Given an integer N, find the smallest N digit number such that the sum of the square of digits (in decimal representation) of the number is also a perfect square. If no such number exists, print -1.
Examples:
Input : N = 2
Output : 34
Explanation:
The smallest possible 2 digit number whose sum of square of digits is a perfect square is 34 because 32 + 42 = 52.
Input : N = 1
Output : 1
Explanation:
The smallest possible 1 digit number is 1 itself.
Method 1:
To solve the problem mentioned above we can use Backtracking. Since we want to find the minimum N digit number satisfying the given condition, the answer will have digits in non-decreasing order. Therefore we generate the possible numbers recursively keeping track of following in each recursive step :
- position: the current position of the recursive step i.e. which position digit is being placed.
- prev: the previous digit placed because the current digit has to be greater than equal to prev.
- sum: the sum of squares of digits placed till now. When digits are placed, this will be used to check whether the sum of squares of all digits placed is a perfect square or not.
- A vector which stores what all digits have been placed till this position.
If placing a digit at a position and moving to the next recursive step leads to a possible solution then return 1, else backtrack.
Below is the implementation of the above approach:
// C++ implementation to find Smallest N // digit number whose sum of square // of digits is a Perfect Square #include <bits/stdc++.h> using namespace std;
// function to check if // number is a perfect square int isSquare( int n)
{ int k = sqrt (n);
return (k * k == n);
} // function to calculate the // smallest N digit number int calculate( int pos, int prev,
int sum, vector< int >& v)
{ if (pos == v.size())
return isSquare(sum);
// place digits greater than equal to prev
for ( int i = prev; i <= 9; i++) {
v[pos] = i;
sum += i * i;
// check if placing this digit leads
// to a solution then return it
if (calculate(pos + 1, i, sum, v))
return 1;
// else backtrack
sum -= i * i;
}
return 0;
} string minValue( int n)
{ vector< int > v(n);
if (calculate(0, 1, 0, v)) {
// create a string representing
// the N digit number
string answer = "" ;
for ( int i = 0; i < v.size(); i++)
answer += char (v[i] + '0' );
return answer;
}
else
return "-1" ;
} // driver code int main()
{ // initialise N
int N = 2;
cout << minValue(N);
return 0;
} |
// Java implementation to find Smallest N // digit number whose sum of square // of digits is a Perfect Square import java.io.*;
import java.util.*;
class GFG{
// function to check if // number is a perfect square static int isSquare( int n)
{ int k = ( int )Math.sqrt(n);
return k * k == n ? 1 : 0 ;
} // Function to calculate the // smallest N digit number static int calculate( int pos, int prev,
int sum, int [] v)
{ if (pos == v.length)
return isSquare(sum);
// Place digits greater than equal to prev
for ( int i = prev; i <= 9 ; i++)
{
v[pos] = i;
sum += i * i;
// Check if placing this digit leads
// to a solution then return it
if (calculate(pos + 1 , i, sum, v) != 0 )
return 1 ;
// Else backtrack
sum -= i * i;
}
return 0 ;
} static String minValue( int n)
{ int [] v = new int [n];
if (calculate( 0 , 1 , 0 , v) != 0 )
{
// Create a string representing
// the N digit number
String answer = "" ;
for ( int i = 0 ; i < v.length; i++)
answer += ( char )(v[i] + '0' );
return answer;
}
else
return "-1" ;
} // Driver code public static void main(String[] args)
{ // Initialise N
int N = 2 ;
System.out.println(minValue(N));
} } // This code is contributed by jrishabh99 |
# Python3 implementation to find Smallest N # digit number whose sum of square # of digits is a Perfect Square from math import sqrt
# function to check if # number is a perfect square def isSquare(n):
k = int (sqrt(n))
return (k * k = = n)
# function to calculate the # smallest N digit number def calculate(pos, prev, sum , v):
if (pos = = len (v)):
return isSquare( sum )
# place digits greater than equal to prev
for i in range (prev, 9 + 1 ):
v[pos] = i
