Smallest N digit number whose sum of square of digits is a Perfect Square
Last Updated :
20 Dec, 2022
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++
// 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>
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++
// 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.
Time Complexity : O(81 * N)
Auxiliary Space: O(81 * 105)
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Given a positive integers N, the task is to find the smallest number whose sum of digits is N.Example: Input: N = 10Output: 19Explanation: 1 + 9 = 10 = N Input: N = 18Output: 99Explanation: 9 + 9 = 18 = N Naive Approach: A Naive approach is to run a loop of i starting from 0 and find Sum of digits o
6 min read
Find a subarray of size K whose sum is a perfect square
Given an array arr[] and an integer K, the task is to find a subarray of length K having a sum which is a perfect square. If no such subarray exists, then print -1. Otherwise, print the subarray. Note: There can be more than one possible subarray. Print any one of them.Examples: Input: arr[] = {20,
15 min read
Find smallest number with given digits and sum of digits
Given two positive integers P and Q, find the minimum integer containing only digits P and Q such that the sum of the digits of the integer is N. Example: Input: N = 11, P = 4, Q = 7 Output: 47Explanation: There are two possible integers that can be formed from 4 and 7 such that their sum is 11 i.e.
9 min read
Smallest number whose product with N has sum of digits equal to that of N
Given an integer N, the task is to find the smallest positive integer, which when multiplied by N, has sum of digits equal to the sum of digits of N. Examples: Input: N = 4Output: 28Explanation: Sum of digits of N = 44 * 28 = 112Sum of digits = 1 + 1 + 2 = 4, which is equal to sum of digits of N. In
6 min read
Find second smallest number from sum of digits and number of digits
Given the sum of digits as S and the number of digits as D, the task is to find the second smallest number Examples: Input: S = 9, D = 2Output: 27Explanation: 18 is the smallest number possible with sum = 9 and total digits = 2, Whereas the second smallest is 27. Input: S = 16, D = 3Output: 178Expla
8 min read
Smallest number whose sum of digits is N and every digit occurring at most K times
Given two positive integers N and K, the task is to find the smallest number whose sum of digits is N and every distinct digit in that number occurs at most K times. If no such number exists, print "-1". Examples: Input: N = 25, K = 3Output: 799Explanation: Sum of digits of the number = (7 + 9 + 9)
15 min read