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# Smallest root of the equation x^2 + s(x)*x – n = 0, where s(x) is the sum of digits of root x.

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++

 // CPP program to find smallest value of root// of an equation under given constraints.#include using namespace std; // function to check if the sum of digits is// equal to the summation assumedbool check(long long a, long long b){    long long int c = 0;     // calculate the sum of digit    while (a != 0) {        c = c + a % 10;        a = a / 10;    }     return (c == b);} // function to find the largest root possible.long long root(long long n){    bool found = 0;    long long mx = 1e18;     // iterate for all possible sum of digits.    for (long long i = 0; i <= 90; i++) {         // check if discriminant is a perfect square.        long long s = i * i + 4 * n;        long long sq = sqrt(s);         // check if discriminant is a perfect square and        // if it as perfect root of the equation        if (sq * sq == s && check((sq - i) / 2, i)) {            found = 1;            mx = min(mx, (sq - i) / 2);        }    }     // function returns answer    if (found)        return mx;    else        return -1;} // driver program to check the above functionint main(){    long long n = 110;    cout << root(n);}

## Java

 // Java program to find smallest value of root// of an equation under given constraints. class GFG{// function to check if the sum of digits is// equal to the summation assumedstatic boolean check(long a, long b){    long c = 0;     // calculate the sum of digit    while (a != 0) {        c = c + a % 10;        a = a / 10;    }     return (c == b);} // function to find the largest root possible.static long root(long n){    boolean found = false;    long mx = (long)1E18;     // iterate for all possible sum of digits.    for (long i = 0; i <= 90; i++) {         // check if discriminant is a perfect square.        long s = i * i + 4 * n;        long sq = (long)Math.sqrt(s);         // check if discriminant is a perfect square and        // if it as perfect root of the equation        if (sq * sq == s && check((sq - i) / 2, i)) {            found = true;            mx = Math.min(mx, (sq - i) / 2);        }    }     // function returns answer    if (found)        return mx;    else        return -1;} // driver program to check the above functionpublic static void main(String[] args){    long n = 110;    System.out.println(root(n));}}// This code is contributed by mits

## Python3

 # Python3 program to find smallest# value of root of an equation# under given constraints.import math # function to check if the sum# of digits is equal to the# summation assumeddef check(a, b):    c = 0;     # calculate the    # sum of digit    while (a != 0):        c = c + a % 10;        a = int(a / 10);     return True if(c == b) else False; # function to find the# largest root possible.def root(n):    found = False;         # float(1E+18)    mx = 1000000000000000001;     # iterate for all    # possible sum of digits.    for i in range(91):         # check if discriminant        # is a perfect square.        s = i * i + 4 * n;        sq = int(math.sqrt(s));         # check if discriminant is        # a perfect square and        # if it as perfect root        # of the equation        if (sq * sq == s and            check(int((sq - i) / 2), i)):            found = True;            mx = min(mx, int((sq-i) / 2));     # function returns answer    if (found):        return mx;    else:        return -1; # Driver Coden = 110;print(root(n)); # This code is contributed by mits

## C#

 //C# program to find smallest value of root// of an equation under given constraints.using System;public class GFG{    // function to check if the sum of digits is    // equal to the summation assumed    static bool check(long a, long b)    {        long c = 0;         // calculate the sum of digit        while (a != 0) {            c = c + a % 10;            a = a / 10;        }         return (c == b);    }     // function to find the largest root possible.    static long root(long n)    {        bool found = false;        long mx = (long)1E18;         // iterate for all possible sum of digits.        for (long i = 0; i <= 90; i++) {             // check if discriminant is a perfect square.            long s = i * i + 4 * n;            long sq = (long)Math.Sqrt(s);             // check if discriminant is a perfect square and            // if it as perfect root of the equation            if (sq * sq == s && check((sq - i) / 2, i)) {                found = true;                mx = Math.Min(mx, (sq - i) / 2);            }        }         // function returns answer        if (found)            return mx;        else            return -1;    }     // driver program to check the above function    public static void Main()    {        long n = 110;        Console.Write(root(n));    }}// This code is contributed by Raput-Ji



## Javascript



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.