sum + = i * i
# check if placing this digit leads
# to a solution then return it
if (calculate(pos + 1 , i, sum , v)):
return 1
# else backtrack
sum - = i * i
return 0
def minValue(n):
v = [ 0 ] * (n)
if (calculate( 0 , 1 , 0 , v)):
# create a representing
# the N digit number
answer = ""
for i in range ( len (v)):
answer + = chr (v[i] + ord ( '0' ))
return answer
else :
return "-1"
# Driver code if __name__ = = '__main__' :
# initialise N
N = 2
print (minValue(N))
# This code is contributed by mohit kumar 29 |
// C# implementation to find Smallest N // digit number whose sum of square // of digits is a Perfect Square using System;
class GFG{
// function to check if // number is a perfect square static int isSquare( int n)
{ int k = ( int )Math.Sqrt(n);
return k * k == n ? 1 : 0;
} // Function to calculate the // smallest N digit number static int calculate( int pos, int prev,
int sum, int [] v)
{ if (pos == v.Length)
return isSquare(sum);
// Place digits greater than equal to prev
for ( int i = prev; i <= 9; i++)
{
v[pos] = i;
sum += i * i;
// Check if placing this digit leads
// to a solution then return it
if (calculate(pos + 1, i, sum, v) != 0)
return 1;
// Else backtrack
sum -= i * i;
}
return 0;
} static string minValue( int n)
{ int [] v = new int [n];
if (calculate(0, 1, 0, v) != 0)
{
// Create a string representing
// the N digit number
string answer = "" ;
for ( int i = 0; i < v.Length; i++)
answer += ( char )(v[i] + '0' );
return answer;
}
else
return "-1" ;
} // Driver code public static void Main()
{ // Initialise N
int N = 2;
Console.Write(minValue(N));
} } |
<script> // Javascript implementation to find Smallest N // digit number whose sum of square // of digits is a Perfect Square // function to check if // number is a perfect square function isSquare(n)
{ let k = Math.floor(Math.sqrt(n));
return k * k == n ? 1 : 0;
} // Function to calculate the // smallest N digit number function calculate(pos, prev, sum, v)
{ if (pos == v.length)
return isSquare(sum);
// Place digits greater than equal to prev
for (let i = prev; i <= 9; i++)
{
v[pos] = i;
sum += i * i;
// Check if placing this digit leads
// to a solution then return it
if (calculate(pos + 1, i, sum, v) != 0)
return 1;
// Else backtrack
sum -= (i * i);
}
return 0;
} function minValue(n)
{ let v = Array.from({length: n}, (_, i) => 0);
if (calculate(0, 1, 0, v) != 0)
{
// Create a string representing
// the N digit number
let answer = "" ;
for (let i = 0; i < v.length; i++)
answer += (v[i] + 0);
return answer;
}
else
return "-1" ;
} // Driver Code // Initialise N
let N = 2;
document.write(minValue(N));
// This code is contributed by sanjoy_62.
</script> |
34
Time Complexity: O(sqrt(n))
Auxiliary Space: O(n)
Method 2:
The above-mentioned problem can also be solved using Dynamic Programming. If we observe the question carefully we see that it can be converted to the standard Coin Change problem. Given N as the number of digits, the base answer will be N 1’s, the sum of the square of whose digits will be N.
- If N itself is a perfect square then the N times 1 will be the final answer.
- Otherwise, we will have to replace some 1’s in the answer with other digits from 2-9. Each replacement in the digit will increase the sum of the square by a certain amount and since 1 can be changed to only 8 other possible digits there are only 8 such possible increments. For example, if 1 is changed to 2, then increment will be 22 – 12 = 3. Similarly, all possible changes are : {3, 8, 15, 24, 35, 48, 63, 80}.
So the problem now can be interpreted as having 8 kinds of coins of the aforementioned values and we can use any coin any number of times to create the required sum. The sum of squares will lie in the range of N (all digits are 1) to 81 * N (all digits are 9). We just have to consider perfect square sums in the range and use the idea of coin change to find the N digits that will be in the answer. One important point we need to take into account is that we have to find the smallest N digit number not the number with the smallest square sum of digits.
