) {
long long s = i * i + 4 * n;
long long sq = sqrt(s);
if (sq * sq == s && check((sq - i) / 2, i)) {
found = 1;
mx = min(mx, (sq - i) / 2);
}
}
if (found)
return mx;
else
return -1;
}
int main()
{
long long n = 110;
cout << root(n);
}
|
Java
class GFG{
static boolean check(long a, long b)
{
long c = 0;
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
static long root(long n)
{
boolean found = false;
long mx = (long)1E18;
for (long i = 0; i <= 90; i++) {
long s = i * i + 4 * n;
long sq = (long)Math.sqrt(s);
if (sq * sq == s && check((sq - i) / 2, i)) {
found = true;
mx = Math.min(mx, (sq - i) / 2);
}
}
if (found)
return mx;
else
return -1;
}
public static void main(String[] args)
{
long n = 110;
System.out.println(root(n));
}
}
|
Python3
import math
def check(a, b):
c = 0;
while (a != 0):
c = c + a % 10;
a = int(a / 10);
return True if(c == b) else False;
def root(n):
found = False;
mx = 1000000000000000001;
for i in range(91):
s = i * i + 4 * n;
sq = int(math.sqrt(s));
if (sq * sq == s and
check(int((sq - i) / 2), i)):
found = True;
mx = min(mx, int((sq-i) / 2));
if (found):
return mx;
else:
return -1;
n = 110;
print(root(n));
|
C#
using System;
public class GFG{
static bool check(long a, long b)
{
long c = 0;
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
static long root(long n)
{
bool found = false;
long mx = (long)1E18;
for (long i = 0; i <= 90; i++) {
long s = i * i + 4 * n;
long sq = (long)Math.Sqrt(s);
if (sq * sq == s && check((sq - i) / 2, i)) {
found = true;
mx = Math.Min(mx, (sq - i) / 2);
}
}
if (found)
return mx;
else
return -1;
}
public static void Main()
{
long n = 110;
Console.Write(root(n));
}
}
|
PHP
<?php
function check($a, $b)
{
$c = 0;
while ($a != 0)
{
$c = $c + $a % 10;
$a = (int)($a / 10);
}
return ($c == $b) ? true : false;
}
function root($n)
{
$found = false;
$mx = 1000000000000000001;
for ($i = 0; $i <= 90; $i++)
{
$s = $i * $i + 4 * $n;
$sq = (int)(sqrt($s));
if ($sq * $sq == $s &&
check((int)(($sq - $i) / 2), $i))
{
$found = true;
$mx = min($mx, (int)(($sq -
$i) / 2));
}
}
if ($found)
return $mx;
else
return -1;
}
$n = 110;
echo root($n);
?>
|
Javascript
<script>
function check(a, b)
{
var c = 0;
while (a != 0)
{
c = c + a % 10;
a = parseInt(a / 10);
}
return (c == b);
}
function root(n)
{
var found = false;
var mx = 1E18;
for(i = 0; i <= 90; i++)
{
var s = i * i + 4 * n;
var sq = Math.sqrt(s);
if (sq * sq == s && check((sq - i) / 2, i))
{
found = true;
mx = Math.min(mx, (sq - i) / 2);
}
}
if (found)
return mx;
else
return -1;
}
var n = 110;
document.write(root(n));
</script>
|
Output:
10
Time Complexity: O(90*log(sqrt(N))), as we are using nested loops to traverse 90*log(sqrt(N)) times.
Auxiliary Space: O(1), as we are not using any extra space.
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