Below is the implementation of the above-mentioned approach:
// C++ implementation to find the Smallest // N digit number whose sum of square // of digits is a Perfect Square #include <bits/stdc++.h> using namespace std;
long long value[8100006];
int first[8100006];
// array for all possible changes int coins[8] = { 3, 8, 15, 24, 35, 48, 63, 80 };
void coinChange()
{ const long long inf = INT_MAX;
// iterating till 81 * N
// since N is at max 10^5
for ( int x = 1; x <= 8100005; x++) {
value[x] = inf;
for ( auto c : coins) {
if (x - c >= 0 && value[x - c] + 1 < value[x]) {
value[x] = min(value[x], value[x - c] + 1);
// least value of coin
first[x] = c;
}
}
}
} // function to find the // minimum possible value string minValue( int n)
{ // applying coin change for all the numbers
coinChange();
string answer = "" ;
// check if number is
// perfect square or not
if (( sqrt (n) * sqrt (n)) == n) {
for ( int i = 0; i < n; i++)
answer += "1" ;
return answer;
}
long long hi = 81 * n;
long long lo = sqrt (n);
// keeps a check whether
// number is found or not
bool found = false ;
long long upper = 81 * n;
long long lower = n;
// sorting suffix strings
string suffix;
bool suf_init = false ;
while ((lo * lo) <= hi) {
lo++;
long long curr = lo * lo;
long long change = curr - n;
if (value[change] <= lower) {
// build a suffix string
found = true ;
if (lower > value[change]) {
// number to be used for updation of lower,
// first values that will be used
// to construct the final number later
lower = value[change];
upper = change;
suffix = "" ;
suf_init = true ;
int len = change;
while (len > 0) {
int k = sqrt (first[len] + 1);
suffix = suffix + char (k + 48);
len = len - first[len];
}
}
else if (lower == value[change]) {
string tempsuf = "" ;
int len = change;
while (len > 0) {
int k = sqrt (first[len] + 1);
tempsuf = tempsuf + char (k + 48);
len = len - first[len];
}
if (tempsuf < suffix or suf_init == false ) {
lower = value[change];
upper = change;
suffix = tempsuf;
suf_init = true ;
}
}
}
}
// check if number is found
if (found) {
// construct the number from first values
long long x = lower;
for ( int i = 0; i < (n - x); i++)
answer += "1" ;
long long temp = upper;
// fill in rest of the digits
while (temp > 0) {
int dig = sqrt (first[temp] + 1);
temp = temp - first[temp];
answer += char (dig + '0' );
}
return answer;
}
else
return "-1" ;
} // driver code int main()
{ // initialise N
int N = 2;
cout << minValue(N);
return 0;
} |
// Java implementation to find the Smallest // N digit number whose sum of square // of digits is a Perfect Square import java.io.*;
import java.util.*;
class GFG {
static long [] value = new long [( int ) 8100006 ];
static int [] first = new int [ 8100006 ];
// array for all possible changes
static int coins[] = { 3 , 8 , 15 , 24 , 35 , 48 , 63 , 80 };
public static void coinChange()
{
final long inf = Integer.MAX_VALUE;
// iterating till 81 * N
// since N is at max 10^5
for ( int x = 1 ; x <= 8100005 ; x++) {
value[x] = inf;
for ( int c : coins) {
if (x - c >= 0
&& value[x - c] + 1 < value[x]) {
value[x] = Math.min(value[x],
value[x - c] + 1 );
// least value of coin
first[x] = c;
}
}
}
}
// function to find the
// minimum possible value
public static String minValue( int n)
{
// applying coin change for all the numbers
coinChange();
String answer = "" ;
// check if number is
// perfect square or not
if ((Math.sqrt(n) * Math.sqrt(n)) == n) {
for ( int i = 0 ; i < n; i++)
answer += "1" ;
return answer;
}
long hi = 81 * n;
long lo = ( long )Math.sqrt(n);
// keeps a check whether
// number is found or not
boolean found = false ;
long upper = 81 * n;
long lower = n;
// sorting suffix strings
String suffix = "" ;
boolean suf_init = false ;
while ((lo * lo) <= hi) {
lo++;
long curr = lo * lo;
long change = curr - n;
if (value[( int )change] <= lower) {
// build a suffix string
found = true ;
if (lower > value[( int )change])
{
// number to be used for updation of
// lower, first values that will be used
// to construct the final number later
lower = value[( int )change];
upper = change;
suffix = "" ;
suf_init = true ;
int len = ( int )change;
while (len > 0 ) {
int k = ( int )Math.sqrt(first[len]
+ 1 );
suffix = suffix + ( char )(k + 48 );
len = len - first[len];
}
}
else if (lower == value[( int )change]) {
String tempsuf = "" ;
int len = ( int )change;
while (len > 0 ) {
int k = ( int )Math.sqrt(first[len]
+ 1 );
tempsuf = tempsuf + ( char )(k + 48 );
len = len - first[len];
}
if ((tempsuf.compareTo(suffix) < 0 )
|| (suf_init == false )) {
lower = value[( int )change];
upper = change;
suffix = tempsuf;
suf_init = true ;
}
}
}
}
// check if number is found
if (found)
{
// construct the number from first values
long x = lower;
for ( int i = 0 ; i < (n - x); i++)
answer += "1" ;
long temp = upper;
// fill in rest of the digits
while (temp > 0 ) {
int dig
= ( int )Math.sqrt(first[( int )temp] + 1 );
temp = temp - first[( int )temp];
answer += ( char )(dig + '0' );
}
return answer;
}
else
return "-1" ;
}
// Driver code
public static void main(String[] args)
{
// initialise N
int N = 2 ;
System.out.println(minValue(N));
}
} // This code is contributed by Palak Gupta |
# Python3 implementation to find the Smallest # N digit number whose sum of square # of digits is a Perfect Square value = [ 0 for _ in range ( 810006 )];
first = [ 0 for _ in range ( 810006 )];
# array for all possible changes coins = [ 3 , 8 , 15 , 24 , 35 , 48 , 63 , 80 ];
def coinChange():
inf = 99999999 ;
# iterating till 81 * N
# since N is at max 10^5
for x in range ( 1 , 810005 + 1 ):
value[x] = inf;
for c in coins:
if (x - c > = 0 and value[x - c] + 1 < value[x]) :
value[x] = min (value[x], value[x - c] + 1 );
# least value of coin
first[x] = c;
# function to find the # minimum possible value def minValue( n):
# applying coin change for all the numbers
coinChange();
answer = "";
# check if number is
# perfect square or not
if (n = = ( int (n * * 0.5 )) * * 0.5 ):
for i in range (n):
answer + = "1"
return answer;
hi = 81 * n;
lo = int ((n * * 0.5 ));
# keeps a check whether
# number is found or not
found = False ;
upper = 81 * n;
lower = n;
# sorting suffix lets
suffix = "";
suf_init = False ;
while ((lo * lo) < = hi) :
lo + = 1
curr = lo * lo;
change = curr - n;
if (value[change] < = lower) :
# build a suffix let
found = True ;
if (lower > value[change]) :
# number to be used for updation of lower,
# first values that will be used
# to construct the final number later
lower = value[change];
upper = change;
suffix = "";
suf_init = true;
len1 = change;
while (len1 > 0 ) :
k = int ((first[len1] + 1 ) * * 0.5 );
suffix = suffix + str (k)
len1 = len1 - first[len1];
elif (lower = = value[change]) :
tempsuf = "";
len1 = change;
while (len1 > 0 ) :
k = int ((first[len1] + 1 ) * * 0.5 );
tempsuf = tempsuf + str (k);
len1 = len1 - first[len1];
if (tempsuf < suffix or suf_init = = False ):
lower = value[change];
upper = change;
suffix = tempsuf;
suf_init = True ;
# check if number is found
if (found) :
# construct the number from first values
x = lower;
for i in range (n - x):
answer + = '1'
temp = upper;
# fill in rest of the digits
while (temp > 0 ) :
dig = int ((first[temp] + 1 ) * * 0.5 );
temp = temp - first[temp];
answer + = str (dig);
return answer;
else :
return "-1" ;
# driver code # initialise N N = 2 ;
print (minValue(N));
# This code is contributed by phasing17. |
// C# implementation to find the Smallest // N digit number whose sum of square // of digits is a Perfect Square using System;
using System.Collections.Generic;
class GFG {
static long [] value = new long [( int )8100006];
static int [] first = new int [8100006];
// array for all possible changes
static int [] coins = { 3, 8, 15, 24, 35, 48, 63, 80 };
public static void coinChange()
{
long inf = Int32.MaxValue;
// iterating till 81 * N
// since N is at max 10^5
for ( int x = 1; x <= 8100005; x++) {
value[x] = inf;
foreach ( int c in coins) {
if (x - c >= 0
&& value[x - c] + 1 < value[x]) {
value[x] = Math.Min(value[x],
value[x - c] + 1);
// least value of coin
first[x] = c;
}
}
}
}
// function to find the
// minimum possible value
public static string minValue( int n)
{
// applying coin change for all the numbers
coinChange();
string answer = "" ;
// check if number is
// perfect square or not
if ((Math.Sqrt(n) * Math.Sqrt(n)) == n) {
for ( int i = 0; i < n; i++)
answer += "1" ;
return answer;
}
long hi = 81 * n;
long lo = ( long )Math.Sqrt(n);
// keeps a check whether
// number is found or not
bool found = false ;
long upper = 81 * n;
long lower = n;
// sorting suffix strings
string suffix = "" ;
bool suf_init = false ;
while ((lo * lo) <= hi) {
lo++;
long curr = lo * lo;
long change = curr - n;
if (value[( int )change] <= lower) {
// build a suffix string
found = true ;
if (lower > value[( int )change])
{
// number to be used for updation of
// lower, first values that will be used
// to construct the final number later
lower = value[( int )change];
upper = change;
suffix = "" ;
suf_init = true ;
int len = ( int )change;
while (len > 0) {
int k = ( int )Math.Sqrt(first[len]
+ 1);
suffix = suffix + ( char )(k + 48);
len = len - first[len];
}
}
else if (lower == value[( int )change]) {
string tempsuf = "" ;
int len = ( int )change;
while (len > 0) {
int k = ( int )Math.Sqrt(first[len]
+ 1);
tempsuf = tempsuf + ( char )(k + 48);
len = len - first[len];
}
if ((tempsuf.CompareTo(suffix) < 0)
|| (suf_init == false )) {
lower = value[( int )change];
upper = change;
suffix = tempsuf;
suf_init = true ;
}
}
}
}
// check if number is found
if (found)
{
// construct the number from first values
long x = lower;
for ( int i = 0; i < (n - x); i++)
answer += "1" ;
long temp = upper;
// fill in rest of the digits
while (temp > 0) {
int dig
= ( int )Math.Sqrt(first[( int )temp] + 1);
temp = temp - first[( int )temp];
answer += ( char )(dig + '0' );
}
return answer;
}
else
return "-1" ;
}
// Driver code
public static void Main( string [] args)
{
// initialise N
int N = 2;
Console.WriteLine(minValue(N));
}
} // This code is contributed by phasing17 |
// JS implementation to find the Smallest // N digit number whose sum of square // of digits is a Perfect Square let value = new Array(8100006).fill(0);
let first = new Array(8100006).fill(0);
// array for all possible changes let coins = [ 3, 8, 15, 24, 35, 48, 63, 80 ]; function coinChange()
{ let inf = 99999999;
// iterating till 81 * N
// since N is at max 10^5
for (let x = 1; x <= 8100005; x++) {
value[x] = inf;
for ( var c of coins) {
if (x - c >= 0 && value[x - c] + 1 < value[x]) {
value[x] = Math.min(value[x], value[x - c] + 1);
// least value of coin
first[x] = c;
}
}
}
} // function to find the // minimum possible value function minValue( n)
{ // applying coin change for all the numbers
coinChange();
let answer = "" ;
// check if number is
// perfect square or not
if ( Math.floor(Math.sqrt(n)) ** 2 == n) {
for (let i = 0; i < n; i++)
answer += "1" ;
return answer;
}
let hi = 81 * n;
let lo = Math.floor(Math.sqrt(n));
// keeps a check whether
// number is found or not
let found = false ;
let upper = 81 * n;
let lower = n;
// sorting suffix lets
let suffix;
let suf_init = false ;
while ((lo * lo) <= hi) {
lo++;
let curr = lo * lo;
let change = curr - n;
if (value[change] <= lower) {
// build a suffix let
found = true ;
if (lower > value[change]) {
// number to be used for updation of lower,
// first values that will be used
// to construct the final number later
lower = value[change];
upper = change;
suffix = "" ;
suf_init = true ;
let len = change;
while (len > 0) {
let k = Math.floor(Math.sqrt(first[len] + 1));
suffix = suffix + (k).toString();
len = len - first[len];
}
}
else if (lower == value[change]) {
let tempsuf = "" ;
let len = change;
while (len > 0) {
let k = Math.floor(Math.sqrt(first[len] + 1));
tempsuf = tempsuf + (k).toString();
len = len - first[len];
}
if (tempsuf < suffix || suf_init == false ) {
lower = value[change];
upper = change;
suffix = tempsuf;
suf_init = true ;
}
}
}
}
// check if number is found
if (found) {
// construct the number from first values
let x = lower;
for (let i = 0; i < (n - x); i++)
answer += "1" ;
let temp = upper;
// fill in rest of the digits
while (temp > 0) {
let dig = Math.floor(Math.sqrt(first[temp] + 1));
temp = temp - first[temp];
answer += (dig).toString();
}
return answer;
}
else
return "-1" ;
} // driver code // initialise N let N = 2; console.log(minValue(N)); // This code is contributed by phasing17. |
34
Time Complexity : O(81 * N)
Auxiliary Space: O(81 * 105)